Integrating $\int_0^{\infty} \frac{\sin^n (ax)}{x^n \cdot (x^2+1)} dx$

Question

I was trying to evaluate an expression for the family of integrals

$$ I(a,n) = \int_0^{\infty} \frac{\sin^n(ax)}{x^n(x^2+1)} dx$$

For positive $a$ and $n$

The expression for the first few $n$ can be computed by hand. For example, for the case when $n=1$ we have:

$$ I(a,1) = \int_0^{\infty} \frac{\sin(ax)}{x(x^2+1)} dx$$

Differentiating, we get:

$$ I'(a,1) = \int_0^{\infty} \frac{\cos(ax)}{x^2+1} dx = \frac{\pi}{2}\cdot e^{-a}$$

Since $I(0,1) = 0$, we can write $I(a,1)$ as:

$$ I(a,1) = \frac{\pi}{2} \cdot (1-e^{-a})$$

For $n=2$, we again differentiate with respect to $a$. We get:

$$I'(a,2) = \int_0^{\infty} \frac{\sin(2ax)}{x(x^2+1)} dx = I(2a,1) = \frac{\pi}{2} \cdot (1-e^{-2a})$$

Again observing that $I(0,2) = 0$, we can write $I(a,2)$ as:

$$I(a,2) = \frac{\pi}{2} \cdot (a + \frac{1}{2}\cdot e^{-2a} - \frac{1}{2})$$

With some rather tedious effort in computing the third derivative, we can use the same strategy to evaluate $I(a,3)$. I obtained:

$$I(a,3) = \frac{\pi}{8} \cdot (3a^2+ 3e^{-a} -e^{-3a}-2)$$

I hypothesize after looking at the pattern that the expression for $I(a,n)$ must be some fraction of $\pi$, multiplied by the sum of a polynomial of degree $n-1$ added to some exponentials of the form $e^{-ka}$ where $k$ ranges from $1$ to $n$.

But I was wondering if we can afford to keep doing this for higher and higher values of $n$. I don't have a surefire way of obtaining an expression for general $n$, except for repeated differentiation, simplification and reverse integration.

Can anyone help me in evaluating an expression that holds for all natural $n$?

Any help is appreciated. Thank you for reading!

Answer 1

Note that $$ \begin{aligned} I(a,n) &=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin^n(ax)}{x^n(1+x^2)}\\ &=\frac{a^n}{2}\int_{-\infty}^{+\infty}\left(\frac{\sin(ax)}{ax}\right)^n\frac{dx}{1+x^2}\\ &=\frac{a^n}{2}\int_{-\infty}^{+\infty}\left(\frac{\sin(ax)}{ax}\right)^n\frac{e^{-2\pi i \xi x}}{1+x^2}dx\Biggl|_{\xi=0}\\ &=\frac{a^n}{2}\mathcal{F}\left[\left(\frac{\sin(ax)}{ax}\right)^n\frac{1}{1+x^2}\right](\xi)\Biggl|_{\xi=0} \end{aligned} $$ where by $\mathcal{F}$ we denote the Fourier transform $$ \mathcal{F}[f(x)](\xi)=\int_{-\infty}^{+\infty}f(x)e^{-2\pi i \xi x}dx $$ Recall that $$ \mathcal{F}\left[\frac{1}{1+x^2}\right](\xi) =\pi e^{-2\pi|\xi|}, \quad\quad \mathcal{F}\left[\frac{\sin(\pi k x)}{\pi k x}\right](\xi)=\frac{1}{|k|}\operatorname{rect}\left(\frac{\xi}{k}\right) $$ where $$ \operatorname{rect}(x) =\frac{1}{2}\left( \operatorname{sign}\left(x+\frac{1}{2}\right) - \operatorname{sign}\left(x-\frac{1}{2}\right) \right). $$ It is well known that $\mathcal{F}[f(x)\cdot g(x)](\xi)=\mathcal{F}[f(x)](\xi)*\mathcal{F}[g(x)](\xi)$, so for $a> 0$ we get $$ \begin{aligned} I(a, n) &=\frac{a^n}{2}\cdot\mathcal{F}\left[\left(\prod\limits_{i=1}^{n}\frac{\sin(ax)}{ax}\right)\cdot\frac{1}{1+x^2}\right](\xi)\Biggl|_{\xi=0}\\ &=\frac{a^n}{2}\cdot\left(\ast_{i=1}^n\mathcal{F}\left[\frac{\sin(ax)}{ax}\right](\xi)\right)*\mathcal{F}\left[\cdot\frac{1}{1+x^2}\right](\xi)\Biggl|_{\xi=0}\\ &=\frac{a^n}{2}\cdot\left(\ast_{i=1}^n\frac{1}{|a/\pi|}\operatorname{rect}\left(\frac{\xi}{a/\pi}\right)\right)*\pi e^{-2\pi |\xi|}\Biggl|_{\xi=0}\\ &=\frac{a^n\pi^{n+1}}{2a^n}\cdot\left(\ast_{i=1}^n\operatorname{rect}\left(\frac{\pi\xi}{a}\right)\right)* e^{-2\pi |\xi|}\Biggl|_{\xi=0}\\ &=\frac{\pi^{n+1}}{2}\int\limits_{ \sum\limits_{i=1}^{n+1}s_i=\xi}\prod_{i=1}^n\operatorname{rect}\left(\frac{\pi s_i}{a}\right)\cdot e^{-2\pi |s_{n+1}|} d\sigma\Biggl|_{\xi=0}\\ &=\frac{\pi^{n+1}}{2}\int\limits_{\mathbb{R}^n}\prod_{i=1}^n\operatorname{rect}\left(\frac{\pi s_i}{a}\right)\cdot e^{-2\pi \left|\xi-\sum\limits_{i=1}^ns_{i}\right|} ds\Biggl|_{\xi=0}\\ &=\frac{\pi^{n+1}}{2}\int\limits_{\mathbb{R}^n}\prod_{i=1}^n\operatorname{rect}\left(\frac{\pi s_i}{a}\right)\cdot e^{-2\pi \left|\sum\limits_{i=1}^ns_{i}\right|} ds\\ &=\left\{t_i=2\pi s_i,i\in\mathbb{N}_n\right\}\\ &=\frac{\pi^{n+1}}{2}\int\limits_{\mathbb{R}^n}\prod_{i=1}^n\operatorname{rect}\left(\frac{t_i}{2a}\right)\cdot e^{-\left|\sum\limits_{i=1}^n t_{i}\right|} \frac{1}{(2\pi)^n} dt\\ &=\frac{\pi}{2^{n+1}}\int\limits_{\left[-a,a\right]^n} e^{- \left|\sum\limits_{i=1}^n t_{i}\right|} dt\\ \end{aligned} $$ The last integral is doable I believe. At least for $n=2$ it gives the same result as yours. If I have time I'll try to give an explicit formula for it.