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formerly Central European Journal of Mathematics

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Volume 15, Issue 1 (Jan 2017)


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


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.

Keywords: Definite integrals; Frullani integrals; Method of brackets

MSC 2010: 33C67; 81T18

1 Introduction

The integral 0eaxebxxdx=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)=0f(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=0xa1dx,(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 ϕn1n2nr,or simply denotes ϕ12. . .r, the product ϕn1ϕn2ϕnr.

Rules for the production of bracket series

Rule P1. If the function ƒ is given by the power series f(x)=n=0anxαn+β1,(6)

with α, β ϵℂ, then the integral of ƒ over [0, ∞) is converted into a bracket series by the procedure 0f(x)dx=nanαn+β.(7)

Rule P2. For α ϵ ℂ, the multinomial power (a1 + a2 + … + ar)α is assigned the r-dimension bracket series n1n2nrϕn1n2nra1n1arnrα+n1++nrΓ(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 n1nrϕn1nrf(n1,,nr){a11n1++a1rnr+c1}{ar1n1++arrnr+cr}=1|det(A)|f(n1nr)Γ(n1)Γ(nr)

where {ni} 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=0ϕkC(k)xαk,forsomeα>0withC(0)0andC(0)<.(1)

Then, S(a,b):=0f(ax)f(bx)xdx=limε01|α|ΓεαCεα(aεbε)=C(0)logba,(2)

independently of α.


Introduce an extra parameter and write S(a,b)=limε00f(ax)f(bx)x1εdx.(3)

Then, (a,b)=limε00k=0ϕkC(k)aαkbαk10xαk+ε1dx=limε0kϕkC(k)aαkbαkαk+ε.

The method of brackets gives S(a,b)=limε01|α|ΓεαCεαaεbε.(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 0eaxebxxdx=logba.(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=0ϕkC(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=0sinaxsinbxxdx,(7)

for a, b > 0. The integral is evaluated directly as I=0sinaxxdx0sinbxxdx,(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 1p0pf(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 sinx=x0F13214x2.(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 n0ϕn(a2n+1b2n+1)(32)n4nx2n,(12)

that yields I=nϕn(a2n+1b2n+1)(32)n4n2n+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=0cosaxcosbxxdx=logba,(14)

for a, b > 0. The expansion cosx=n=0ϕnn!(2n)!x2n,(15)

and C(n)=n!(2n)!=Γ(n+1)Γ(2n+1) in (1). Then C(0) = 1 and the integral is I=logba, as claimed.

Example 3.6

The integral I=0tan1(eax)tan1(ebx)xdx,(16)

is evaluated next. The expansion of the integrand is tan1(et)=et2F112132e2t =12n=0ϕnΓ(n+12)Γ(n+1)Γ(n+32)k=0ϕk(2n+1)ktk=k=0ϕk12n=0ϕnΓ(n+12)Γ(n+1)Γ(n+32)(2n+1)ktk.

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

and from here it follows that C(0)=12n=0ϕnΓ(n+12)Γ(n+1)Γ(n+32)=tan1(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)=k0ϕkC(k)xαk+β,(20)

with β ≠ 0.

Theorem 4.1

Assume f(x) admits an expansion of the form (20). Then, S(a,b)=0f(ax)f(bx)xdx=limε01|α|Γβα+εαCβαεα[aεbε].(21)


The method of brackets gives S(a,b;ε)=0f(ax)f(bx)x1ε=k0ϕkC(k)aαk+βbαk+β0xαk+β+ε1dx=kϕkC(k)aαk+βbαk+βαk+β+ε=1|α|Γ(k)C(k)aαk+βbαk+β(22)

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

Example 4.2

The integral 0tan1axtan1bxx=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 tan1x=x2F112132x2=k0ϕk12k1k32kx2k+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 0tan1axtan1bxx=limε012Γ1+ε2C1+ε2[aεbε]=limε012Γ1+ε2Γ1ε2[aεbε]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)=01xk=1Nfk(x)dx,(1)

subject to the condition k=1Nfk(0)=0, required for convergence.

The functions {fk(x)}are assumed to admit a series representation of the form fk(x)=n=0ϕnCk(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)


The proof begins with the expansion fk(x)x1ε=n=0ϕnCk(n)xαn1+ε(5)

and the bracket series for the integral is I=limε0nϕnk=1NCk(n)αn+ε=limε01|α|Γϵαk=1NCkεα.(6)

The result follows by letting ε → 0.

Example 5.2

Entry 3.429 in [12] states that I=0ex1+xμdxx=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)=ex=n=0ϕnxn,(8)

and f2(x)=(1+x)μ=n=0ϕn(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, C1(0)=1andC2(0)=Γ(μ)Γ(μ). 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)=π22F112121x2(11)

The reader will find information about these integrals in [4, 17].

Theorem 5.1 is now used to establish the value 0πeax2K(bx)E(cx)xdx=π2logbcaγ4log2+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: πeax2=πn1=0(ax2)n1n1!=n1ϕn1an1x2n1,(13)

and this gives C1(n) = an.For the first elliptic integral, K(bx)=π22F112121b2x2 =π2n2=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)Γn12Γn+12Γ(n+1)c2n.(15)

A direct calculation gives C1(0)=loga,C2(0)=γ2logbψ12andC3(0)=γ2logcψ12.

The result now comes from the values ψ12=2log2γandψ12=2log2γ+2.(16)

Example 5.4

Let a, b ϵ ℝ with a > 0. Then 0exp(ax2)cosbxxdx=γloga+2logb2.(17)

To apply Theorem 5.1 start with the series f1(x)=eax2=nϕnanx2n(18)

and f2(x)=cosbx=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, C1(0)=loga and C2(n)=b2nΓ(n+1)Γ(2n+1)[2logb+ψ(n+1)ψ(2n+1)]yieldC2(0)=2logbψ(1)=2logb+γ. 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 0J0(x)cosaxxdx=log2a.(22)

This appears as entry 6.693.8 in [12]. The expansions J0(x)=n=0ϕn1n!22nx2nandcosax=n=0ϕnn!(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)=2ln2+ψ(n+1)22nΓ(n+1),(25)

and C2(n)=a2nΓ(n+1)(2loga+ψ(n+1)2ψ(2n+1))Γ(2n+1).(26)

Then, C1(0)=γ2log2andC2(0)=(γ+2loga),(27)

and the result now follows from Theorem 5.1. The reader is invited to use the representation J02(x)=1F21211x2,(28)

to verify the identity 0J02(x)cosxxdx=log2.(29)

Example 5.6

The final example in this section is I=0J02(x)ex2cosxxdx.(30)

The evaluation begins with the expansions J0(x)=k=0ϕkx2k4kΓ(k+1)andcosx=k=0Xϕkπ4kΓ(k+12).(31)

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

and ex2cosx=k,nϕk,nπ4kΓ(k+12)x2k+2n.(33)

Integration yields I=0J02(x)ex2cosxx1εdx=k,nϕk,n14k+nΓ(k+1)Γ(n+1)π4kΓ(k+12)0x2k+2n+ε1dx=k,nϕk,n14k+nΓ(k+1)Γ(n+1)π4kΓ(k+12)2k+2n+ϵ.

The method of brackets now gives I=limε012k=0(1)kΓ(k+ε2)k!12εΓ(k+1)Γ(1kε/2)π22kΓ(k+12).

The term corresponding to k = 0 gives limε012Γϵ212εΓ(1ε2)1=log2γ2(34)

and the terms with k ≥ 1 as ε → 0 give π2k=1ϕkΓ(k)22kΓk+12=142F21132214.(35)

Therefore, 0J02(x)ex2cosxxdx=144log22γ+2F21132214.(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)=1,,n=0(1)11!(1)nn!C(1,,n)x1γ1xnγn,(1)

where the γj are linear functions of the indices given by γ1=α111++α1nn+β1γn=αn11++αnnn+βn.(2)

Then, I=00f(b1x1,,bnxn)f(a1x1,,anxn)x11+ρ1xn1+ρndx1dxn=1|detA|limε0[b1ρ1εbnρnεa1ρ1εanρnε]Γ(1)Γ(n)C(1,,n),

where A = (αij) is the matrix of coefficients in (2) and j, 1 ≤ j ≤ n is the solution to the linear system α111++α1nn+β1ρ1+ε=0αn11++αnnn+βnρn+ε=0.(3)


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

Example 6.2

The evaluation of the integral I=00eμst2cos(ast)eμst2cos(bst)sdsdt(4)

uses the expansion f(s,t)=est2cos(st)=n1n2ϕ1,2πΓ(n2+12)4n2sn1+2n2t2n1+2n2,(5)

with parameters ρ1=12,ρ2=1,b1=a2/μ,b2=μ/a,a1=b2/μ,a2=μ/b. The solution to the linear system is n1=12andn2=ε2 and |det A| = 2. Then I=12limε0a2μ1/2εμa1εb2μ1/2εμb1ε×Γ12Γε2πΓ1ε24ε/2=πμlimε0bεaεε×Γ(1+ε)cosπε2(ab)E=πμlogba.

The double integral (4) has been evaluated.

Example 6.3

The method is now used to evaluate 00sin(μxy2)cos(axy)sin(μxy2)cos(bxy)xy=π2logba.(6)

The evaluation begins with the expansion f(x,y)=sin(xy2)cos(xy)=xy2n10ϕn1Γ32(xy2)2n1Γn1+324n1n20ϕn2Γ12(xy)2n2Γn2+124n2=n1n2ϕn1ϕn2π2Γn1+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 n1=12andn2=ε2 and |det A| = 4. Then, I=limε0aεbε4Γ12Γε2π2Γ(1)Γ1ε24ε1)/2=limε0π3/24ε/24bεaε(ab)E212επΓ(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.


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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About the article

Received: 2016-09-13

Accepted: 2016-12-27

Published Online: 2017-01-21

Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0001.

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© 2017 Bravo et al.. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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