Compatibility of the method of brackets with classical integration rules

Example 1.1

For n ∈ N , the evaluation of

is elementary. The only special function required is the primitive of x ↦ x n . In this case, the class of special functions reduces to F = x ↦ x n + 1 n + 1 and the special constants are C = { a , b } . A simple extension of (1.2) shows that the class of polynomials F = f ( x ) = ∑ k = 0 m a k x k : a k ∈ R , m ∈ { 0 , 1 , … } is closed under integration.

Theorem 1.2

Assume g is a differentiable function on the interval [ a , b ] , with derivative g ′ . Then,

Example 1.3

In order to evaluate the integral,

observe that g ( x ) = − e − x is the primitive of f ( x ) = e − x and Theorem 1.2 gives

A simple generalization of this example shows that every polynomial P in the variable f ( x ) = e t x , with t ∈ R a fixed parameter, has a primitive of the same form. Therefore, the integral of P ( e t x ) between a and b can be evaluated using Theorem 1.2.

Example 1.4

The symbolic language Mathematica gives the elementary evaluation

as well as the nonelementary one

where SinIntegral is the Mathematica notation for the sine integral function defined by:

Therefore, (1.7) simply comes from the introduction of a new function. The question of whether Si may be expressed in terms of more elementary functions is more difficult to decide.

Example

The massless sunset diagram is presented in Figure 1. The diagram is parametrized by the integral (in momentum space)

where k and q are the D -dimensional vectors. For reasons of convergence, it is traditional to let D = 4 − 2 ε and then ε → 0 + . The Schwinger parametrization is a systematic procedure to express the integral G in terms of some parameters (see [4] for details). It produces the integral

with P being the square of the norm of the 4-vector p .

Remark 2.1

Observe that this evaluation requires the evaluation of the coefficients C ( n ) outside its original domain, that is, for n ∗ ∉ N . The conditions on the function f given in Theorem 2.2, guarantee the existence of a unique extension of C . It turns out that Rule 4 is precisely a result of Ramanujan, known these days as Ramanujan’s master theorem. This refers to the formal identity

stated by S. Ramanujan’s in his Quarterly Reports [8, p. 298]. It was widely used by him as a tool in computing definite integrals and infinite series. In fact, as Hardy puts it in [9], he “was particularly fond of them, and used them as one of his commonest tools.”

Theorem 2.2

(Ramanujan’s Master Theorem). Let φ ( z ) be an analytic (single-valued) function, defined on a half-plane H ( δ ) = { z ∈ C : Re z ≥ − δ } for some 0 < δ < 1 . Suppose that, for some A < π , φ satisfies the growth condition ∣ φ ( v + i w ) ∣ < C e P v + A ∣ w ∣ for all z = v + i w ∈ H ( δ ) . Then, for all 0 < Re s < δ ,

Note

In the multidimensional evaluation of a bracket series, in the special case when the number of sums and brackets is the same, write

The multiple sum is declared to be

where A = { a i j } and the values { n i ∗ } are the solutions of the linear system obtained from the vanishing of the brackets:

An inductive proof follows directly from the one-dimensional case.

Theorem 3.1

The value of ∫ a b f ′ ( x ) d x computed by the method of brackets gives a consistent result with the analytic procedure, that is,

Proof

Assume f ( x ) = ∑ k = 0 ∞ ϕ k C ( k ) x k , then

and the last expression can be written as:

The change of variables t = x − a b − x maps ( a , b ) to ( 0 , ∞ ) , and it gives

The system

coming from the vanishing of the brackets gives ( m ∗ , n ∗ ) = ( − ℓ − 1 , ℓ − k − 1 ) . Taking k and ℓ as free variables, it gives

Simplifying the previous summand gives

The proof is complete. The solutions corresponding to other combination of free variables yield the same result.□

Theorem 4.1

The evaluation of the Laplace transform (4.1) by the method of brackets also produces (4.3).

Proof

The method of brackets gives

The vanishing of the bracket gives m ∗ = − α n − β , and therefore,

confirming (4.3).□

Theorem 6.1

The method of brackets is consistent with the Cauchy-Schlömilch transformation. That is, using the method of brackets to evaluate the integrals, we have

Proof

Let f e ( x ) = ( f ( x ) + f ( − x ) ) / 2 and f o ( x ) = ( f ( x ) − f ( − x ) ) / 2 be the even and odd parts of f , respectively. Then, f ( x ) = f e ( x ) + f o ( x ) , and we have

Similarly,

Thus, we have reduced (6.1) to the identity

Put another way, it suffices to consider even functions, for the odd part vanishes. Without loss of generality, let us assume that f is even. Suppose now that f ( x ) = ∑ n = 0 ∞ ϕ n g ( n ) x α n + β − 1 . Let ε > 0 . Then, we have

and

Define

so that we have

Now applying the method of brackets, we find

The vanishing of the brackets produces three series, one for each choice of free index. These are

Since f is even, f ′ is odd. Therefore, we have

Applying the method of brackets to this form of d d ε I ( ε ) , we have

The vanishing of the brackets produces three series, one for each choice of free index. These are

But these series exactly cancel the series arising from the first form of d d ε I ( ε ) . Thus, summing the solutions, we find that d d ε I ( ε ) = 0 . Then, I ( 1 ) = I ( 0 ) , and this completes the proof.□

Remark 6.2

An alternative form of the Schlömilch transformation on the half-line is usually written as:

Starting with the expansion

the usual procedure of the method of brackets evaluated the right-hand side as

and this is equivalent to

In order to evaluate the left-hand side of (6.23), start with

Rule 3 (sum expansion) of the method of brackets is now used to produce two expressions for the term ( a x − b x − 1 ) 2 k , namely,