A direct computation of a certain family of integrals

Abstract

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 $n\ge q\ge 1$.

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.

Citation

Fornari, L., Laeng, E. and Pata, V. (2021), "A direct computation of a certain family of integrals", Arab Journal of Mathematical Sciences, Vol. 27 No. 2, pp. 249-252. https://doi.org/10.1108/AJMS-01-2021-0019

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

$$I_{n,q}={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{{\left(\sin \hspace{0.17em}x\right)}^n}{x^q}\hspace{0.17em}dx\text{.}}$$

Theorem 1.

The following formulae hold

• (i) If $n+q$ is even, then

$$I_{n,q}=\frac{{\left(-1\right)}^{\frac{q-n}{2}}\pi }{2^n\left(q-1\right)!}{\displaystyle \sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\left(n-2k\right)}^{q-1}}\text{.}$$

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

$$I_{n,q}=\frac{{\left(-1\right)}^{\frac{q-n+1}{2}}}{2^{n-1}\left(q-1\right)!}{\displaystyle \sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\left(n-2k\right)}^{q-1}\hspace{0.17em}\log \left(n-2k\right)}\text{.}$$

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 $m\ge 0$, let

$$P_m(x)={\displaystyle \sum _{k=0}^m\frac{x^k}{k!}}$$

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 ${\left(\sin x\right)}^n$ of order $q-2$, with $Q=0$ if $q=1$. Since ${\left(\sin x\right)}^n$ has a zero of order n at $x=0$, it follows that $Q(x)\equiv 0$ for all $n\ge q\ge 1$. On the other hand, as

$${\left(\sin x\right)}^n=\frac{1}{{\left(2i\right)}^n}{\displaystyle \sum _{k=0}^n{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right)e^{i\left(n-2k\right)x}\text{,}}$$

we immediately conclude that

(1)$$Q(x)=\frac{1}{{\left(2i\right)}^n}{\displaystyle \sum _{k=0}^n{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right)P_{q-2}(i\left(n-2k\right)x)=0}\text{.}$$

Subtracting the two sums, we obtain

(2)$$I_{n,q}=\frac{1}{{\left(2i\right)}^n}\sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\displaystyle {\int }_0^{\infty }\frac{e^{i\left(n-2k\right)x}-P_{q-2}\left(i(n-2k)x\right)}{x^q}dx}+\frac{{\left(-1\right)}^n}{{\left(2i\right)}^n}\sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\displaystyle {\int }_0^{\infty }\frac{e^{-i\left(n-2k\right)x}-P_{q-2}(-i\left(n-2k\right)x)}{x^q}dx\text{.}}$$

Remark 2.

From (1), we also deduce that the equality

(3) $${\displaystyle \sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\left(n-2k\right)}^{q-1}=0}\text{,}$$

holds for every $n>q\ge 2$ , whenever $n+q$ is odd.

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

$${\mathsf{E}}_{\epsilon }(\alpha )={\displaystyle {\int }_{\epsilon }^{\infty }\frac{e^{i\alpha x}}{x}dx}\text{.}$$

Lemma 3.

For every $q\ge 1$, every $\epsilon >0$ and every $\alpha \ne 0$, we have

$${\displaystyle {\int }_{\epsilon }^{\infty }\frac{e^{i\alpha x}-P_{q-2}\left(i\alpha x\right)}{x^q}dx}=c_q{\alpha }^{q-1}+\frac{{\left(i\alpha \right)}^{q-1}}{\left(q-1\right)!}{\mathsf{E}}_{\epsilon }(\alpha )+\omega (\epsilon )\text{,}$$

where $c_q=\frac{i^{q-1}}{\left(q-1\right)!}{{{\displaystyle \sum }}^{}}_{k=0}^{q-2}\frac{1}{k+1}$ for $q\ge 2$ and $c_1=0$.

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

$${\displaystyle {\int }_{\epsilon }^{\infty }\frac{e^{i\alpha x}-P_{q-1}\left(i\alpha x\right)}{x^{q+1}}\hspace{0.17em}dx}=\frac{e^{i\alpha \epsilon }-P_{q-1}\left(i\alpha \epsilon \right)}{q{\epsilon }^q}+\frac{i\alpha }{q}{\displaystyle {\int }_{\epsilon }^{\infty }\frac{e^{i\alpha x}-P_{q-2}\left(i\alpha x\right)}{x^q}\hspace{0.17em}dx}\text{.}$$

By the inductive hypothesis,

$$\frac{i\alpha }{q}{\displaystyle {\int }_{\epsilon }^{\infty }\frac{e^{i\alpha x}-P_{q-2}\left(i\alpha x\right)}{x^q}dx}=\frac{i{c}_q}{q}{\alpha }^q+\frac{{\left(i\alpha \right)}^q}{q!}\hspace{0.17em}{\mathsf{E}}_{\epsilon }(\alpha )+{\omega }_q(\epsilon )\text{,}$$

for some function ${\omega }_q$ vanishing at 0. Noting that

$${\varpi }_q(\epsilon )=-\frac{{\left(i\alpha \right)}^q}{q!q}+\frac{e^{i\alpha \epsilon }-P_{q-1}\left(i\alpha \epsilon \right)}{q{\epsilon }^q}\to 0\text{as }\epsilon \to 0\text{,}$$

we end up with the equality

$${\displaystyle {\int }_{\epsilon }^{\infty }\frac{e^{i\alpha x}-P_{q-1}\left(i\alpha x\right)}{x^{q+1}}\hspace{0.17em}dx}=\left[\frac{i^q}{q!q}+\frac{i{c}_q}{q}\right]{\alpha }^q+\frac{{\left(i\alpha \right)}^q}{q!}{\mathsf{E}}_{\epsilon }(\alpha )+{\omega }_q(\epsilon )+{\varpi }_q(\epsilon )\text{.}$$

The final observation that $\frac{i^q}{q!q}+\frac{i{c}_q}{q}=c_{q+1}$ completes the proof. □

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

$${\mathsf{E}}_{\epsilon }\left(n-2k\right)-{\mathsf{E}}_{\epsilon }(-\left(n-2k\right))=2i\mathsf{Si}((n-2k)\epsilon )\text{,}$$

where

$$\mathsf{Si}(t)={\displaystyle {\int }_t^{\infty }\frac{\sin x}{x}\hspace{0.17em}dx}$$

is the SinIntegral function, we obtain

$$I_{n,q}=\frac{{\left(-1\right)}^{\frac{q-n}{2}}}{2^{n-1}\left(q-1\right)!}{\displaystyle \sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\left(n-2k\right)}^{q-1}\mathsf{Si}\left((n-2k)\epsilon \right)+\omega (\epsilon )}\text{.}$$

Since

$$\mathsf{Si}(\left(n-2k\right)\epsilon )\to \mathsf{Si}(0)=\frac{\pi }{2}\text{as }\epsilon \to 0,$$

the result follows. □

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

$${\mathsf{E}}_{\epsilon }\left(n-2k\right)+{\mathsf{E}}_{\epsilon }(-(n-2k))=2\mathsf{Ci}((n-2k)\epsilon )\text{,}$$

where

$$\mathsf{Ci}(t)={\displaystyle \underset{t}{\overset{\infty }{\int }}\frac{\cos x}{x}dx}$$

is the CosIntegral function, we obtain

$$I_{n,q}=\frac{{\left(-1\right)}^{\frac{q-n-1}{2}}}{2^{n-1}\left(q-1\right)!}{\displaystyle \sum _{k=0}^{\lfloor \frac{n-1}{2}\rfloor }{\left(-1\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\left(n-2k\right)}^{q-1}\mathsf{Ci}\left((n-2k)\epsilon \right)+\omega (\epsilon )}\text{.}$$

By a further use of (3), we can replace $\mathsf{Ci}((n-2k)\epsilon )$ with

$$\mathsf{Ci}((n-2k)\epsilon )-\mathsf{Ci}(\epsilon )\to -\log \left(n-2k\right)\text{as} \epsilon \to 0\text{,}$$

and a final limit $\epsilon \to 0$ completes the argument. □