A direct computation of a certain family of integrals

Fornari, Lorenzo; Laeng, Enrico; Pata, Vittorino

doi:10.1108/AJMS-01-2021-0019

Purpose

The authors propose a rather elementary method to compute a family of integrals on the half line, involving positive powers of sin x and negative powers of x, depending on the integer parameters $n \geq q \geq 1$ ⁠.

Design/methodology/approach

Combinatorics, sine and cosine integral functions.

Findings

The authors prove an explicit formula to evaluate sinc-type integrals.

Originality/value

The proof is not present in the current literature, and it could be of interest for a large audience.

In this note, let $n \geq q \geq 1$ be any two given integers. The symbol $⌊ . ⌋$ will stand, as usual, for the integer part. We consider the family of integrals

I_{n, q} = \int_{0}^{\infty} \frac{{(\sin x)}^{n}}{x^{q}} d x .

Theorem 1.

The following formulae hold

(i) If $n + q$ is even, then

I_{n, q} = \frac{{(- 1)}^{\frac{q - n}{2}} π}{2^{n} (q - 1)!} \sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) {(n - 2 k)}^{q - 1} .

(ii) If $n + q$ is odd and $q \geq 2$ ⁠, then

I_{n, q} = \frac{{(- 1)}^{\frac{q - n + 1}{2}}}{2^{n - 1} (q - 1)!} \sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) {(n - 2 k)}^{q - 1} \log (n - 2 k) .

The formulae above are recorded in the Wolfram MathWorld web page titled Sinc Function [1], which refers to the result as “amazing” and “spectacular”. However, the web page omits the proof, citing a 20-year-old online paper that seems not to be available any longer. Nor the proof is reported anywhere else, to the best of our knowledge. Nonetheless, particular instances of $I_{n, q}$ are discussed in several textbooks, typically by means of complex analysis tools (see, e.g. Ref. [2]).

The remaining of the paper is devoted to our proof of Theorem 1. To this end, for $m \geq 0$ ⁠, let

P_{m} (x) = \sum_{k = 0}^{m} \frac{x^{k}}{k!}

denote the Maclaurin polynomial of $e^{x}$ of order m. We agree to set $P_{- 1} = 0$ ⁠. Let $Q (x)$ be the Maclaurin polynomial of ${(\sin x)}^{n}$ of order $q - 2$ ⁠, with $Q = 0$ if $q = 1$ ⁠. Since ${(\sin x)}^{n}$ has a zero of order n at $x = 0$ ⁠, it follows that $Q (x) \equiv 0$ for all $n \geq q \geq 1$ ⁠. On the other hand, as

{(\sin x)}^{n} = \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) e^{i (n - 2 k) x},

we immediately conclude that

Q (x) = \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) P_{q - 2} (i (n - 2 k) x) = 0 .

(1)

Subtracting the two sums, we obtain

I_{n, q} = \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) \int_{0}^{\infty} \frac{e^{i (n - 2 k) x} - P_{q - 2} (i (n - 2 k) x)}{x^{q}} d x + \frac{{(- 1)}^{n}}{{(2 i)}^{n}} \sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) \int_{0}^{\infty} \frac{e^{- i (n - 2 k) x} - P_{q - 2} (- i (n - 2 k) x)}{x^{q}} d x .

(2)

Remark 2.

From (1), we also deduce that the equality

\sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) {(n - 2 k)}^{q - 1} = 0,

(3)

holds for every

n > q \geq 2

, whenever

n + q

is odd.

We now start from formula (2) but considering the integral on $(ε, \infty)$ and only at the end we will take the limit $ε \to 0$ ⁠. This allows us to move the integral inside the sum. In what follows $ω (ε)$ will denote a generic function of ε, vanishing at 0 as $ε \to 0$ ⁠. Moreover, for $α \neq 0$ ⁠, let us define

E_{ε} (α) = \int_{ε}^{\infty} \frac{e^{i α x}}{x} d x .

Lemma 3.

For every $q \geq 1$ ⁠, every $ε > 0$ and every $α \neq 0$ ⁠, we have

\int_{ε}^{\infty} \frac{e^{i α x} - P_{q - 2} (i α x)}{x^{q}} d x = c_{q} α^{q - 1} + \frac{{(i α)}^{q - 1}}{(q - 1)!} E_{ε} (α) + ω (ε),

where

c_{q} = \frac{i^{q - 1}}{(q - 1)!} {\sum^{}}_{k = 0}^{q - 2} \frac{1}{k + 1}

for

q \geq 2

and

c_{1} = 0

⁠.

Proof: The proof goes by induction on q. If $q = 1$ ⁠, equality holds with $ω (ε) = 0$ ⁠. Then, we prove the formula for $q + 1$ ⁠, assuming it true for $q \geq 1$ ⁠. Since $P ′_{q - 1} = P_{q - 2}$ ⁠, an integration by parts yields

\int_{ε}^{\infty} \frac{e^{i α x} - P_{q - 1} (i α x)}{x^{q + 1}} d x = \frac{e^{i α ε} - P_{q - 1} (i α ε)}{q ε^{q}} + \frac{i α}{q} \int_{ε}^{\infty} \frac{e^{i α x} - P_{q - 2} (i α x)}{x^{q}} d x .

By the inductive hypothesis,

\frac{i α}{q} \int_{ε}^{\infty} \frac{e^{i α x} - P_{q - 2} (i α x)}{x^{q}} d x = \frac{i c_{q}}{q} α^{q} + \frac{{(i α)}^{q}}{q!} E_{ε} (α) + ω_{q} (ε),

for some function

ω_{q}

vanishing at 0. Noting that

ϖ_{q} (ε) = - \frac{{(i α)}^{q}}{q! q} + \frac{e^{i α ε} - P_{q - 1} (i α ε)}{q ε^{q}} \to 0 as ε \to 0,

we end up with the equality

\int_{ε}^{\infty} \frac{e^{i α x} - P_{q - 1} (i α x)}{x^{q + 1}} d x = [\frac{i^{q}}{q! q} + \frac{i c_{q}}{q}] α^{q} + \frac{{(i α)}^{q}}{q!} E_{ε} (α) + ω_{q} (ε) + ϖ_{q} (ε) .

The final observation that

\frac{i^{q}}{q! q} + \frac{i c_{q}}{q} = c_{q + 1}

completes the proof. □

Proof of Theorem 1 for the case $n + q$ even. Substituting the expression given by Lemma 3 into (2) and noting that

E_{ε} (n - 2 k) - E_{ε} (- (n - 2 k)) = 2 i Si ((n - 2 k) ε),

where

Si (t) = \int_{t}^{\infty} \frac{\sin x}{x} d x

is the SinIntegral function, we obtain

I_{n, q} = \frac{{(- 1)}^{\frac{q - n}{2}}}{2^{n - 1} (q - 1)!} \sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) {(n - 2 k)}^{q - 1} Si ((n - 2 k) ε) + ω (ε) .

Since

Si ((n - 2 k) ε) \to Si (0) = \frac{π}{2} as ε \to 0,

the result follows. □

Proof of Theorem 1 for the case $n + q$ odd. Again, we substitute the expression given by Lemma 3 into (2). Using (3) and noting that

E_{ε} (n - 2 k) + E_{ε} (- (n - 2 k)) = 2 Ci ((n - 2 k) ε),

where

Ci (t) = \int_{t}^{\infty} \frac{\cos x}{x} d x

is the CosIntegral function, we obtain

I_{n, q} = \frac{{(- 1)}^{\frac{q - n - 1}{2}}}{2^{n - 1} (q - 1)!} \sum_{k = 0}^{⌊ \frac{n - 1}{2} ⌋} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) {(n - 2 k)}^{q - 1} Ci ((n - 2 k) ε) + ω (ε) .

By a further use of (3), we can replace $Ci ((n - 2 k) ε)$ with

Ci ((n - 2 k) ε) - Ci (ε) \to - \log (n - 2 k) as ε \to 0,

and a final limit

ε \to 0

completes the argument. □

References

[1]

Weisstein

ES

.

Sinc function

.

Available from:

https://mathworld.wolfram.com/SincFunction.html.

[2]

Ahlfors

LV

.

Complex analysis

.

New York

:

McGraw-Hill

;

1978

.

Google Scholar

2021

Lorenzo Fornari, Enrico Laeng and Vittorino Pata

Published in Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) licence. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this licence may be seen at http://creativecommons.org/licences/by/4.0/legalcode

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