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# Integrals of Frullani type and the method of brackets

Sergio Bravo
/ Ivan Gonzalez
/ Karen Kohl
/ Victor H. Moll
Published Online: 2017-01-21 | DOI: https://doi.org/10.1515/math-2017-0001

## Abstract

The method of brackets is a collection of heuristic rules, some of which have being made rigorous, that provide a flexible, direct method for the evaluation of definite integrals. The present work uses this method to establish classical formulas due to Frullani which provide values of a specific family of integrals. Some generalizations are established.

MSC 2010: 33C67; 81T18

## 1 Introduction

The integral $∫0∞e−ax−e−bxxdx=logba$(1)

appears as entry 3.434.2 in [12]. It is one of the simplest examples of the so-called Frullani integrals. These are examples of the form $S(a,b)=∫0∞f(ax)−f(bx)xdx,$(2)

and Frullani’s theorem states that $S(a,b)=[f(0)−f(∞)]logba.$(3)

The identity (3) holds if, for example, f′ is a continuous function and the integral in (3) exists. Other conditions for the validity of this formula are presented in [3, 13, 16]. The reader will find in [1] a systematic study of the Frullani integrals appearing in [12].

The goal of the present work is to use the method of bractets, a new procedure for the evaluation of definite integrals, to compute a variety of integrals similar to those in (1). The method itself is described in Section 2. This is based on a small number of heuristic rules, some of which have been rigorously established [2, 8]. The point to be stressed here is that the application of the method of brackets is direct and it reduces the evaluation of a definite integral to the solution of a linear system of equations.

## 2 The method of brackets

A method to evaluate integrals over the half-line [0, ∞), based on a small number of rules has been developed in [6, 9-11]. This method of brackets is described next. The heuristic rules are currently being placed on solid ground [2]. The reader will find in [5, 7, 8] a large collection of evaluations of definite integrals that illustrate the power and flexibility of this method.

For a ϵ ℝ, the symbol $〈a〉=∫0∞xa−1dx,$(4)

is the bracket associated to the (divergent) integral on the right. The symbol $ϕn=(−1)nΓ(n+1),$(5)

is called the indicator associated to the index n. The notation ${\varphi }_{{n}_{1}{n}_{2}\cdots {n}_{r}},$or simply denotes ϕ12. . .r, the product ${\varphi }_{{n}_{1}}{\varphi }_{{n}_{2}}\cdots {\varphi }_{{n}_{r}}.$

## Rules for the production of bracket series

Rule P1. If the function ƒ is given by the power series $f(x)=∑n=0∞anxαn+β−1,$(6)

with α, β ϵℂ, then the integral of ƒ over [0, ∞) is converted into a bracket series by the procedure $∫0∞f(x)dx=∑nan〈αn+β〉.$(7)

Rule P2. For α ϵ ℂ, the multinomial power (a1 + a2 + … + ar)α is assigned the r-dimension bracket series $∑n1∑n2⋯∑nrϕn1n2⋯nra1n1⋯arnr〈−α+n1+⋯+nr〉Γ(−α).$(8)

## Rules for the evaluation of a bracket series

Rule E1. The one-dimensional bracket series is assigned the value $∑nϕnf(n)〈an+b〉=1|a|f(n∗)Γ(−n∗),$(9)

where n* is obtained from the vanishing of the bracket; that is, n* solves an + b = 0. This is precisely the Ramanujan’s Master Theorem.

The next rule provides a value for multi-dimensional bracket series of index 0, that is, the number of sums is equal to the number of brackets.

Rule E2. Assume the matrix A = (aij) is non-singular, then the assignment is $∑n1⋯∑nrϕn1⋯nrf(n1,⋯,nr){a11n1+⋯+a1rnr+c1}⋯{ar1n1+⋯+arrnr+cr}=1|det(A)|f(n1∗⋯nr∗)Γ(−n1∗)⋯Γ(−nr∗)$

where $\left\{{n}_{i}^{\ast }\right\}$ is the (unique) solution of the linear system obtained from the vanishing of the brackets.

Rule E3. The value of a multi-dimensional bracket series of positive index is obtained by computing all the contributions of maximal rank by Rule E2. These contributions to the integral appear as series in the free parameters. Series converging in a common region are added and divergent series are discarded.

## 3 The formula in one dimension

The goal of this section is to establish Frullani’s evaluation (3) by the method of brackets. The notation φk = (–1)k / Г(k + 1) is used in the statement of the next theorem.

#### Theorem 3.1

Assume f(x) admits an expansion of the form $f(x)=∑k=0∞ϕkC(k)xαk,forsomeα>0withC(0)≠0andC(0)<∞.$(1)

Then, $S(a,b):=∫0∞f(ax)−f(bx)xdx=limε→01|α|ΓεαC−εα(a−ε−b−ε)=C(0)logba,$(2)

independently of α.

#### Proof

Introduce an extra parameter and write $S(a,b)=limε→0∫0∞f(ax)−f(bx)x1−εdx.$(3)

Then, $(a,b)=limε→0∫0∞∑k=0∞ϕkC(k)aαk−bαk1∫0∞xαk+ε−1dx=limε→0∑kϕkC(k)aαk−bαk〈αk+ε〉.$

The method of brackets gives $S(a,b)=limε→01|α|ΓεαC−εαa−ε−b−ε.$(4)

The result follows from the expansions Γ (ε/α) = α/ε – γ + O(ϵ),

C(–ε/α) = C(0) + Ο (ε)and αε – bε = (log b – log a) ε + O(ε2).

In the examples given below, observe that C(0) = f(0) and that f(∞)= 0 is imposed as a condition on the integrand.

#### Example 3.2

Entry 3.434.2 of [12] states the value $∫0∞e−ax−e−bxxdx=log⁡ba.$(5)

This follows directly from (2).

#### Note 3.3

The method of brackets gives a direct approach to Frullani style problems if the expansion (1) is replaced by the more general one $f(x)=∑k=0∞ϕkC(k)xαk+β,$(6)

with β ≠ 0 and if the function ƒ does not necessarily have a limit at infinity.

#### Example 3.4

Consider the evaluation of $I=∫0∞sin⁡ax−sin⁡bxxdx,$(7)

for a, b > 0. The integral is evaluated directly as $I=∫0∞sin⁡axxdx−∫0∞sin⁡bxxdx,$(8)

and since a, b > 0, both integrals are π/2, giving I = 0. The classical version of Frullani theorem does not apply, since ƒ (x) does not have a limit as x → ∞. Ostrowski [15] shows that in the case ƒ (x) is periodic of period p, the value ƒ (∞) might be replaced by $1p∫0pf(x)dx.$(9)

In the present case, ƒ (x) = sin χ has period 2π and mean 0. This yields the vanishing of the integral. The computation of (7) by the method of brackets begins with the expansion $sin⁡x=x⋅0F1−32−14x2.$(10)

Here $pFqb1,⋯,bqa1,⋯,ap|z=∑n=0X(a1)n⋯(ap)n(b1)n⋅(bq)nznn!,$(11)

with (a) n = a (a + 1) ··· (a +n – 1), is the classical hypergeometric function. The integrand has the series expansion $∑n≥0ϕn(a2n+1−b2n+1)(32)n4nx2n,$(12)

that yields $I=∑nϕn(a2n+1−b2n+1)(32)n4n〈2n+1〉.$(13)

The vanishing of the bracket gives n* = —1/2 and the bracket series vanishes in view of the factor a2n+1b2n+1.

#### Example 3.5

The next example is the evaluation of $I=∫0∞cos⁡ax−cos⁡bxxdx=logba,$(14)

for a, b > 0. The expansion $cos⁡x=∑n=0ϕnn!(2n)!x2n,$(15)

and $C\left(n\right)=\frac{n!}{\left(2n\right)!}=\frac{\mathrm{\Gamma }\left(n+1\right)}{\mathrm{\Gamma }\left(2n+1\right)}$ in (1). Then C(0) = 1 and the integral is $I=\mathrm{log}\left(\frac{b}{a}\right),$ as claimed.

#### Example 3.6

The integral $I=∫0∞tan−1⁡(e−ax)−tan−1⁡(e−bx)xdx,$(16)

is evaluated next. The expansion of the integrand is $tan−1(e−t)=e−t⋅2F112132−e−2t$ $=12∑n=0∞ϕnΓ(n+12)Γ(n+1)Γ(n+32)∑k=0∞ϕk(2n+1)ktk=∑k=0∞ϕk12∑n=0∞ϕnΓ(n+12)Γ(n+1)Γ(n+32)(2n+1)ktk.$

Therefore, $C(k)=12∑n=0∞ϕnΓ(n+12)Γ(n+1)Γ(n+32)(2n+1)k,$(17)

and from here it follows that $C(0)=12∑n=0∞ϕnΓ(n+12)Γ(n+1)Γ(n+32)=tan−1⁡(1)=π4.$(18)

Thus, the integral is $I=C(0)logba=π4logba.$(19)

## 4 A first generalization

This section describes examples of Frullani-type integrals that have an expansion of the form $f(x)=∑k≥0ϕkC(k)xαk+β,$(20)

with β ≠ 0.

#### Theorem 4.1

Assume f(x) admits an expansion of the form (20). Then, $S(a,b)=∫0∞f(ax)−f(bx)xdx=limε→01|α|Γβα+εαC−βα−εα[a−ε−b−ε].$(21)

#### Proof

The method of brackets gives $S(a,b;ε)=∫0∞f(ax)−f(bx)x1−ε=∑k⩾0ϕkC(k)aαk+β−bαk+β∫0∞xαk+β+ε−1dx=∑kϕkC(k)aαk+β−bαk+β〈αk+β+ε〉=1|α|Γ(−k)C(k)aαk+β−bαk+β$(22)

with k = – (β + ϵ) / α in the last line. The result follows by taking ϵ 0.

#### Example 4.2

The integral $∫0∞tan−1⁡ax−tan−1⁡bxx=−π2logba$(23)

appears as entry 4.536.2 in [12]. It is evaluated directly by the classical Frullani theorem. Its evaluation by the method of brackets comes from the expansion $tan−1x=x⋅2F112132−x2=∑k⩾0ϕk12k1k32kx2k+1.$(24)

Therefore, a = 2, β = 1 and $C(k)=Γ(12+k)Γ(1+k)2Γ(32+k)=Γ(1+k)2k+1.$(25)

Then $∫0∞tan−1ax−tan−1bxx=limε→012Γ1+ε2C−1+ε2[a−ε−b−ε]=limε→012Γ1+ε2Γ1−ε2[a−ε−b−ε]−ε=−π2logba.$

## 5 A second class of Frullani type integrals

Let f1, ···, fN be a family of functions. This section uses the method of brackets to evaluate $I=I(f1,⋯,fN)=∫0∞⁡1x∑k=1Nfk(x)dx,$(1)

subject to the condition $\sum _{k=1}^{N}{f}_{k}\left(0\right)=0,$ required for convergence.

The functions {fk(x)}are assumed to admit a series representation of the form $fk(x)=∑n=0∞ϕnCk(n)xαn,$(2)

where α > 0 is independent of k and Ck (0) ≠ 0. The coefficients Ck are assumed to admit a meromorphic extension from n ϵ ℕ to n ϵ ℂ.

#### Theorem 5.1

The integral I is given by $I=−1|α|∑k=1XCk′(0),$(3)

where $Ck′(0)=dCk(ε)dε|ε=0.$(4)

#### Proof

The proof begins with the expansion $fk(x)x1−ε=∑n=0∞ϕnCk(n)xαn−1+ε$(5)

and the bracket series for the integral is $I=limε→0∑nϕn∑k=1NCk(n)〈αn+ε〉=limε→01|α|Γ−ϵα∑k=1NCk−εα.$(6)

The result follows by letting ε → 0.

#### Example 5.2

Entry 3.429 in [12] states that $I=∫0∞e−x−1+x−μdxx=ψμ,$(7)

where μ > 0 and ψ(x) = Г (x) / Г(x) is the digamma function. This is one of many integral representation for this basic function. The reader will find a classical proof of this identity in [14]. The method of brackets gives a direct proof.

The functions appearing in this example are $f1(x)=e−x=∑n=0∞ϕnxn,$(8)

and $f2(x)=−(1+x)−μ=−∑n=0∞ϕn(μ)nxn,$(9)

where (μ)n = μ(μ + 1) (μ + n – 1) is the Pochhammer symbol (this comes directly from the binomial theorem). The condition ƒ1(0) + ƒ2(0) = 0 is satisfied and the coefficients are identified as $C1(n)=1andC2(n)=−(μ)n=−Γ(μ+n)Γ(μ).$(10)

Then, ${C}_{1}^{\prime }\left(0\right)=1\phantom{\rule{0.056em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{0.056em}{0ex}}{C}_{2}^{\prime }\left(0\right)=-\frac{{\mathrm{\Gamma }}^{\prime }\left(\mu \right)}{\mathrm{\Gamma }\left(\mu \right)}.$ This gives the evaluation.

#### Example 5.3

The elliptic integrals K(x) and E(x) may be expressed in hypergeometric form as $K(x)=π22F112121x2andE(x)=π22F1−12121x2$(11)

Theorem 5.1 is now used to establish the value $∫0∞πe−ax2−K(bx)−E(cx)xdx=π2logbca−γ−4log⁡2+1.$(12)

Here γ = –Γ′(1) is Euler’s constant.

The first step is to compute series expansions of each of the terms in the integrand. The exponential term is easy: $πe−ax2=π∑n1=0∞(−ax2)n1n1!=∑n1ϕn1an1x2n1,$(13)

and this gives C1(n) = an.For the first elliptic integral, $K(bx)=π22F112121b2x2$ $=π2∑n2=0∞(12)n2(12)n2(1)n2n2!b2n2x2n2=∑n2ϕn2π2(−1)n2b2n2n2!12n22x2n2.$

Therefore, $C2(n)=π2cos⁡(πn)Γ2(n+12)Γ(n+1)b2n,$(14)

where the term (–1)n has been replaced by cos(πn). A similar calculation gives $C3(n)=π4cos⁡(πn)Γn−12Γn+12Γ(n+1)c2n.$(15)

A direct calculation gives $C1′(0)=log⁡a,C2′(0)=−γ2−log⁡b−ψ12andC3′(0)=−γ2−log⁡c−ψ−12.$

The result now comes from the values $ψ12=−2log⁡2−γandψ−12=−2log⁡2−γ+2.$(16)

#### Example 5.4

Let a, b ϵ ℝ with a > 0. Then $∫0∞exp⁡(−ax2)−cos⁡bxxdx=γ−log⁡a+2log⁡b2.$(17)

To apply Theorem 5.1 start with the series $f1(x)=e−ax2=∑nϕnanx2n$(18)

and $f2(x)=cos⁡bx=∑nϕnΓ(n+1)Γ(2n+1)b2nx2n.$(19)

In both expansions α = 2 and the coefficients are given by $C1(n)=anandC2(n)=Γ(n+1)Γ(2n+1)b2n.$(20)

Then, ${C}_{1}^{\prime }\left(0\right)=\mathrm{log}a$ and ${C}_{2}^{\prime }\left(n\right)=\frac{{b}^{2n}\mathrm{\Gamma }\left(n+1\right)}{\mathrm{\Gamma }\left(2n+1\right)}\left[2\mathrm{log}b+\psi \left(n+1\right)-\psi \left(2n+1\right)\right]\phantom{\rule{thinmathspace}{0ex}}\mathit{y}\mathit{i}\mathit{e}\mathit{l}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}{C}_{2}^{\prime }\left(0\right)=2\mathrm{log}b-\psi \left(1\right)=2\mathrm{log}b+\gamma .$ The value (17) follows from here.

#### Example 5.5

The next example in this section involves the Bessel function of order 0 $J0(x)=∑n=0∞(−1)nn!2x22n$(21)

and Theorem 5.1 is used to evaluate $∫0∞J0(x)−cos⁡axxdx=log⁡2a.$(22)

This appears as entry 6.693.8 in [12]. The expansions $J0(x)=∑n=0∞ϕn1n!22nx2nandcos⁡ax=∑n=0∞ϕnn!(2n)!a2nx2n,$(23)

show a = 2 and $C1(n)=1Γ(n+1)22nandC2(n)=−Γ(n+1)Γ(2n+1)a2n.$(24)

Differentiation gives $C1′(n)=−2ln⁡2+ψ(n+1)22nΓ(n+1),$(25)

and $C2′(n)=−a2nΓ(n+1)(2log⁡a+ψ(n+1)−2ψ(2n+1))Γ(2n+1).$(26)

Then, $C1′(0)=γ−2log⁡2andC2′(0)=−(γ+2log⁡a),$(27)

and the result now follows from Theorem 5.1. The reader is invited to use the representation $J02(x)=1F21211−x2,$(28)

to verify the identity $∫0∞J02(x)−cos⁡xxdx=log⁡2.$(29)

#### Example 5.6

The final example in this section is $I=∫0∞J02(x)−e−x2cos⁡xxdx.$(30)

The evaluation begins with the expansions $J0(x)=∑k=0∞ϕkx2k4kΓ(k+1)andcos⁡x=∑k=0Xϕkπ4kΓ(k+12).$(31)

Then, $J02(x)=∑k,nϕk,n14k+nΓ(k+1)Γ(n+1)x2k+2n,$(32)

and $e−x2cos⁡x=∑k,nϕk,nπ4kΓ(k+12)x2k+2n.$(33)

Integration yields $I=∫0∞J02(x)−e−x2cos⁡xx1−εdx=∑k,nϕk,n14k+nΓ(k+1)Γ(n+1)−π4kΓ(k+12)∫0∞x2k+2n+ε−1dx=∑k,nϕk,n14k+nΓ(k+1)Γ(n+1)−π4kΓ(k+12)〈2k+2n+ϵ〉.$

The method of brackets now gives $I=limε→012∑k=0∞(−1)kΓ(k+ε2)k!12−εΓ(k+1)Γ(1−k−ε/2)−π22kΓ(k+12).$

The term corresponding to k = 0 gives $limε→012Γϵ212−εΓ(1−ε2)−1=log⁡2−γ2$(34)

and the terms with k ≥ 1 as ε → 0 give $−π2∑k=1∞ϕkΓ(k)22kΓk+12=142F211322−14.$(35)

Therefore, $∫0∞J02(x)−e−x2cos⁡xxdx=144log⁡2−2γ+2F211322−14.$(36)

No further simplification seems to be possible.

## 6 A multi-dimensional extension

The method of brackets provides a direct proof of the following multi-dimensional extension of Frullani’s theorem.

#### Theorem 6.1

Let aj , bj ϵ ℝ+. Assume the function ƒ has an expansion of the form $f(x1,⋯,xn)=∑ℓ1,⋯,ℓn=0∞(−1)ℓ1ℓ1!⋯(−1)ℓnℓn!C(ℓ1,⋯,ℓn)x1γ1⋯xnγn,$(1)

where the γj are linear functions of the indices given by $γ1=α11ℓ1+⋯+α1nℓn+β1…………………γn=αn1ℓ1+⋯+αnnℓn+βn.$(2)

Then, $I=∫0∞…∫0∞f(b1x1,⋯,bnxn)−f(a1x1,⋯,anxn)x11+ρ1⋯xn1+ρndx1⋯dxn=1|detA|limε→0[b1ρ1−ε⋯bnρn−ε−a1ρ1−ε⋯anρn−ε]Γ(−ℓ1∗)⋯Γ(−ℓn∗)C(ℓ1∗,⋯,ℓn∗),$

where A = (αij) is the matrix of coefficients in (2) and ${\ell }_{j}^{\ast },$ 1 ≤ j ≤ n is the solution to the linear system $α11ℓ1+⋯+α1nℓn+β1−ρ1+ε=0⋯⋯⋯⋯⋯⋯⋯⋯αn1ℓ1+⋯+αnnℓn+βn−ρn+ε=0.$(3)

#### Proof

The proof is a direct extension of the one-dimensional case, so it is omitted.

#### Example 6.2

The evaluation of the integral $I=∫0∞∫0∞e−μst2cos⁡(ast)−e−μst2cos⁡(bst)sdsdt$(4)

uses the expansion $f(s,t)=e−st2cos⁡(st)=∑n1∑n2ϕ1,2πΓ(n2+12)4n2sn1+2n2t2n1+2n2,$(5)

with parameters ${\rho }_{1}=-\frac{1}{2},{\rho }_{2}=-1,{b}_{1}={a}^{2}/\mu ,{b}_{2}=\mu /a,{a}_{1}={b}^{2}/\mu ,{a}_{2}=\mu /b.$ The solution to the linear system is ${n}_{1}^{\ast }=-\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{n}_{2}^{\ast }=-\frac{\epsilon }{2}\phantom{\rule{thinmathspace}{0ex}}$ and |det A| = 2. Then $I=12limε→0a2μ−1/2−εμa−1−ε−b2μ−1/2−εμb−1−ε×Γ12Γε2πΓ1−ε24−ε/2=πμlimε→0bε−aεε×Γ(1+ε)cosπε2(ab)E=πμlogba.$

The double integral (4) has been evaluated.

#### Example 6.3

The method is now used to evaluate $∫0∞∫0∞sin⁡(μxy2)cos⁡(axy)−sin⁡(μxy2)cos⁡(bxy)xy=π2log⁡ba.$(6)

The evaluation begins with the expansion $f(x,y)=sin⁡(xy2)cos⁡(xy)=xy2∑n1≥0ϕn1Γ32(xy2)2n1Γn1+324n1∑n2≥0ϕn2Γ12(xy)2n2Γn2+124n2=∑n1∑n2ϕn1ϕn2π2Γn1+32Γn2+124n1+n2x2n1+2n2+1y4n1+2n2.$

The parameters are b1 = a2/μ, b2 = μ/α, a1 = b2/μ, a2 = μ/b and ρ1 = ρ2 = 0. The solution to the linear system is ${n}_{1}^{\ast }=-\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\mathit{a}\mathit{n}\mathit{d}\phantom{\rule{thinmathspace}{0ex}}{n}_{2}^{\ast }=-\frac{\epsilon }{2}$ and |det A| = 4. Then, $I=limε→0a−ε−b−ε4Γ12Γε2π2Γ(1)Γ1−ε24−ε−1)/2=limε→0π3/24ε/24bε−aε(ab)E21−2επΓ(ε)πcsc12+ε2=π2logba,$

as claimed.

## 7 Conclusions

The method of brackets consists of a small number of heuristic rules that reduce the evaluation of a definite integral to the solution of a linear system of equations. The method has been used to establish a classical theorem of Frullani and to evaluate, in an algorithmic manner, a variety of integrals of Frullani type. The flexibility of the method yields a direct and simple solution to these evaluations.

## Acknowledgement

The second author wishes to thank the partial support of the Centro de Astrofísica de Valparaiso (CAV). The last author acknowledges the partial support of NSF-DMS 1112656.

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Accepted: 2016-12-27

Published Online: 2017-01-21

Citation Information: Open Mathematics, ISSN (Online) 2391-5455,

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