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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 15, Issue 1

# An extension of the method of brackets. Part 1

Ivan Gonzalez
/ Karen Kohl
/ Lin Jiu
/ Victor H. Moll
Published Online: 2017-09-27 | DOI: https://doi.org/10.1515/math-2017-0100

## Abstract

The method of brackets is an efficient method for the evaluation of alarge class of definite integrals on the half-line. It is based on a small collection of rules, some of which are heuristic. The extension discussed here is based on the concepts of null and divergent series. These are formal representations of functions, whose coefficients an have meromorphic representations for n ∈ ℂ, but might vanish or blow up when n ∈ ℕ. These ideas are illustrated with the evaluation of a variety of entries from the classical table of integrals by Gradshteyn and Ryzhik.

MSC 2010: 33C05; 33F10

## 1 Introduction

The evaluation of definite integrals have a long history dating from the work of Eudoxus of Cnidus (408-355 BC) with the creation of the method of exhaustion. The history of this problem is reported in [17]. A large variety of methods developed for the evaluations of integrals may be found in older Calculus textbooks, such as those by J. Edwards [4, 5]. As the number of examples grew, they began to be collected in tables of integrals. The table compiled by I. S. Gradshteyn and I. M. Ryzhik [16] is the most widely used one, now in its 8th-edition.

The interest of the last author in this topic began with entry 3.248.5 in [14] $I=∫0∞(1+x2)−3/2[φ(x)+φ(x)]−1/2dx$(1)

where φ(x) = 1 + $\frac{4}{3}$x2(1 + x2)−2. The value $\pi /2\sqrt{6}$ given in the table is incorrect, as a direct numerical evaluation will confirm. Since an evaluation of the integral still elude us, the editors of the table found an ingenious temporary solution to this problem: it does not appear in [15] nor in the latest edition [16]. This motivated an effort to present proofs of all entries in Gradshteyn-Ryzhik. It began with [19] and has continued with several short papers. These have appeared in Revista Scientia, the latest one being [1].

The work presented here deals with the method of brackets. This is a new method for integration developed in [1113] in the context of integrals arising from Feynman diagrams. It consists of a small number of rules that converts the integrand into a collection of series. These rules are reviewed in Section 2, it is important to emphasize that most of these rules are still not rigorously justified and currently should be considered a collection of heuristic rules.

The success of the method depends on the ability to give closed-form expressions for these series. Some of these heuristic rules are currently being placed on solid ground [2]. The reader will find in [810] a large collection of examples that illustrate the power and flexibility of this method.

The operational rules are described in Section 2. The method applies to functions that can be expanded in a formal power series $f(x)=∑n=0∞a(n)xαn+β−1,$(2)

where α, β ∈ ℂ and the coefficients a(n) ∈ ℂ. (The extra -1 in the exponent is for a convenient formulation of the operational rules). The adjective formal refers to the fact that the expansion is used to integrate over [0, ∞), even though it might be valid only on a proper subset of the half-line.

#### Note 1.1

There is no precise description of the complete class of functions f for which the method can be applied. At the moment, it is a working assumption, that the coefficients a(n) in (2) are expressions that admit a unique meromorphic continuation to n ∈ ℂ. This is required, since the method involves the evaluation of a(n) for n not a natural number, hence an extension is needed. For example, the Bessel function $I0(x)=∑n=0∞1n!2(x2)2n$(3)

has α = 2, β = 1 and a(n) = 1/22nn!2 can be written as a(n) = 1/22nΓ2(n + 1) and now the evaluation, say at n = $\frac{1}{2}$, is possible. The same observation holds for the Bessel function $J0(x)=∑n=0∞(−1)nn!2(x2)2n.$(4)

The goal of the present work is to produce non-classical series representations for functions f, which do not have expansions like (2). These representations are formally of the type (2) but some of the coefficients a(n) might be null or divergent. The examples show how to use these representations in conjunction with the method of brackets to evaluate definite integrals. The examples presented here come from the table [16]. This process is, up to now, completely heuristic. These non-classical series are classified according to the following types:

1. Totally (partially) divergent series. Each term (some of the terms) in the series is a divergent value. For example, $∑n=0∞Γ(−n)xn and ∑n=0∞Γ(n−3)n!xn.$(5)

2. Totally (partially) null series. Each term (some of the terms) in the series vanishes. For example, $∑n=0∞1Γ(−n)xn and ∑n=0∞1Γ(3−n)xn.$(6)

This type includes series where all but finitely many terms vanish. These are polynomials in the corresponding variable.

3. Formally divergent series. This is a classical divergent series: the terms are finite but the sum of the series diverges. For example, $∑n=0∞n!2(n+1)(2n)!5n.$(7)

In spite of the divergence of these series, they will be used in combination with the method of brackets to evaluate a variety of definite integrals. Examples of these type of series are given next.

Some examples of functions that admit non-classical representations are given next.

• The exponential integral with the partially divergent series $Ei(−x)=−∫1∞t−1e−xtdt=∑n=0∞(−1)nxnnn!.$(8)

• The Bessel K0-function $K0(x)=∫0∞cosxtdt(t2+1)1/2$(9)

with totally null representation $K0(x)=1x∑n=0∞(−1)nΓ2(n+12)n!Γ(−n)(4x2)n$(10)

and the totally divergent one $K0(x)=12∑n=0∞(−1)nΓ(−n)n!(x24)n.$(11)

Section 2 presents the rules of the method of brackets. Section 3 shows that the bracket series associated to an integral is independent of the presentation of the integrand. The remaining sections use the method of brackets and non-classical series to evaluate definite integrals. Section 4 contains the exponential integral Ei(−x) in the integrand, Section 5 has the Tricomi function U(a, b; x) (as an example of the confluent hypergeometric function), Section 6 is dedicated to integrals with the Airy function Ai(x) and then Section 7 has the Bessel function Kν(x), with special emphasis on K0(x). Section 8 gives examples of definite integral whose value contains the Bessel function Kν(x). The final section has a new approach to the evaluation of bracket series, based on a differential equation involving parameters.

The examples presented in the current work have appeared in the literature, where the reader will find proofs of these formulas by classical methods. One of the goals of this work is to illustrate the flexibility of the method of brackets to evaluate these integrals.

## 2 The method of brackets

The method of brackets evaluates integrals over the half line [0, ∞). It is based on a small number of rules reviewed in this section.

#### Definition 2.1

For a ∈ ℂ, the symbol $〈a〉=∫0∞xa−1dx$(12)

is the bracket associated to the (divergent) integral on the right. The symbol $ϕn=(−1)nΓ(n+1)$(13)

is called the indicator associated to the index n. The notation ϕn1n2nr, or simply ϕ12…r, denotes the product ϕn1ϕn2ϕnr.

#### Note 2.2

The indicator ϕn will be used in the series expressions used in the method of brackets. For instance (8) is written as $Ei(−x)=∑nϕnxnn$(14)

and (11) as $K0(x)=12∑nϕnΓ(−n)(x24)n.$(15)

In the process of implementing the method of brackets, these series will be evaluated for n ∈ ℂ, not necessarily positive integers. Thus the notation for the indices does not include its range of values.

## Rules for the production of bracket series

The first part of the method is to associate to the integral $I(f)=∫0∞f(x)dx$(16)

a bracket series. This is done following two rules:

Rule P1. Assume f has the expansion $f(x)=∑n=0∞ϕna(n)xαn+β−1.$(17)

Then I(f) is assigned the bracket series $I(f)=∑nϕna(n)〈αn+β〉.$(18)

#### Note 2.3

The series including the indicator ϕn have indices without limits, since its evaluation requires to take n outside ℕ.

Rule P2. For α ∈ ℂ, the multinomial power (u1 + u2 + … + ur)α is assigned the r-dimension bracket series $∑n1,n2,…,nrϕn1n2⋯nru1n1⋯urnr〈−α+n1+⋯+nr〉Γ(−α).$(19)

The integer r is called the dimension of the bracket series.

## Rules for the evaluation of a bracket series

The next set of rules associates a complex number to a bracket series.

Rule E1. The one-dimensional bracket series is assigned the value $∑nϕna(n)〈αn+b〉=1|α|a(n∗)Γ(−n∗),$(20)

where n* is obtained from the vanishing of the bracket; that is, n* solves an + b = 0.

#### Note 2.4

The rule E1 is a version of the Ramanujan’s Master Theorem. This theorem requires an extension of the coefficients a(n) from n ∈ ℕ to n ∈ ℂ. The assumptions imposed on the function f is precisely for the application of this result. A complete justification of this rule is provided in [2]. Making the remaining rules rigorous is the subject of active research.

The next rule provides a value for multi-dimensional bracket series where the number of sums is equal to the number of brackets.

Rule E2. Assume the matrix B = (bij) is non-singular, then the assignment is $∑n1,n2,…,nrϕn1⋯nra(n1,⋯,nr)〈b11n1+⋯+b1rnr+c1〉⋯〈br1n1+⋯+brrnr+cr〉=1|det(B)|a(n1∗,⋯nr∗)Γ(−n1∗)⋯Γ(−nr∗)$

where $\left\{{n}_{i}^{\ast }\right\}$ is the (unique) solution of the linear system obtained from the vanishing of the brackets. There is no assignment if B is singular.

Rule E3. Each representation of an integral by a bracket series has associated an index of the representation via $index = number of sums − number of brackets.$(21)

In the case of a multi-dimensional bracket series of positive index, the system generated by the vanishing of the coefficients has a number of free parameters. The solution is obtained by computing all the contributions of maximal rank in the system by selecting these free parameters. Series expressed in the same variable (or argument) are added.

#### Example 2.5

A generic bracket series of index 1 has the form $∑n1,n2ϕn1,n2C(n1,n2)An1Bn2〈a11n1+a12n2+c1〉,$(22)

where a11, a12, c1 are fixed coefficients, A, B are parameters and C(n1, n2) is a function of the indices.

The Rule E3 is used to generate two series by leaving first n1 and then n2 as free parameters. The Rule E1 is used to assign a value to the corresponding series:

n1 as a free parameter produces $T1=B−c1/a12|a12|∑n1=0∞ϕn1Γ(a11n1+c1a12)C(n1,−a11n1+c1a12)(AB−a11/a12)n1;$

n2 as a free parameter produces $T2=A−c1/a11|a11|∑n2=0∞ϕn2Γ(a12n2+c1a11)C(−a12n2+c1a11,n2)(BA−a12/a11)n2.$

The series T1 and T2 are expansions of the solution in terms of different parameters $x1=AB−a11/a12andx2=BA−a12/a11.$(23)

Observe that $\begin{array}{}{x}_{2}={x}_{1}^{{a}_{12}/{a}_{11}}.\end{array}$ Therefore the bracket series is assigned the value T1 or T2. If one of the series is a null-series or divergent, it is discarded. If both series are discarded, the method of brackets does not produce a value for the integral that generates the bracket series.

Some special cases will clarify the rules to follow in the use of the series T1 and T2. Suppose a12 = −a11, then $T1=B−c1/a11|a11|∑n1=0∞ϕn1Γn1+c1a11Cn1,−n1−c1a11(AB)n1$(24)

and $T2=A−c1/a11|a11|∑n2=0∞ϕn2Γn2+c1a11C−n2−c1a11,n2(AB)n2$(25)

and since both series are expansions in the same parameter (AB), their values must be added to compute the value associated to the bracket series. On the other hand, if a12 = −2a11, then $T1=Bc1/2a112|a11|∑n1=0∞ϕn1Γ−12n1−c12a11Cn1,12n1+c12a11(AB1/2)n1$

and $T2=A−c1/a11|a11|∑n2=0∞ϕn2Γ−2n2+c1a11C2n2−c1a11,n2(A2B)n2.$

Splitting the sum in T1 according to the parity of the indices produces a power series in A2B when n1 = 2n3 is even and for n1 odd a second power series in the same argument A2B times an extra factor AB1/2. Since these are expansions in the same argument, they have to be added to count their contribution to the bracket series.

#### Note 2.6

It is important to observe that the index is attached to a specific representation of the integral and not just to integral itself. The experience obtained by the authors using this method suggests that, among all representations of an integral as a bracket series, the one with minimal index should be chosen.

#### Note 2.7

The extension presented in this work shows how to use these divergent series in the evaluation of definite integrals. Example 9.3 illustrates this procedure.

Rule E4. In the evaluation of a bracket series, repeated series are counted only once. For instance, a convergent series appearing repeated in the same region of convergence should be counted only once. The same treatment should be given to null and divergent series.

#### Note 2.8

Example 8.1 in Section 4 illustrates the use of this rule.

#### Note 2.9

A systematic procedure in the simplification of the series has been used throughout the literature: express factorials in terms of the gamma function and the transform quotients of gamma terms into Pochhammer symbols, defined by $(a)k=a(a+1)⋯(a+k−1)=Γ(a+k)Γ(a).$(26)

Any presence of a Pochhammer with a negative index k is transformed by the rule $(a)−k=(−1)k(1−a)k,fork∈N.$(27)

In the special case when a is also a negative integer, the rule $(−km)−m=kk+1⋅(−1)m(km)!((k+1)m)!$(28)

holds. This value is justified in [7]. The duplication formula $(a)2n=22n(a2)n(a+12)n$(29)

is also used in the simplifications.

Many of the evaluations are given as values of the hypergeometric functions $pFq(a1,⋯,apb1,…,bq|z)=∑n=0∞(a1)n⋯(ap)n(b1)n⋯(bq)nznn!,$(30)

with (a)n as in (26). It is often that the value of 2F1 at z = 1 is required. This is given by the classical formula of Gauss: $2F1(abc|1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b).$(31)

#### Note 2.10

The extension considered here is to use the method of brackets to functions that do not admit a series representation as described in Rule P1. For example, the Bessel function K0(x) has a singular expansion of the form $K0(x)=−(γ−ln⁡2+ln⁡x)I0(x)+∑j=0∞Hjj!2x2j22j$(32)

(see [21, 10.31.2]). Here I0(x) is the Bessel function given in (3), $\begin{array}{}{H}_{j}=\sum _{k=1}^{j}\frac{1}{k}\end{array}$ is the harmonic number and $\begin{array}{}\gamma =\underset{j\to \mathrm{\infty }}{lim}\left({H}_{j}-\mathrm{ln}j\right)\end{array}$ is Euler’s constant. The presence of the logarithm term in (32) does not permit a direct application of the method of brackets. An alternative is presented in Section 7.

## 3 Independence of the factorization

The evaluation of a definite integral by the method of brackets begins with the association of a bracket series to the integral. It is common that the integrand contains several factors from which the bracket series is generated. This representation is not unique. For example, the integral $I=∫0∞e−axJ0(x)dx$(33)

is associated the bracket series $∑n1,n2ϕn1,n2an122n2Γ(n2+1)〈n1+2n2+1〉,$(34)

and rewriting (33) as $I=∫0∞e−ax/2e−ax/2J0(x)dx,$(35)

provides the second bracket series $∑n1,n2n3 ϕn1,n2,n3an1+n22n1+n2+2n3Γ(n3+1)〈n1+n2+2n3+1〉$(36)

associated to (33). It is shown next that all such bracket series representations of an integral produce the same value.

#### Theorem 3.1

Assume f(x) = g(x)h(x), where f, g and h have expansions as in (2). Then, the method of brackets assigns the same value to the integrals $I1=∫0∞f(x)dxandI2=∫0∞g(x)h(x)dx.$(37)

#### Proof

Suppose that $f(x)=∑nϕna(n)xαn+βg(x)=∑n1ϕn1b(n1)xαn1+β1h(x)=∑n2ϕn2c(n2)xαn2+β2.$

Then $I1=∫0∞f(x)dx=∑nϕna(n)〈αn+β+1〉=1|α|a(−s)Γ(s),$(38)

with s = (1 + β)/α.

To evaluate the second integral, observe that $g(x)h(x)=xβ1+β2∑n1=0∞(−1)n1n1!b(n1)xαn1∑n2=0∞(−1)n2n2!c(n2)xαn2=xβ1+β2∑n=0∞F(n)xαn,$

with $F(n)=∑k=0n(−1)kk!b(k)(−1)n−k(n−k)!c(n−k)=(−1)nn!∑k=0nnkb(k)c(n−k).$(39)

This yields $f(x)=∑n=0∞(−1)nn![∑k=0nnkb(k)c(n−k)]xαn+β1+β2$(40)

and matching this with (38) gives β = β1 + β2 and $a(n)=∑k=0nnkb(k)c(n−k)=∑k=0∞(−1)kk!(−n)kb(k)c(n−k).$(41)

Now, the method of brackets gives $I2=∫0∞g(x)h(x)dx=∑n1,n2ϕn1,n2b(n1)c(n2)〈αn1+αn2+β+1〉$(42)

and it yields two series as solutions $T1=1|α|∑nϕnΓ(n+s)b(n)c(−n−s)T2=1|α|∑nϕnΓ(n+s)b(−n−s)c(n),$(43)

with s = (β + 1)/α. Comparing with (38) shows that I1 = I2 is equivalent to $Γ(s)a(−s)=∑nϕnΓ(n+s)b(n)c(−s−n),$(44)

that is, $a(−s)=∑nϕn(s)nb(n)c(−s−n).$(45)

The identity (45) is the extension of (41) from n ∈ ℕ to s ∈ ℂ. This extension is part of the requirements on the functions f explained in Note 1.1. The proof is complete. □

It is direct to extend the result to the case of a finite number of factors.

#### Theorem 3.2

Assume f admits a representation of the form $\begin{array}{}f\left(x\right)=\prod _{i=1}^{r}{f}_{i}\left(x\right).\end{array}$ Then the value of the integral, obtained by method of brackets, is the same for both series representations.

## 4 The exponential integral

The exponential integral function is defined by the integral formula $Ei(−x)=−∫1∞exp⁡(−xt)tdt, for x>0.$(46)

(See [16, 8.211.1]). The method of brackets is now used to produce a non-classical series for this function. Start by replacing the exponential function by its power series to obtain $Ei(−x)=−∑n1ϕn1xn1∫1∞tn1−1dt$(47)

and then use the method of brackets to produce $∫1∞tn1−1dt=∫1∞(y+1)n1−1dy=∑n2,n3ϕn2n3〈−n1+1+n2+n3〉〈n2+1〉Γ(−n1+1).$

Replace this in (47) to obtain $Ei(−x)=−∑n1,n2,n3ϕn1n2n3xn1〈−n1+1+n2+n3〉〈n2+1〉Γ(−n1+1).$(48)

The evaluation of this series by the method of brackets generates two identical terms for Ei(−x): $Ei(−x)=∑n=0∞(−1)nnΓ(n+1)xn.$(49)

Only one of them is kept, according to Rule E4. This is a partially divergent series (from the value at n = 0), written as $Ei(−x)=∑nϕnxnn.$(50)

The next example illustrates how to use this partially divergent series in the evaluation of an integral.

#### Example 4.1

Entry 6.223 of [16] gives the Mellin transform of the exponential integral as $∫0∞xμ−1Ei(−bx)dx=−b−μμΓ(μ).$(51)

To verify this, use the partially divergent series (50) and the method of brackets to obtain $∫0∞xμ−1Ei(−bx)dx=∑nϕnbnn∫0∞xμ+n−1dx=∑nϕnbnn〈μ+n〉=−b−μμΓ(μ),$(52)

as claimed.

#### Example 4.2

Entry 6.228.2 in [16] is $G(v,μ,β)=∫0∞xv−1e−μxEi(−βx)dx=−Γ(v)v(β+v)v2F1(1vv+1|μβ+μ).$(53)

The partially divergent series (50) is now used to establish this formula. First form the bracket series $G(v,μ,β)=∑n1,n2ϕn1,n2βn1μn2n1〈n1+n2+v〉.$(54)

Rule E1 yields two cases from the equation n1 + n2 + v = 0:

Case 1: n2 = −n1v produces $T1=μ−v∑n1=0∞(−1)n1n1!Γ(n1+v)n1(βμ)n1,$(55)

which is discarded since it is partially divergent (due to the term n1 = 0).

Case 2: n1 = −n2v gives $T2=−β−v∑n2=0∞(−1)n2n2!(μβ)n2Γ(n2+v)n2+v,$(56)

and using $Γ(n2+v)=(v)n2Γ(v)andn2+v=Γ(n2+v+1)Γ(n2+v)=(v+1)n2Γ(v+1)(v)n2Γ(v)$(57)

equation (56) becomes $T2=−Γ(v)vβv∑n2=0∞(v)n2(v)n2n2!(v+1)n2(−μβ)n2=−Γ(v)vβv2F1(vvv+1|−μβ).$(58)

The condition |μ|<|β| is imposed to guarantee the convergence of the series. Finally, the transformation rule (see entry 9. 131. 1 in [16]) $2F1(αβγ|z)=(1−z)−α2F1(αγ−βγ|zz−1)$(59)

with α = β = v, y = v + 1 and z = −μ/β yields (53).

#### Example 4.3

The next evaluation is entry 6.232.2 in [16]: $G(a,b)=∫0∞Ei(−ax)cosbxdx=−1btan−1⁡(ba).$(60)

A direct application of the method of brackets using $cos⁡x=0F1(−12|−x24)$(61)

gives $G(a,b)=π∑n1,n2ϕn1,n2b2n1an222n1Γ(n1+12)n2〈2n1+n2+1〉.$(62)

This produces two series for G(a, b): $T1=πb∑n2=0∞(−1)n2n2!Γ(12(n2+1))n2Γ(−12n2)(2ab)n2,$(63)

and $T2=−πa∑n1=0∞(−1)n1n1!Γ(2n1+1)(2n1+1)Γ(n1+12)(b24a2)n1.$(64)

The analysis begins with a simplification of T2. Use the duplication formula for the gamma function $Γ(2u)Γ(u)=22u−1πΓ(u+12)$(65)

and write $12n1+1=(1)n1(12)n1n1!(32)n1$(66)

to obtain $T2=−1a2F1(11232|−b2a2),$(67)

provided |b| < |a| to guarantee convergence. The form (60) comes from the identity $2F1(11232|−z2)=tan−1⁡zZ$(68)

(see 9.121.27 in [16]).

The next step is the evaluation of T1. Separating the sum (63) into even and odd indices yields $T1=π2b∑n=0∞1(2n)!Γ(n+12)nΓ(−n)(4a2b2)n−πb∑n=0∞1(2n+1)!Γ(n+1)(2n+1)Γ(−n−12)2ab2n+1,$(69)

and in hypergeometric form $T1=−π2b2F1(01212|−a2b2)+ab22F1(12132|−a2b2)=−π2b+1btan−1⁡(ab).$(70)

and this is the same as (60).

The evaluation of entry 6.232.1 in [16] $∫0∞Ei(−ax)sin⁡bxdx=−12bln1+b2a2$(71)

is obtained in a similar form.

#### Example 4.4

Entry 6.782.1 in [16] is $B(z)=∫0∞Ei(−x)J0(2zx)dx=e−z−1z.$(72)

Here $J0(x)=0F1(−1|−x24)$(73)

is the classical Bessel function defined in (4). Therefore $J0(2zx)=∑n2ϕn2zn2Γ(n2+1)xn2.$(74)

The standard procedure using the partially divergent series (49) now gives $B(z)=∑n1,n2ϕn1,n21n1zn2Γ(n2+1)〈n1+n2+1〉,$(75)

which gives the convergent series $T1=−∑n1=0∞(−1)n1n1!(1)n1(2)n1zn1=−1F1(21|−z)=e−z−1z,$(76)

and the series $T2=−1z∑n2=0∞(−z)−n2Γ(1−n2).$(77)

Observe that the expression T2 contains a single non-vanishing term, so it is of the partially null type. An alternative form of T2 is to write $T2=−1z∑n2=0∞(−z−1)n2Γ(1−n2)=−1z∑n2=0∞(−z−1)n2Γ(1)(1)−n2=−1z∑n2=0∞(z−1)n2(0)n2(1)n2(z−1)n2n2!=−1z2F001−|1z.$(78)

The series $\begin{array}{}{}_{2}{F}_{0}\left(\begin{array}{c}a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}b\\ -\end{array}|z\right)\end{array}$ diverges, unless one of the parameters a or b is a non-positive integer in which case the series terminates and it reduces to a polynomial. This is precisely what happens here: only the term for n2 = 0 is non-vanishing and T2 reduces to $T2=−1z.$(79)

This gives the asymptotic behavior B(z)∼−1/z, consistent with the value of T1 for large z. This phenomena occurs every time one obtains a series of the form pFq(z) with pq+2 when the series diverges. The truncation represents an asymptotic approximation of the solution.

## 5 The Tricomi function

The confluent hypergeometric function, denoted by $\begin{array}{}{}_{1}{F}_{1}\left(\begin{array}{c}a\\ c\end{array}|z\right),\end{array}$ defined in (30), arises when two of the regular singular points of the differential equation for the Gauss hypergeometric function $\begin{array}{}{}_{2}{F}_{1}\left(\begin{array}{c}a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}b\\ c\end{array}|z\right),\end{array}$ given by $z(1−z)y″+(c−(a+b+1)z)y′−aby=0,$(80)

are allowed to merge into one singular point. More specifically, if we replace z by z/b in $\begin{array}{}{}_{2}{F}_{1}\left(\begin{array}{c}a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}b\\ c\end{array}|z\right),\end{array}$ then the corresponding differential equation has singular points at 0, b and ∞. Now let b → ∞ so as to have infinity as a confluence of two singularities. This results in the function $\begin{array}{}{}_{1}{F}_{1}\left(\begin{array}{c}a\\ c\end{array}|z\right)\end{array}$ so that $1F1ac|z=limb→∞2F1abc|zb,$(81)

and the corresponding differential equation $zy″+(c−z)y′−ay=0,$(82)

known as the confluent hypergeometric equation. Evaluation of integrals connected to this equation are provided in [3].

The equation (82) has two linearly independent solutions: $M(a,b;x)=1F1ab|x,$(83)

known as the Kummer function and the Tricomi function with integral representation $U(a,b;x)=1Γ(a)∫0∞ta−1exp⁡(−xt)(1+t)b−a−1dt,$(84)

and hypergeometric form $U(a,b;x)=Γ(b−1)Γ(a)x1−b1F1(1+a−b2−b|x)+Γ(1−b)Γ(1+a−b)1F1(ab|x).$(85)

A direct application of the method of brackets gives $U(a,b;x)=1Γ(a)∫0∞ta−1∑n1ϕn1xn1tn1∑n2,n3ϕn2,n3tn3〈1+a−b+n2+n3〉Γ(1+a−b)dt=1Γ(a)∑n1,n2,n3ϕn1,n2,n3xn1〈1+a−b+n2+n3〉Γ(1+a−b)〈a+n1+n3〉.$

This is a bracket series of index 1 and its evaluation produces three terms: $U1(a,b;x)=Γ(1−b)Γ(1+a−b)1F1(ab|x),U2(a,b;x)=Γ(b−1)Γ(a)x1−b1F1(1+a−b2−b|x),U3(a,b;x)=x−a2F0(a1+a−b−|−1x).$

The first two are convergent in the region |x| < 1 and their sum yields (85). The series U3 is formally divergent, the terms are finite but the series is divergent.

#### Example 5.1

The Mellin transform of the Tricomi function is given by $I(a,b;β)=∫0∞xβ−1U(a,b,x)dx.$(86)

Entry 7.612. 1 of [16] $∫0∞xβ−11F1(ab|x−x)dx=Γ(β)Γ(a−β)Γ(b)Γ(b−β)Γ(a)$(87)

is used in the evaluation of I(a, b, β). A proof of (87) appears in [3].

The first evaluation of (86) uses the hypergeometric representation (85) and the formula (87). This is a traditional computation. Direct substitution gives $I(a,b,β)=Γ(b−1)Γ(a)∫0∞xβ−b1F1(1+a−b2−b|x)dx+Γ(1−b)Γ(1+a−b)∫0∞xβ−11F1∞(ab|x)dx=−(−1)−β+bΓ(b−1)Γ(a)Γ(β−b+1)Γ(a−β)Γ(2−b)Γ(1+a−b)Γ(1−β)+(−1)−βΓ(1−b)Γ(1+a−b)Γ(β)Γ(a−β)Γ(b)Γ(b−β)Γ(a).$

The result $∫0∞xβ−1U(a,b,x)dx=Γ(a−β)Γ(β−b+1)Γ(β)Γ(a)Γ(a−b+1)$(88)

follows from simplification of the previous expression.

The second evaluation of (86) uses the method of brackets and the divergent series U3. It produces the result directly. Start with $I(a,b,β)=∫0∞xβ−1U(a,b,x)dx=∫0∞xβ−a−12F0(a1+a−b−|−1x)dx=∑nϕn(a)n(1+a−b)n〈β−a−n〉.$

A standard evaluation by the method of brackets now reproduces (88).

#### Example 5.2

The evaluation of $J(a,b;μ)=∫0∞e−μxU(a,b,x)dx$(89)

is given next. Start with the expansions $exp⁡(−μx)=∑n1ϕn1μn1xn1$(90)

and $U(a,b,x)=x−a2F0(a1+a−b−|−1x)=x−aΓ(a)Γ(1+a−b)∑n2ϕn2Γ(a+n2)Γ(1+a−b+n2)x−n2,$

to write $J(a,b;μ)=1Γ(a)Γ(1+a−b)∑n1,n2ϕn1,n2μn1Γ(a+n2)Γ(1+a−b+n2)〈n1−a−n2+1〉.$

This yields the two series $J1(a,b;μ)=1Γ(a)Γ(1+a−b)∑nϕnΓ(a−1−n)Γ(n+1)Γ(2−b+n)μn=Γ(2−b)(a−1)Γ(1+a−b)2F1(12−b2−a|μ),$

and $J2(a,b;μ)=μa−1Γ(a)Γ(1+a−b)∑nϕnΓ(−a+1−n)Γ(a+n)Γ(1+a−b+n)μn=μa−1Γ(1−a)1F0(1+a−b−|μ)=μa−1Γ(1−a)(1−μ)1+a−b.$

In the case |μ| < 1, both J1 and J2 are convergent. Therefore $∫0∞exp⁡(−μx)U(a,b,x)dx=Γ(2−b)(a−1)Γ(1+a−b)2F1(12−b2−a|μ)+μa−1Γ(1−a)(1−μ)1+a−b.$

In the case |μ| = 1, the series J2 diverges, so it is discarded. This produces $∫0∞e−xU(a,b,x)dx=Γ(2−b)(a−1)Γ(1+a−b)2F1(12−b2−a|1).$(91)

Gaussvalue (31) gives $∫0∞e−xU(a,b,x)dx=Γ(2−b)Γ(2−b+a).$(92)

In particular, if a is a positive integer, say a = k, then $∫0∞e−xU(k,b,x)dx=1(b−2)k.$(93)

This result is summarized next.

#### Proposition 5.3

Let $J(a,b;μ)=∫0∞e−μxU(a,b,x)dx.$(94)

Then, for |μ| < 1, $J(a,b,μ)=Γ(2−b)(a−1)Γ(1+a−b)2F1(12−b2−a|μ)+μa−1Γ(1−a)(1−μ)1+a−b,$(95)

and for μ = 1, $J(a,b;1)=Γ(2−b)Γ(2−b+a).$(96)

In the special case a = k ∈ ℕ, $J(k,b;1)=1(b−2)k.$(97)

## 6 The Airy function

The Airy function, defined by the integral representation $Ai(x)=1π∫0∞cos⁡(t33+xt)dt$(98)

satisfies the equation $d2ydx2−xy=0,$(99)

and the condition y → 0 as x → ∞. A second linearly independent solution of (99) is usually taken to be $Bi(x)=1π∫0∞[exp⁡(−t33+xt)+sin⁡(t33+xt)]dt.$(100)

Using (61) produces $Ai(x)=1π∑n1ϕn1(12)n122n1∫0∞(t33+xt)2n1dt=1π∑n1,n2;n3ϕn1;n2;n3xn2〈−2n1+n2+n3〉(12)n122n1Γ(−2n1)3n3∫0∞t3n3+n2dt=∑n1;n2;n3ϕn1;n2;n3xn2πΓ(−2n1)Γ(12+n1)22n13n3〈−2n1+n2+n3〉〈3n3+n2+1〉.$

The usual resolution of this bracket series gives three cases: $T1=123π∑n=0∞(−1)nn!Γ(−12−3n)Γ(−2n)(34)nx3n+1/2$(101)

a totally null series, $T2=162/3π∑n=0∞(−1)nn!Γ(16−n3)Γ(13−2n3)(34)n/3xn$(102)

a partially divergent series (at the index n = 18), and $T3=1π∑n=0∞(−1)nn!Γ(3n+1)Γ(n+12)Γ(−n)Γ(2n+1)(43)nx−3n−1$(103)

a totally null series, as T1 was.

#### Example 6.1

The series for Ai(x) are now used to evaluate the Mellin transform $I(s)=∫0∞xs−1Ai(x)dx.$(104)

This integral is now computed using the three series Tj given above. Using first the value of T1 and the formulas $Γ(2u)=22u−1πΓ(u)Γ(u+12)andΓ(3u)=33u−122πΓ(u)Γ(u+13)Γ(u+23)$(105)

(these appear as 8.335.1 and 8.335.2 in [16], respectively), give $I(s)=123π∑nϕn(34)nΓ(−12−3n)Γ(−2n)〈s+3n+12〉=163π(34)−s/3−1/6Γ(2s+16)Γ(s)Γ(2s+13).=3−(s+2)/3Γ(s)Γ(s+23)=3(4s−7)/62πΓ(s+13)Γ(s3).$(106)

Similar calculations, using T2 or T3, give the same result. This result is stated next.

#### Lemma 6.2

The Mellin transform of the Airy function is given by $∫0∞xs−1Ai(x)dx=12π3(4s−7)/6Γ(s+13)Γ(s3).$(107)

## 7 The Bessel function Kν

This section presents series representations for the Bessel function Kν(x) defined by the integral representation $Kv(x)=2vΓ(v+12)Γ(12)xv∫0∞cos⁡tdt(x2+t2)v+12;$(108)

given as entry 8.432.5 in [16]. Using the representation (61) of cos t as $\begin{array}{}{0}^{{F}_{1}}\left(\begin{array}{c}-\\ \frac{1}{2}\end{array}|-\frac{{t}^{2}}{4}\right)\end{array}$ and using Rule P2 in Section 2 to expand the binomial in the integrand as a bracket series gives $Kv(x)=2v∑n1;n2;n3ϕn1;n2;n3x2n3+v22n1Γ(n1+12)〈v+12+n2+n3〉〈2n1+2n2+1〉.$(109)

The usual procedure to evaluate this bracket series gives three expressions: $T1=2v−1x−v∑nϕnΓ(v−n)(x24)n,T2=2−1−vxv∑nϕnΓ(−v−n)(x24)n,T3=2v∑nϕn22nΓ(−n)Γ(n+v+12)Γ(n+12)x−2n−v−1.$(110)

The series T3 is a totally null series for Kv. In the case v ∉ ℕ, the series T1 and T2 are finite and Kv(x) = T1+T2 gives the usual expression in terms of the Bessel Iv function $Kv(x)=π2I−v(x)−Iv(x)sin⁡πv;$(111)

as given in entry 8.485 in [16].

In the case v = k ∈ ℕ, the series T1 is partially divergent (the terms n = 0,1,…, k have divergent coefficients) and the series T2 is totally divergent (every coefficient is divergent). In the case v = 0, both the series T1 and T2 become $Totally divergent series for K0(x)=12∑nϕnΓ(−n)(x24)n,$(112)

using Rule E4 to keep a single copy of the divergent series. This complements the $Totally null series for K0(x)=∑nϕn22nΓ(−n)Γ2(n+12)x−2n−1.$(113)

The examples presented below illustrate the use of these divergent series in the computation of definite integrals with the Bessel function K0 in the integrand. Entries in [16] with K0 as the result of an integral have been discussed in [6].

#### Example 7.1

Entry 6.511.12 of [16] states that $∫0∞K0(x)dx=π2.$(114)

To verify this result, use the totally null representation (113) to obtain $∫0∞K0(x)dx=∑nϕnΓ(n+12)2Γ(−n)4n∫0∞x−2n−1dx=∑nϕnΓ(n+12)2Γ(−n)4n〈−2n〉.$(115)

The value of the bracket series is $∫0∞K0(x)dx=12Γ(n+12)24n|n=0=π2.$(116)

#### Example 7.2

The Mellin transform $G(β,s)=∫0∞xs−1K0(βx)dx$(117)

is evaluated next. Example 7.1 corresponds to the special case s = β = 1. The totally divergent series (112) yields $G(β,s)=12∑nϕnΓ(−n)β2n22n〈2n+s〉$(118)

and a direct evaluation of the brackets series using Rule E1 gives $G(β,s)=2s−2βsΓ2(s2).$(119)

Now using the totally null representation (113) gives the bracket series $G(β,s)=∑nϕn22nΓ2(n+12)β2n+1Γ(−n)〈s−1−2n〉.$(120)

One more application of Rule E1 gives (119) again.

#### Example 7.3

Entry 6.611.9 of [16] is $∫0∞e−axK0(bx)dx=1b2−a2cos−1⁡(ab),$(121)

for Re(a+b) > 0. This is a generalization of Example 7.1. The totally divergent representation (112) and the series for the exponential function (90) give the bracket series $∫0∞e−axK01(bx)dx=12∑n1,n2ϕn1n2Γ(−n2)an1b2n222n2〈n1+2n2+1〉.$(122)

The usual procedure gives two expressions: $T1=12a∑nϕnΓ(2n+1)Γ(−n)(b24a2)n,$(123)

which is discarded since it is divergent and $T2=12b∑n=0∞Γ(n+12)2n!(−2ab)n.$(124)

Separating the series according to the parity of the index n yields $T2=12b[π∑n=0∞(12)nn!(a2b2)n−2ab∑n=0∞(1)n2n!(32)n(a2b2)n].$(125)

The identity [16, 9. 121. 1] $2F1(−n,bb|−z)=(1+z)n,$(126)

with $\begin{array}{}n=-\frac{1}{2}\end{array}$ gives $π2b∑n=0∞(12)nn!(a2b2)n=π21b2−a2.$(127)

The identity $−ab2∑n=0∞(1)n2n!(32)n(ab)2n=−1b2−a2sin−1⁡(ab)$(128)

comes from the Taylor series $2xsin−1⁡x1−x2=∑n=1∞22nx2nn(2nn).$(129)

(See Theorem 7.6.2 in [20] for a proof). The usual argument now gives $T2=∫0∞e−axK0(bx)dx=1b2−a2[π2−sin−1⁡(ab)],$(130)

an equivalent form of (121).

#### Example 7.4

The next example, $∫0∞xsin⁡(bx)K0(ax)dx=πb2(a2+b2)−3/2,$(131)

appears as entry 6.691 in [16]. The factor sin bx in integrand is expressed as a series: $sin⁡(bx)=bx0F1(−32|−b2x24)=bΓ(32)∑n2ϕn2(b24)n2Γ(n2+12)x2n2+1$(132)

and the Bessel factor is replaced by its totally-null representation (113) $K0(ax)=1a∑n1ϕn1Γ(n1+12)2Γ(−n1)(4a2)n1x−2n1−1.$(133)

This yields $∫0∞xsin⁡(bx)K01(ax)dx=Γ(32)∑n1,n2ϕn1,n2Γ(n1+12)2Γ(n2+32)Γ(−n1)4n1−n2b2n2+1a2n1+1〈2+2n2−2n1〉.$(134)

These representation produces two solutions S1 and S2, one perfree index, that are identical. The method of brackets rules state that one only should be taken. This is: $S1=πba3∑k=0∞Γ(k+32)(−1)kb2kk!a2k.$(135)

The result nowfollows from the identity $∑k=0∞(32)kk!(−ba)k=1F0(32−|−ba)$(136)

and the binomial theorem obtaining $1F0(32−|x)=1(1−x)3/2.$(137)

#### Example 7.5

The next example in this section evaluates $G(a,b)=∫0∞J0(ax)K0(bx)dx.$(138)

From the representation $J0(ax)=∑n1ϕn1a2n1x2n122n1Γ(n1+1)$(139)

and the null-series (10) it follows that $G(a,b)=∑n1,n2ϕn1,n2a2n122(n2−n1)Γ2(n2+12)Γ(n1+1)Γ(−n2)b2n2+1〈2n1−2n2〉.$(140)

This bracket series generates two identical series, so only one is kept to produce $G(a,b)=12b∑nϕnΓ2(n+12)Γ(n+1)(a2b2)n=π2b2F1(12121|−a2b2)=1bK(iab).$(141)

Here K(z) is the elliptic integral of the first kind. Using the identity $K(iz)=1z2+1K(Zz2+1)$(142)

yields $G(a,b)=1a2+b2K(aa2+b2).$(143)

#### Example 7.6

The next example evaluates $H(a)=∫0∞K02(ax)dx.$(144)

Naturally H(a) = H(1)/a, but it is convenient to keep a as a parameter. The problem is generalized to $H1(a,b)=∫0∞K0(ax)K0(bx)dx,$(145)

and H(a) = H1(a,a) . The evaluation uses the totally divergent series (112) $K0(ax)=∑n1ϕn1a2n1Γ(−n1)22n1+1x2n1$(146)

as well as the integral representation (see 8.432.6 [16]) and the corresponding bracket series $K0(bx)=12∫0∞exp(−t−b2x24t)dtt=∑n2,n3ϕn2,n3b2n3x2n322n3+1〈n2−n3〉.$(147)

Then $H1(a,b)=∑n1,n2,n3ϕn1,n2,n3a2n1b2n3Γ(−n1)22n1+2n3+2〈n2−n3〉〈2n1+2n3+1〉.$(148)

The evaluation of this bracket series requires an extra parameter ϵ and to consider $H2(a,b,ϵ)=∑n1,n2,n3ϕn1,n2,n3a2n1b2n3Γ(−n1)22n1+2n3+2〈n2−n3+ϵ〉〈2n1+2n3+1〉.$(149)

Evaluating this brackets series produces three values, one divergent, which is discarded, and two others: $T2=14acϵ∑nϕnΓ(−n−ϵ)Γ2(ϵ+n+12)cnT3=14a∑nϕnΓ(−n+ϵ)Γ2(n+12)cn,$(150)

with c = b2/a2. Converting the Γ-factors into Pochhammer symbols produces $T2=14acϵΓ(−ϵ)Γ2(12+ϵ)2F112+ϵ12+ϵ1+ϵ|cT3=π4aΓ(ϵ)2F112121−ϵ|c.$(151)

This yields $H2(a,b,ϵ)=π4a[Γ(ϵ)2F112121−ϵ|c−cϵΓ2(12+ϵ)ϵΓ(ϵ)sin⁡πϵ2F112+ϵ12+ϵ1+ϵ|c].$

Let c → 1 (ba) and use Gaussformula (31) to obtain $2F112121−ϵ|1=Γ(1−ϵ)Γ(−ϵ)Γ2(12−ϵ)and2F112+ϵ12+ϵ1+ϵ|1=Γ(1+ϵ)Γ(−ϵ)Γ2(12),$

and this produces $H2(a,a,ϵ)=Γ(−ϵ)2Γ2(ϵ+12)Γ(ϵ+1)4πa+πΓ(1−ϵ)Γ(−ϵ)Γ(ϵ)4aΓ2(12−ϵ)=π4a[Γ2(−ϵ)Γ(ϵ+1)Γ2(ϵ+12)π2+Γ(1−ϵ)Γ(−ϵ)Γ(ϵ)Γ2(12−ϵ)].$

Expanding H2(a, a, ϵ) in powers of ϵ gives $H(a,a,ϵ)=π24a−π24a(γ+4ln⁡2)ϵ+o(ϵ).$(152)

Letting ϵ → 0 gives $∫0∞K02(ax)dx=π24a.$(153)

#### Example 7.7

The final example in this section is the general integral $I(a,b;v,λ;ρ)=∫0∞xρ−1Kv(ax)Kλ(bx)dx.$(154)

The case a = b appears in [18].

The evaluation uses the integral representation $Kv(ax)=(ax)v2v+1∫0∞exp(−t−a2x24t)dttv+1$(155)

appearing in [16, 8.432.6]. This produces the bracket series representation $Kv(ax)=12v+1∑n1,n2ϕn1,n2a2n2+v22n2x2n2+v〈n1−n2−v〉.$(156)

The second factor uses the totally null representation (10) $Kλ(bx)=2λ∑n3ϕn322n3Γ(n3+λ+12)Γ(n3+12)Γ(−n3)b2n3+λ+11x2n3+λ+1.$(157)

Replacing in (154) produces the bracket series $I(a,b;v,λ;ρ)=∑n1,n2,n3ϕn1,n2,n3a2n2+v2λ−v−1+2n3−2n2Γ(n3+λ+12)Γ(n3+12)b2n3+λ+1Γ(−n3)×〈n1−n2−v〉〈ρ+v−λ+2n2−2n3−1〉.$(158)

The vanishing of the brackets gives the system of equations $n1−n2=v2n2−2n3=−ρ−v+λ+1.$(159)

The matrix of coefficients is of rank 2, so it produces three series as candidates for values of the integral one per free index.

Case 1: n1 free. Then n2 = n1v and ${n}_{3}=\frac{\rho -v-\lambda -1}{2}+{n}_{1}.$ This gives $T1=2ρ−3bv−ρavΓ(ρ−v+λ2)Γ(ρ−v−λ2)Γ(v)2F1ρ−v+λ2ρ−v−λ21−v|a2b2.$

Case 2: n2 free. Then n1 = n2 + v and ${n}_{3}=\frac{\rho +v-\lambda -1}{2}+{n}_{2}.$ This gives $T2=2ρ−3avbv+ρΓ(−v)Γ(ρ+v+λ2)Γ(ρ+v−λ2)Γ(v)2F1ρ+v+λ2ρ+v−λ21+v|a2b2.$

Case 3: n3 free. Then ${n}_{2}={n}_{3}+\frac{\lambda -\rho -v+1}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{n}_{1}={n}_{3}+\frac{\lambda -\rho +v+1}{2}.$ This produces $T3=2ρ−3a−ρ+λ+1bλ+1∑nϕnΓ(−n)Γ(ρ+v−λ−12−n)Γ(ρ−v−λ−12−n)Γ(n+λ+12)Γ(n+12)(a2b2)n.$

This series has the value zero. This proves the next statement:

#### Proposition 7.8

The integral $I(a,b;v,λ;ρ)=∫0∞xρ−1Kv(ax)Kλ(bx)dx$(160)

is given by $I(a,b;v,λ;ρ)=2ρ−3bv−ρavΓ(v)Γ(ρ−v+λ2)Γ(ρ−v−λ2)2F1ρ−v+λ2ρ−v−λ21−v|a2b2+2ρ−3avbv+0Γ(−v)Γ(ρ+v+λ2)Γ(ρ+v−λ2)2F1ρ+v+λ2ρ+v−λ21+v|a2b2.$

Some special cases of this evaluation are interesting in their own right. Consider first the case a = b. Using Gausstheorem (31) it follows that $T1=2ρ−3Γ(v)Γ(ρ+λ−v2)Γ(ρ−λ−v2)Γ(1−v)Γ(1−ρ)aρΓ(1−ρ+v+λ2)Γ(1−ρ+v−λ2)$(161)

and $T2=2ρ−3Γ(−v)Γ(ρ+λ+v2)Γ(v+ρ−λ2)Γ(v+1)Γ(1−ρ)aρΓ(1−ρ−v−λ2)Γ(1−ρ−v+λ2).$(162)

#### Proposition 7.9

The integral $J(a;v,λ;ρ)=∫0∞xρ−1Kv(ax)Kλ(ax)dx$(163)

is given by $J(a;v,λ;ρ)=2ρ−3Γ(v)Γ(ρ+λ−v2)Γ(ρ−λ−v2)Γ(1−v)Γ(1−ρ)aρΓ(1−ρ+v+λ2)Γ(1−ρ+v−λ2)+2ρ−3Γ(−v)Γ(ρ+λ+v2)Γ(v+ρ−λ2)Γ(v+1)Γ(1−ρ)aρΓ(1−ρ−v−λ2)Γ(1−ρ−v+λ2).$

The next special case is to take a = b and λ = v. Then $T1=2ρ−3aρΓ(v)Γ(ρ2)Γ(ρ2−v)Γ(1−v)Γ(1−ρ)Γ(1−ρ2−v)Γ(1−ρ2)$(164)

and $T2=2ρ−3aρΓ(−v)Γ(ρ2)Γ(ρ2+v)Γ(v+1)Γ(1−ρ)Γ(1−ρ2+v)Γ(1−ρ2).$(165)

This proves the next result:

#### Proposition 7.10

The integral $L(a;v,ρ)=∫0∞xρ−1Kv2(ax)dx$(166)

is given by $L(a;v,ρ)=2ρ−3aρ[Γ(v)Γ(1−v)Γ(ρ2−v)Γ(1−ρ2−v)+Γ(−v)Γ(1+v)Γ(ρ2+v)Γ(1−ρ2+v)].$

The last special case is ρ = 1; that is, the integral $M(a,b;v,λ)=∫0∞Kv(ax)Kλ(bx)dx.$(167)

It is shown that the usual application of the method of brackets yield only divergent series, so a new approach is required.

The argument begins with converting the brackets series in (158) to $M(a,b;v,λ)=∑n1,n2,n3ϕn1,n2,n3a2n2+v2λ−v−1+2n3−2n2Γ(n3+λ+12)Γ(n3+12)b2n3+λ+1Γ(−n3)×〈n1−n2−v〉〈v−λ+2n2−2n3〉.$(168)

A routine application of the method of brackets gives three series $T1=bv−14avΓ(1−v+λ2)Γ(1−v−λ2)Γ(v)2F11−v+λ21−v−λ21−v|a2b2T2=av4bv+1Γ(1+v+λ2)Γ(1+v−λ2)Γ(−v)2F11+v+λ2v−λ+121+v|a2b2$

and a totally null series T3. Gaussvalue (31) shows that T1 and T2 diverge when ab. Therefore (168) is replaced by $M(a,b;v,λ)=limϵ→0∑n1,n2,n3ϕn1,n2,n3a2n2+v2λ−v−1+2n3−2n2Γ(n3+λ+12)Γ(n3+12)b2n3+λ+1Γ(−n3)×〈n1−n2−v+ϵ〉〈v−λ+2n2−2n3〉.$(169)

Proceeding as before produces a null series that is discarded and also $T1=a−v+2ϵ4b1−v+2ϵΓ(v−ϵ)Γ(1+λ−v2+ϵ)Γ(1−λ−v2+ϵ)×2F11−v+λ2+ϵ1−v−λ2+ϵ1−v+ϵ|a2b2T2=av4b1+vΓ(−v+ϵ)Γ(1+λ+v2)Γ(1−λ+v2)×2F11+v−λ21+v+λ21+v−ϵ|a2b2.$

In the limit as ba, these become $T1=Γ(v−ϵ)Γ(1+λ−v2+ϵ)Γ(1−λ−v2+ϵ)Γ(1−v+ϵ)Γ(−ϵ)4aΓ(1−v−λ2)Γ(1−v+λ2)T2=Γ(−v+ϵ)Γ(1+λ+v2)Γ(1−λ+v2)Γ(1+v−ϵ)Γ(−ϵ)4aΓ(1+v+λ2−ϵ)Γ(1+v−λ2−ϵ).$

Passing to the limit as ϵ → 0 gives $∫0∞Kv(ax)Kλ(ax)dx=π24asin⁡πv[tan⁡(π2(λ+v))−tan⁡(π2(λ−v))].$(170)

In the special case λ = v, it follows that $∫0∞Kv2(ax)dx=π24acos⁡πv,validfor|v|<12.$(171)

This value generalizes (153). It appears in Prudnikov et al. [22] as entries.2.16.28.3 and 2.16.33.2.

## 8 An example with an integral producing the Bessel function

The evaluation of integrals in Section 7 contain the Bessel function Kv in the integrand. This section uses the method developed in the current work to evaluate some entries in [16] where the answer involves K0.

#### Example 8.1

The first example is entry 6.532.4 in [16] $∫0∞xJ0(ax)x2+b2dx=K0(ab).$(172)

The analysis begins with the series $J0(ax)=∑n=0∞1n!2(−a2x24)n=∑n1=0∞ϕn1a2n122n1Γ(n1+1)x2n1$(173)

Rule P2 gives $1x2+b2=∑n2,n3ϕn2,n3x2n2b2n3〈1+n2+n3〉.$(174)

Therefore $∫0∞xJ0(ax)x2+b2dx=∑n1,n2,n3ϕn1,n2,n3a2n1b2n322n1Γ(n1+1)〈1+n2+n3〉〈2+2n1+2n2〉.$(175)

The method of brackets produces three series as candidates for solutions, one perfree index n1, n2, n3: $T1=12∑n=0∞ϕnΓ(−n)(a2b24)nT2=2a2b2∑n=0∞ϕnΓ2(n)Γ(−n)(4a2b2)nT3=12∑n=0∞ϕnΓ(−n)(a2b24)n.$(176)

The fact that T1 = T3 and using Rule E4 shows that only one of these series has to be counted. Since T1 and T2 are non-classical series of distinct variables, both are representations of the value of the integral. Observe that T2 is the totally null representation of K0 (ab) given in (11). This confirms (172). The fact that T3 is also a value for the integral gives another totally divergent representation for K0: $K0(x)=2x2∑n=0∞ϕnΓ2(n+1)Γ(−n)(4x2)n.$(177)

To test its validity, the integral in Example 7.1 is evaluated again, this time using (177): $∫0∞K0(x)dx=∫0∞2x2∑nϕnΓ2(n+1)Γ(−n)22nX−2n=∑nϕn22n+1Γ2(n+1)Γ(−n)∫0∞x−2n−2dx=∑nϕn22n+1Γ2(n+1)Γ(−n)〈−2n−1〉.$(178)

The bracket series is evaluated using Rule E1 to confirm (114).

#### Example 8.2

Entry 6.226.2 in [16] is $∫0∞Ei(−a24x)e−μxdx=−2μK0(aμ).$(179)

The evaluation starts with the partially divergent series (50) $Ei(−a24x)=∑n1=0∞ϕn1a2n1n122n11xn1$(180)

and this yields $∫0∞Ei(−a24x)e−μxdx=∑n1,n2ϕn1n2a2n1μn2n122n1〈n2−n1+1〉.$(181)

The method of brackets gives two series. The first one $T1=1μ∑n1ϕn1Γ(1−n1)n122n1(a2μ)n1=−1μ∑n1ϕn1Γ(−n1)22n1(a2μ)n1=−2μK0(aμ),$(182)

using (11). The second series is $T2=∑n2ϕn2a2n2+2μn2(n2+1)22(n2+1)Γ(−n2−1).$(183)

Now shift the index by m = n2+1 to obtain $T2=∑mϕm−1a2mμm−1m22mΓ(−m).=−1μ∑mϕmΓ(−m)a2mμm22m.$

This is the same sum as T1 in the second line of (182). Recall that the summation indices are placed after the conversion of the indicator ϕn2 to its expression in terms of the gamma function. According to Rule E4, the sum T2 is discarded. This establishes (179).

## 9 A new use of the method of brackets

This section introduces a procedure to evaluate integrals of the form $I(a1,a2)=∫0∞f1(a1x)f2(a2x)dx.$(184)

Differentiating with respect to the parameters leads to $a1∂I(a1,a2)∂a1+a2∂I(a1,a2)∂a2=∫0∞xddx[f1(a1x)f2(a2x)]dx.$(185)

Integration by parts produces $I(a1,a2)=xf1(a1x)f2(a2x)|0∞−(a1∂I(a1,a2)∂a1+a2∂I(a1,a2)∂a2).$(186)

A direct extension to many parameters leads to the following result.

#### Theorem 9.1

Let $I(a1,⋯,an)=∫0∞∏j=1nf(ajx)dx.$(187)

Then $I(a1,⋯,an)=x∏j=1nfj(ajx)|0∞−∑j=1naj∂I(a1,⋯,an)∂aj.$(188)

#### Example 9.2

The integral $I(a,b)=∫0∞e−axJ0(bx)dx$(189)

is evaluated first by a direct application of the method of brackets and then using Theorem 9.1.

The bracket series for I(a, b) $I(a,b)=∑n1,n2ϕn1,n2an1b2n222n2Γ(n2+1)〈n1+2n2+1〉$(190)

is obtained directly from (90) $e−ax=∑n1ϕn1an1xn1$(191)

and $J0(bx)=0F1(−1|−(bx)24)=∑n2ϕn2b2n2Γ(n2+1)22n2x2n2.$(192)

Solving for n1 in the equation coming from the vanishing of the bracket gives n1 = −2n2−1, which yields $T1=∑n2=0∞(−1)n2n2!a−2n2−1b2n222n2Γ(2n2+1)Γ(n2+1).$(193)

To simplify this sum transform the gamma factors via (26) and use the duplication formula (29) to produce $T1=1a∑n2=0∞(12)n2n2!(−b2a2)n=1a1F0(12−|−b2a2).$(194)

The identity $\begin{array}{}{}_{1}{F}_{0}\left(\begin{array}{c}c\\ -\end{array}|z\right)=\left(1-z{\right)}^{-c}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}gives\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{T}_{1}=\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}.\end{array}$ A direct calculation shows that the series obtained from solving for n2 yields the same solution, so it is discarded. Therefore $∫0∞e−axJ0(bx)dx=1a2+b2.$(195)

The evaluation of this integral using Theorem 9.1 begins with checking that the boundary terms vanish. This comes from the asymptotic behavior J0(x) ∼ 1 as x → 0 and $\begin{array}{}{J}_{0}\left(x\right)\sim \sqrt{\frac{2}{\pi x}}\mathrm{cos}x\end{array}$ as x → ∞. The term $a∂I(a,b)∂a=∑n1,n2ϕn1n2n1an1b2n222n2Γ(n2+1)〈n1+2n2+1〉.$(196)

This generates two series $T1=1b∑n=0∞(−1)nn!nΓ(1+n2)Γ(1−n2)(2ab)n$(197)

and $T2=−1a∑n=0∞(−1)nn!Γ(2n+2)Γ(n+1)(b24a2)n.$(198)

Similarly $b∂I(a,b)∂b=2∑n1,n2ϕn1,n2n2an1b2n222n2Γ(n2+1)〈n1+2n2+1〉$(199)

which yields the two series $T~1=2b∑n=0∞ϕnΓ(n+12)Γ(−n+12)Γ(1−n2)(2ab)nT~2=2a∑n=0∞2a∑n=0∞ϕnnΓ(2n+1)Γ(n+1)(b24a2)n$

Since the boundary terms vanish, the relation (186) gives $I(a,b)=−T1−T~1,|4a2|<|b2|−T2−T~2,|b2|<|4a2|.$(200)

The form T2 + ${\stackrel{~}{T}}_{2}$ is simplified by converting them to hypergeometric form to produce $T2=−1a∑n=0∞(−1)nn!Γ(2n+2)Γ(n+1)(b24a2)n=−a2(a2+b2)3/2T~2=2a∑n=0∞ϕnnΓ(2n+1)Γ(n+1)(b24a2)n=−b2(a2+b2)3/2.$(201)

Then $I(a,b)=−T2−T~2=a2(a2+b2)3/2+b2(a2+b2)3/2=1a2+b2.$(202)

This gives $I(a,b)=∫0∞e−axJ0(bx)dx=1a2+b2.$(203)

The option T1 + ${\stackrel{~}{T}}_{1}$ gives the same result.

#### Example 9.3

Entry 6.222 in [16] is $I(a1,a2)=∫0∞Ei(−a1x)Ei(−a2x)dx=1a1+1a2In(a1+a2)−ln⁡a1a2−ln⁡a2a1.$(204)

In particular $∫0∞Ei2(−ax)dx=2ln⁡2a.$(205)

The evaluation of this integral by the method of brackets begins with the partially divergent series for Ei(−x) which yields (using (14) = (50)): $I(a1,a2)=∑n1,n2ϕn1,n2a1n1a2n2n1n2〈n1+n2+1〉.$(206)

The usual procedure requires the relation n1 + n2 + 1 = 0 and taking n1 as the free parameter gives $I1(a1,a2)=−1a2∑n1=0∞(−1)n1n1(n1+1)(a1a2)n1,$(207)

and when n2 as free parameter one obtains the series $I2(a1,a2)=−1a1∑n2=0∞(−1)n2n2(n2+1)(a2a1)n2.$(208)

These two series correspond to different expansions: the first one in x = a1/a2 and the second one in x−1 = a2/a1. Both series are partially divergent, so the Rule E3 states that these sums must be discarded. The usual method of brackets fails for this problem.

The solution using Theorem 9.1 is described next. An elementary argument shows that xEi(−x) → 0 as x → 0 or ∞. Then (186) becomes $I(a1,a2)=−a1∂I∂a1−a2∂I∂a2=−∑n1,n2ϕn1,n2a1n1a2n2n2〈n1+n2+1〉−∑n1,n2ϕn1,n2a1n1a2n2n1〈n1+n2+1〉,≡S1+S2,$(209)

using (206) to compute the partial derivatives. The method of brackets gives two series for each of the sums S1 and S2: $T1,1=1a2∑n=0∞(−1)nn+1(a1a2)n$(210) $T1,2=−1a1∑n=0∞(−1)nn(a2a1)n$(211) $T2,1=−1a2∑n=0∞(−1)nn(a1a2)n$(212) $T2,2=1a1∑n=0∞(−1)nn+1(a2a1)n,$(213)

the series T1,1 and T1,2 come from the first sum S1 and T2,1, T2,2 from S2. Rule E3 indicates that the value of the integral is either $I(a1,a2)=T1,1+T2,1orI(a1,a2)=T1,2+T2,2;$(214)

the first form is an expression in a1/a2 and the second one in a2/a1.

The series T1, 1 is convergent when |a1|<|a2| and it produces the function $f(a1,a2)=1a1log⁡(1+a1a2)$(215)

and T2,2 is also convergent and is gives $g(a1,a2)=1a2log⁡(1+a2a1).$(216)

Observe that, according to (214) to complete the evaluation of I(a1, a2), some of the series required are partially divergent series. The question is how to make sense of these divergent series. The solution proposed here is, for instance, to interpret T2,1 as a partially divergent series attached to the function g(a1, a2). Therefore, the sum in (214), the term T2, 1 is replaced by g(a1, a2) to produce $I(a1,a2)=f(a1,a2)+g(a1,a2)=1a1log⁡(1+a1a2)+1a2log⁡(1+a2a1),$(217)

and this confirms (204). A similar interpretation of T1,2 + T2,2 gives the same result.

## 10 Conclusions

The method of brackets consists of a small number of heuristic rules used for the evaluation of definite integrals on [0, +∞). The original formulation of the method applied to functions that admit an expansion of the form $\begin{array}{}\sum _{n=0}^{\mathrm{\infty }}\end{array}$a(n)xαn+β−1. The results presented here extend this method to functions, like the Bessel function Kv and the exponential integral Ei, where the expansions have expansions of the form $\begin{array}{}\sum _{n=0}^{\mathrm{\infty }}\end{array}$ Γ(−n)xn (where all the coefficients are divergent) or $\begin{array}{}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\mathrm{\Gamma }\left(-n\right)}{x}^{n}\end{array}$ (where all the coefficients vanish). A variety of examples illustrate the validity of this formal procedure.

## Acknowledgement

The authors wish to thank a referee for a careful reading of the original version of the paper. The first author thanks the support of the Centro de Astrofísica de Valparaiso. The last author acknowledges the partial support of NSF-DMS 1112656.

## References

• [1]

Amdeberhan T., Dixit A., Guan X., Jiu L., Kuznetsov A., Moll V., Vignat C., The integrals in Gradshteyn and Ryzhik. Part 30: Trigonometric integrals. Scientia, 2016, 27:47–74. Google Scholar

• [2]

Amdeberhan T., Espinosa O., Gonzalez I., Harrison M., Moll V., Straub A., Ramanujan Master Theorem. The Ramanujan Journal, 2012, 29:103–120.

• [3]

Dixit A., Moll V., The integrals in Gradshteyn and Ryzhik. Part 28: The confluent hypergeometric function and Whittaker functions. Scientia, 2015, 26:49–61. Google Scholar

• [4]

Edwards J., A treatise on the Integral Calculus, volume I. MacMillan, New York, 1922. Google Scholar

• [5]

Edwards J., A treatise on the Integral Calculus, volume II. MacMillan, New York, 1922. Google Scholar

• [6]

Glasser M.L., Kohl K., Koutschan C., Moll V., Straub A., The integrals in Gradshteyn and Ryzhik. Part 22: Bessel-K functions. Scientia, 2012, 22:129–151. Google Scholar

• [7]

Gonzalez I., Jiu L., Moll V., Pochhammer symbols with negative indices. a new rule for the method of brackets. Open Mathematics, 2016, 14:681–686. Google Scholar

• [8]

Gonzalez I., Kohl K., Moll V., Evaluation of entries in Gradshteyn and Ryzhik employing the method of brackets. Scientia, 2014, 25:65–84. Google Scholar

• [9]

Gonzalez I., Moll V., Definite integrals by the method of brackets. Part 1. Adv. Appl. Math., 2010, 45:50–73.

• [10]

Gonzalez I., Moll V., Straub A., The method of brackets. Part 2: Examples and applications. In: Amdeberhan T., Medina L., Moll V.H. (Eds.), Gems in Experimental Mathematics, 2010, vol. 517 of Contemporary Mathematics, 157–172. American Mathematical Society, 2010. Google Scholar

• [11]

Gonzalez I., Schmidt I., Optimized negative dimensional integration method (NDIM) and multiloop Feynman diagram calculation. Nuclear Physics B, 2007, 769:124–173.

• [12]

Gonzalez I., Schmidt I., Modular application of an integration by fractional expansion (IBFE) method to multiloop Feynman diagrams. Phys. Rev. D, 2008, 78:086003.

• [13]

Gonzalez I., Schmidt I., Modular application of an integration by fractional expansion (IBFE) method to multiloop Feynman diagrams II. Phys. Rev. D, 2009, 79:126014.

• [14]

Gradshteyn I.S., Ryzhik I.M., Table of Integrals, Series, and Products. Edited by A. Jeffrey and D. Zwillinger. Academic Press, New York, 6th edition, 2000. Google Scholar

• [15]

Gradshteyn I.S., Ryzhik I.M., Table of Integrals, Series, and Products. Edited by A. Jeffrey and D. Zwillinger. Academic Press, New York, 7th edition, 2007. Google Scholar

• [16]

Gradshteyn I.S., Ryzhik I.M., Table of Integrals, Series, and Products. Edited by D. Zwillinger and V. Moll. Academic Press, New York, 8th edition, 2015. Google Scholar

• [17]

Kallio B.V., A history of the definite integral. Master’s thesis, University of British Columbia, 1966. Google Scholar

• [18]

Kölbig K.S., Two infinite integrals of products of modified Bessel functions and powers of logarithms. J. Comp. Appl. Math., 1995, 62:41–65.

• [19]

Moll V., The integrals in Gradshteyn and Ryzhik. Part 1: A family of logarithmic integrals. Scientia, 2007, 14:1–6. Google Scholar

• [20]

Moll V., Numbers and Functions. Special Functions for Undergraduates. Student Mathematical Library. American Mathematical Society, 2012. Google Scholar

• [21]

Olver F.W.J., Lozier D.W., Boisvert R.F., Clark C.W. (Eds.), NIST Handbook of Mathematical Functions. Cambridge University Press, 2010. Google Scholar

• [22]

Prudnikov A.P., Brychkov Yu.A., Marichev O.I., Integrals and Series, volume 2: Special Functions. Gordon and Breach Science Publishers, 1986. Google Scholar

Accepted: 2017-08-16

Published Online: 2017-09-27

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1181–1211, ISSN (Online) 2391-5455,

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