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How to evaluate

$$ \int_0^\infty \frac{\sin(\varphi_1x)}{x}\frac{\sin\varphi_2x}{x} \cdots \frac{\sin\varphi_nx}{x} \frac{\sin(ax)}{x}\cos(a_1x) \cdots \cos(a_mx) \, dx \text{ ?} $$

For small $n$ and $m$ it's simple (setting $\sin(kx)=\dfrac{e^{ikx}-e^{-ikx}}{2i}$ and using Jordan's lemma, but for arbitrary $n$ the calculation is too tedious. Surely there must be some nice trick here?
Thanks

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    $\begingroup$ Are there any restrictions on the $\phi_k, a_j?$ $\endgroup$ – saulspatz Feb 5 '18 at 17:08
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    $\begingroup$ This is an exercise in Whittaker and Watson, page 122; under conditions they give the answer as the product of the $\phi_i$ (including $\phi_0:=a$). They ascribe the result to Stormer, Acta Mathematica XIX. See archive.org/stream/courseofmodernan00whit#page/122/mode/2up $\endgroup$ – ancientmathematician Feb 5 '18 at 17:08
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    $\begingroup$ @saulspatz if $\phi_1,\cdots,\phi_n$, $a_1,\cdots,a_m$ be real and $a$ be positive and $a > |\phi_1| + \cdots |\phi_n|+|a_1|+\cdots+|a_m|$. $\endgroup$ – achille hui Feb 5 '18 at 17:18
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    $\begingroup$ @ancientmathematician I have a hard copy ;-) BTW, the integral is $\frac{\pi}{2} \phi_1\cdots \phi_n$ assume above condition is satisfied. $\endgroup$ – achille hui Feb 5 '18 at 17:20
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    $\begingroup$ @ancientmathematician I found a copy that was easier to read at books.google.com/… $\endgroup$ – saulspatz Feb 5 '18 at 17:25
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We will assume the parameters $\phi_1,\ldots,\phi_n$, $a_1,\ldots, a_m$ satisfy the condition given in page 122 of Whittaker and Waston's classic "A Course of Modern Analysis".

$\phi_1,\ldots,\phi_n$, $a_1, \ldots, a_m$ are real, $a$ is positive and $a > \sum_{p=1}^m |\phi_p| + \sum_{q=1}^m | a_q|$

Under this condition, the integral at hand evaluates to $\frac{\pi}{2}\prod_{p=1}^n \phi_p$.

We will further assume all $\phi_p \ne 0$. Otherwise, the integral trivially evaluates to zero.

When $\phi_p \ne 0$, the singularity of $\frac{\sin(\phi_p x)}{x}$ at $x = 0$ is removable, if we define the value of $\frac{\sin(\phi_p x)}{x}$ at $x = 0$ to be $\phi_p$, we will turn this into an entire function.

Let $f(x)$ be the product $\prod\limits_{p=1}^n\frac{\sin(\phi_p x)}{x}\prod\limits_{q=1}^m \cos(a_q x)$. With above interpretation in mind, this is an entire function in $x$. For large $z \in \mathbb{C}$, we can bound the growth of $f(z)$ as $$|f(z)| = O( e^{K|\Im z|} )\quad\text{ where }\quad K = \sum_{p=1}^n|\phi_p| + \sum_{q=1}^m|a_m|\tag{*1}$$

Notice $\frac{\sin(a x)}{x}$ is finite at $x = 0$ and $f(x)$ is an even function in $x$. Our integral equals to

$$\int_0^\infty f(x)\frac{\sin a x}{x} dx = \lim_{\substack{R\to\infty\\ \epsilon\to 0}} \int_{\epsilon}^R f(x)\frac{\sin( a x)}{x} dx = \frac12 \lim_{\substack{R\to\infty\\ \epsilon\to 0}} \left( \int_{-R}^{-\epsilon} + \int_{\epsilon}^R\right) f(x)\frac{\sin(a x)}{x} dx \\= \frac{1}{2i} \lim_{\substack{R\to\infty\\ \epsilon\to 0}} \left( \int_{-R}^{-\epsilon} + \int_{\epsilon}^R\right) f(x)\frac{e^{iax}}{x} dx $$ To evaluate this integral, consider following contour $C$ consists of 4 segments:

  • a line segment from $-R$ to $-\epsilon$.
  • $C_\epsilon$ a circular segment $\epsilon e^{i\theta}$ with $\theta$ from $\pi$ to $0$.
  • a line segment from $\epsilon$ to $R$.
  • $C_R$ a circular segment $R e^{i\phi}$ with $\phi$ from $0$ to $\pi$.

Since $C$ doesn't contain any singularity of the integrand, we have

$$\oint_C f(x)\frac{e^{iax}}{x} dx = 0 \implies \left( \int_{-R}^{-\epsilon} + \int_{\epsilon}^R\right) f(x)\frac{e^{iax}}{x} dx = -\left( \int_{C_R} + \int_{C_\epsilon}\right) f(x)\frac{e^{iax}}{x} dx $$ Under the assumption $a > K = \sum_{p=1}^n|\phi_p| + \sum_{q=1}^m |a_q|$, the bound $(*1)$ tell us $$\lim_{R\to\infty}\int_{C_R} f(x)\frac{e^{iax}}{x} dx = 0$$ Since $f(x)$ is regular near $x = 0$, as $\epsilon \to 0$, the integral over $C_{\epsilon}$ gives us $-\pi i = (-\frac12)(2\pi i)$ of the residue of $f(x) \frac{e^{iax}}{x}$ at $x = 0$ ($-\pi i$ instead of $\pi i$ because $\theta$ varies from $\pi$ to $0$). As a result, the integral at hand equals to

$$\frac{i}{2}\lim_{\epsilon\to 0}\int_{C_\epsilon}f(x)\frac{e^{iax}}{x}dx = \frac{i}{2} (-\pi i) f(0) e^{i0} = \frac{\pi}{2}f(0) = \frac{\pi}{2} \prod_{p=1}^n \phi_p$$

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  • $\begingroup$ Perfect:) Thanks $\endgroup$ – Alex Feb 5 '18 at 18:25
  • $\begingroup$ What a masterful demonstration $\endgroup$ – Biggs Feb 5 '18 at 20:17
  • $\begingroup$ Excellent, @achille hui. But have you any idea why Whittaker and Watson chose to call the $\sin$ parameters $\phi_i$ but the $\cos$ ones $a_i$? $\endgroup$ – ancientmathematician Feb 6 '18 at 9:00
  • $\begingroup$ @ancientmathematician Whittaker and Waston just follow what Carl Störmer does who attempt to generalize the formula $\frac{\phi}{2} = \frac{\sin\phi}{1} - \frac{\sin(2\phi)}{2} + \frac{\sin(3\phi)}{3} - \cdots$. Störmer seems to have derived an identity in Acta Math. xix $\endgroup$ – achille hui Feb 6 '18 at 9:32
  • $\begingroup$ $$\begin{align} \frac{\phi_1\cdots\phi_n}{2} =&\phantom{+} \frac{\sin\phi_1}{1}\frac{\sin\phi_2}{1}\cdots\frac{\sin\phi_n}{1}\cos\alpha_1\cos\alpha_2\cdots\cos\alpha_m\\ &- \frac{\sin 2\phi_1}{2}\frac{\sin 2\phi_2}{2}\cdots\frac{\sin 2\phi_n}{2}\cos 2 \alpha_1\cos 2\alpha_2\cdots\cos 2\alpha_m\\ &+ \frac{\sin 3\phi_1}{3}\frac{\sin 3\phi_2}{3}\cdots\frac{\sin 3\phi_n}{3}\cos 3\alpha_1\cos 3\alpha_2\cdots\cos 3\alpha_m\\ &- \cdots\cdots \end{align} $$ Since I don't know German, I can't tell exactly what happens. $\endgroup$ – achille hui Feb 6 '18 at 9:32

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