BY-NC-ND 3.0 license Open Access Published by De Gruyter Open Access January 21, 2017

Integrals of Frullani type and the method of brackets

Sergio Bravo, Ivan Gonzalez, Karen Kohl and Victor H. Moll
From the journal Open Mathematics

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

(1) 0 e a x e b x x d x = log b a

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

(2) S ( a , b ) = 0 f ( a x ) f ( b x ) x d x ,

and Frullani’s theorem states that

(3) S ( a , b ) = [ f ( 0 ) f ( ) ] log b a .

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

(4) a = 0 x a 1 d x ,

is the bracket associated to the (divergent) integral on the right. The symbol

(5) ϕ n = ( 1 ) n Γ ( n + 1 ) ,

is called the indicator associated to the index n. The notation ϕ n 1 n 2 n r , or simply denotes ϕ12. . .r, the product ϕ n 1 ϕ n 2 ϕ n r .

Rules for the production of bracket series

Rule P1. If the function ƒ is given by the power series

(6) f ( x ) = n = 0 a n x α n + β 1 ,

with α, β ϵℂ, then the integral of ƒ over [0, ∞) is converted into a bracket series by the procedure

(7) 0 f ( x ) d x = n a n α n + β .

Rule P2. For α ϵ ℂ, the multinomial power (a1 + a2 + … + ar)α is assigned the r-dimension bracket series

(8) n 1 n 2 n r ϕ n 1 n 2 n r a 1 n 1 a r n r α + n 1 + + n r Γ ( α ) .

Rules for the evaluation of a bracket series

Rule E1. The one-dimensional bracket series is assigned the value

(9) n ϕ n f ( n ) a n + b = 1 | a | f ( n ) Γ ( n ) ,

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

n 1 n r ϕ n 1 n r f ( n 1 , , n r ) { a 11 n 1 + + a 1 r n r + c 1 } { a r 1 n 1 + + a r r n r + c r } = 1 | det ( A ) | f ( n 1 n r ) Γ ( n 1 ) Γ ( n r )

where { n i } 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

(1) f ( x ) = k = 0 ϕ k C ( k ) x α k , f o r s o m e α > 0 w i t h C ( 0 ) 0 a n d C ( 0 ) < .

Then,

(2) S ( a , b ) := 0 f ( a x ) f ( b x ) x d x = lim ε 0 1 | α | Γ ε α C ε α ( a ε b ε ) = C ( 0 ) log b a ,

independently of α.

Proof

Introduce an extra parameter and write

(3) S ( a , b ) = lim ε 0 0 f ( a x ) f ( b x ) x 1 ε d x .

Then,

( a , b ) = lim ε 0 0 k = 0 ϕ k C ( k ) a α k b α k 1 0 x α k + ε 1 d x = lim ε 0 k ϕ k C ( k ) a α k b α k α k + ε .

The method of brackets gives

(4) S ( a , b ) = lim ε 0 1 | α | Γ ε α C ε α a ε b ε .

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

(5) 0 e a x e b x x d x = log b a .

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

(6) f ( x ) = k = 0 ϕ k C ( k ) x α k + β ,

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

Example 3.4

Consider the evaluation of

(7) I = 0 sin a x sin b x x d x ,

for a, b > 0. The integral is evaluated directly as

(8) I = 0 sin a x x d x 0 sin b x x d x ,

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

(9) 1 p 0 p f ( x ) d x .

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

(10) sin x = x 0 F 1 3 2 1 4 x 2 .

Here

(11) p F q b 1 , , b q a 1 , , a p | z = n = 0 X ( a 1 ) n ( a p ) n ( b 1 ) n ( b q ) n z n n ! ,

with (a) n = a (a + 1) ··· (a +n – 1), is the classical hypergeometric function. The integrand has the series expansion

(12) n 0 ϕ n ( a 2 n + 1 b 2 n + 1 ) ( 3 2 ) n 4 n x 2 n ,

that yields

(13) I = n ϕ n ( a 2 n + 1 b 2 n + 1 ) ( 3 2 ) n 4 n 2 n + 1 .

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

(14) I = 0 cos a x cos b x x d x = log b a ,

for a, b > 0. The expansion

(15) cos x = n = 0 ϕ n n ! ( 2 n ) ! x 2 n ,

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

Example 3.6

The integral

(16) I = 0 tan 1 ( e a x ) tan 1 ( e b x ) x d x ,

is evaluated next. The expansion of the integrand is

tan 1 ( e t ) = e t 2 F 1 1 2 1 3 2 e 2 t

= 1 2 n = 0 ϕ n Γ ( n + 1 2 ) Γ ( n + 1 ) Γ ( n + 3 2 ) k = 0 ϕ k ( 2 n + 1 ) k t k = k = 0 ϕ k 1 2 n = 0 ϕ n Γ ( n + 1 2 ) Γ ( n + 1 ) Γ ( n + 3 2 ) ( 2 n + 1 ) k t k .

Therefore,

(17) C ( k ) = 1 2 n = 0 ϕ n Γ ( n + 1 2 ) Γ ( n + 1 ) Γ ( n + 3 2 ) ( 2 n + 1 ) k ,

and from here it follows that

(18) C ( 0 ) = 1 2 n = 0 ϕ n Γ ( n + 1 2 ) Γ ( n + 1 ) Γ ( n + 3 2 ) = tan 1 ( 1 ) = π 4 .

Thus, the integral is

(19) I = C ( 0 ) log b a = π 4 log b a .

4 A first generalization

This section describes examples of Frullani-type integrals that have an expansion of the form

(20) f ( x ) = k 0 ϕ k C ( k ) x α k + β ,

with β ≠ 0.

Theorem 4.1

Assume f(x) admits an expansion of the form (20). Then,

(21) S ( a , b ) = 0 f ( a x ) f ( b x ) x d x = lim ε 0 1 | α | Γ β α + ε α C β α ε α [ a ε b ε ] .

Proof

The method of brackets gives

(22) S ( a , b ; ε ) = 0 f ( a x ) f ( b x ) x 1 ε = k 0 ϕ k C ( k ) a α k + β b α k + β 0 x α k + β + ε 1 d x = k ϕ k C ( k ) a α k + β b α k + β α k + β + ε = 1 | α | Γ ( k ) C ( k ) a α k + β b α k + β

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

Example 4.2

The integral

(23) 0 tan 1 a x tan 1 b x x = π 2 log b a

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

(24) tan 1 x = x 2 F 1 1 2 1 3 2 x 2 = k 0 ϕ k 1 2 k 1 k 3 2 k x 2 k + 1 .

Therefore, a = 2, β = 1 and

(25) C ( k ) = Γ ( 1 2 + k ) Γ ( 1 + k ) 2 Γ ( 3 2 + k ) = Γ ( 1 + k ) 2 k + 1 .

Then

0 tan 1 a x tan 1 b x x = lim ε 0 1 2 Γ 1 + ε 2 C 1 + ε 2 [ a ε b ε ] = lim ε 0 1 2 Γ 1 + ε 2 Γ 1 ε 2 [ a ε b ε ] ε = π 2 log b a .

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

(1) I = I ( f 1 , , f N ) = 0 1 x k = 1 N f k ( x ) d x ,

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

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

(2) f k ( x ) = n = 0 ϕ n C k ( n ) x α n ,

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

(3) I = 1 | α | k = 1 X C k ( 0 ) ,

where

(4) C k ( 0 ) = d C k ( ε ) d ε | ε = 0 .

Proof

The proof begins with the expansion

(5) f k ( x ) x 1 ε = n = 0 ϕ n C k ( n ) x α n 1 + ε

and the bracket series for the integral is

(6) I = lim ε 0 n ϕ n k = 1 N C k ( n ) α n + ε = lim ε 0 1 | α | Γ ϵ α k = 1 N C k ε α .

The result follows by letting ε → 0.

Example 5.2

Entry 3.429 in [12] states that

(7) I = 0 e x 1 + x μ d x x = ψ μ ,

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

(8) f 1 ( x ) = e x = n = 0 ϕ n x n ,

and

(9) f 2 ( x ) = ( 1 + x ) μ = n = 0 ϕ n ( μ ) n x n ,

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

(10) C 1 ( n ) = 1 a n d C 2 ( n ) = ( μ ) n = Γ ( μ + n ) Γ ( μ ) .

Then, C 1 ( 0 ) = 1 a n d C 2 ( 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

(11) K ( x ) = π 2 2 F 1 1 2 1 2 1 x 2 a n d E ( x ) = π 2 2 F 1 1 2 1 2 1 x 2

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

Theorem 5.1 is now used to establish the value

(12) 0 π e a x 2 K ( b x ) E ( c x ) x d x = π 2 log b c a γ 4 log 2 + 1 .

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:

(13) π e a x 2 = π n 1 = 0 ( a x 2 ) n 1 n 1 ! = n 1 ϕ n 1 a n 1 x 2 n 1 ,

and this gives C1(n) = an.For the first elliptic integral,

K ( b x ) = π 2 2 F 1 1 2 1 2 1 b 2 x 2

= π 2 n 2 = 0 ( 1 2 ) n 2 ( 1 2 ) n 2 ( 1 ) n 2 n 2 ! b 2 n 2 x 2 n 2 = n 2 ϕ n 2 π 2 ( 1 ) n 2 b 2 n 2 n 2 ! 1 2 n 2 2 x 2 n 2 .

Therefore,

(14) C 2 ( n ) = π 2 cos ( π n ) Γ 2 ( n + 1 2 ) Γ ( n + 1 ) b 2 n ,

where the term (–1)n has been replaced by cos(πn). A similar calculation gives

(15) C 3 ( n ) = π 4 cos ( π n ) Γ n 1 2 Γ n + 1 2 Γ ( n + 1 ) c 2 n .

A direct calculation gives

C 1 ( 0 ) = log a , C 2 ( 0 ) = γ 2 log b ψ 1 2 a n d C 3 ( 0 ) = γ 2 log c ψ 1 2 .

The result now comes from the values

(16) ψ 1 2 = 2 log 2 γ a n d ψ 1 2 = 2 log 2 γ + 2.

Example 5.4

Let a, b ϵ ℝ with a > 0. Then

(17) 0 exp ( a x 2 ) cos b x x d x = γ log a + 2 log b 2 .

To apply Theorem 5.1 start with the series

(18) f 1 ( x ) = e a x 2 = n ϕ n a n x 2 n

and

(19) f 2 ( x ) = cos b x = n ϕ n Γ ( n + 1 ) Γ ( 2 n + 1 ) b 2 n x 2 n .

In both expansions α = 2 and the coefficients are given by

(20) C 1 ( n ) = a n a n d C 2 ( n ) = Γ ( n + 1 ) Γ ( 2 n + 1 ) b 2 n .

Then, C 1 ( 0 ) = log a and C 2 ( n ) = b 2 n Γ ( n + 1 ) Γ ( 2 n + 1 ) [ 2 log b + ψ ( n + 1 ) ψ ( 2 n + 1 ) ] y i e l d C 2 ( 0 ) = 2 log b ψ ( 1 ) = 2 log b + γ . The value (17) follows from here.

Example 5.5

The next example in this section involves the Bessel function of order 0

(21) J 0 ( x ) = n = 0 ( 1 ) n n ! 2 x 2 2 n

and Theorem 5.1 is used to evaluate

(22) 0 J 0 ( x ) cos a x x d x = log 2 a .

This appears as entry 6.693.8 in [12]. The expansions

(23) J 0 ( x ) = n = 0 ϕ n 1 n ! 2 2 n x 2 n a n d cos a x = n = 0 ϕ n n ! ( 2 n ) ! a 2 n x 2 n ,

show a = 2 and

(24) C 1 ( n ) = 1 Γ ( n + 1 ) 2 2 n a n d C 2 ( n ) = Γ ( n + 1 ) Γ ( 2 n + 1 ) a 2 n .

Differentiation gives

(25) C 1 ( n ) = 2 ln 2 + ψ ( n + 1 ) 2 2 n Γ ( n + 1 ) ,

and

(26) C 2 ( n ) = a 2 n Γ ( n + 1 ) ( 2 log a + ψ ( n + 1 ) 2 ψ ( 2 n + 1 ) ) Γ ( 2 n + 1 ) .

Then,

(27) C 1 ( 0 ) = γ 2 log 2 a n d C 2 ( 0 ) = ( γ + 2 log a ) ,

and the result now follows from Theorem 5.1. The reader is invited to use the representation

(28) J 0 2 ( x ) = 1 F 2 1 2 1 1 x 2 ,

to verify the identity

(29) 0 J 0 2 ( x ) cos x x d x = log 2.

Example 5.6

The final example in this section is

(30) I = 0 J 0 2 ( x ) e x 2 cos x x d x .

The evaluation begins with the expansions

(31) J 0 ( x ) = k = 0 ϕ k x 2 k 4 k Γ ( k + 1 ) a n d cos x = k = 0 X ϕ k π 4 k Γ ( k + 1 2 ) .

Then,

(32) J 0 2 ( x ) = k , n ϕ k , n 1 4 k + n Γ ( k + 1 ) Γ ( n + 1 ) x 2 k + 2 n ,

and

(33) e x 2 cos x = k , n ϕ k , n π 4 k Γ ( k + 1 2 ) x 2 k + 2 n .

Integration yields

I = 0 J 0 2 ( x ) e x 2 cos x x 1 ε d x = k , n ϕ k , n 1 4 k + n Γ ( k + 1 ) Γ ( n + 1 ) π 4 k Γ ( k + 1 2 ) 0 x 2 k + 2 n + ε 1 d x = k , n ϕ k , n 1 4 k + n Γ ( k + 1 ) Γ ( n + 1 ) π 4 k Γ ( k + 1 2 ) 2 k + 2 n + ϵ .

The method of brackets now gives

I = lim ε 0 1 2 k = 0 ( 1 ) k Γ ( k + ε 2 ) k ! 1 2 ε Γ ( k + 1 ) Γ ( 1 k ε / 2 ) π 2 2 k Γ ( k + 1 2 ) .

The term corresponding to k = 0 gives

(34) lim ε 0 1 2 Γ ϵ 2 1 2 ε Γ ( 1 ε 2 ) 1 = log 2 γ 2

and the terms with k ≥ 1 as ε → 0 give

(35) π 2 k = 1 ϕ k Γ ( k ) 2 2 k Γ k + 1 2 = 1 4 2 F 2 1 1 3 2 2 1 4 .

Therefore,

(36) 0 J 0 2 ( x ) e x 2 cos x x d x = 1 4 4 log 2 2 γ + 2 F 2 1 1 3 2 2 1 4 .

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

(1) f ( x 1 , , x n ) = 1 , , n = 0 ( 1 ) 1 1 ! ( 1 ) n n ! C ( 1 , , n ) x 1 γ 1 x n γ n ,

where the γj are linear functions of the indices given by

(2) γ 1 = α 11 1 + + α 1 n n + β 1 γ n = α n 1 1 + + α n n n + β n .

Then,

I = 0 0 f ( b 1 x 1 , , b n x n ) f ( a 1 x 1 , , a n x n ) x 1 1 + ρ 1 x n 1 + ρ n d x 1 d x n = 1 | det A | lim ε 0 [ b 1 ρ 1 ε b n ρ n ε a 1 ρ 1 ε a n ρ n ε ] 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

(3) α 11 1 + + α 1 n n + β 1 ρ 1 + ε = 0 α n 1 1 + + α n n n + β n ρ n + ε = 0.

Proof

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

Example 6.2

The evaluation of the integral

(4) I = 0 0 e μ s t 2 cos ( a s t ) e μ s t 2 cos ( b s t ) s d s d t

uses the expansion

(5) f ( s , t ) = e s t 2 cos ( s t ) = n 1 n 2 ϕ 1 , 2 π Γ ( n 2 + 1 2 ) 4 n 2 s n 1 + 2 n 2 t 2 n 1 + 2 n 2 ,

with parameters ρ 1 = 1 2 , ρ 2 = 1 , b 1 = a 2 / μ , b 2 = μ / a , a 1 = b 2 / μ , a 2 = μ / b . The solution to the linear system is n 1 = 1 2 a n d n 2 = ε 2 and |det A| = 2. Then

I = 1 2 lim ε 0 a 2 μ 1 / 2 ε μ a 1 ε b 2 μ 1 / 2 ε μ b 1 ε × Γ 1 2 Γ ε 2 π Γ 1 ε 2 4 ε / 2 = π μ lim ε 0 b ε a ε ε × Γ ( 1 + ε ) cos π ε 2 ( a b ) E = π μ log b a .

The double integral (4) has been evaluated.

Example 6.3

The method is now used to evaluate

(6) 0 0 sin ( μ x y 2 ) cos ( a x y ) sin ( μ x y 2 ) cos ( b x y ) x y = π 2 log b a .

The evaluation begins with the expansion

f ( x , y ) = sin ( x y 2 ) cos ( x y ) = x y 2 n 1 0 ϕ n 1 Γ 3 2 ( x y 2 ) 2 n 1 Γ n 1 + 3 2 4 n 1 n 2 0 ϕ n 2 Γ 1 2 ( x y ) 2 n 2 Γ n 2 + 1 2 4 n 2 = n 1 n 2 ϕ n 1 ϕ n 2 π 2 Γ n 1 + 3 2 Γ n 2 + 1 2 4 n 1 + n 2 x 2 n 1 + 2 n 2 + 1 y 4 n 1 + 2 n 2 .

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 = 1 2 a n d n 2 = ε 2 and |det A| = 4. Then,

I = lim ε 0 a ε b ε 4 Γ 1 2 Γ ε 2 π 2 Γ ( 1 ) Γ 1 ε 2 4 ε 1 ) / 2 = lim ε 0 π 3 / 2 4 ε / 2 4 b ε a ε ( a b ) E 2 1 2 ε π Γ ( ε ) π c s c 1 2 + ε 2 = π 2 log b a ,

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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Received: 2016-9-13
Accepted: 2016-12-27
Published Online: 2017-1-21

© 2017 Bravo et al.

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