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

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

• Are there any restrictions on the $\phi_k, a_j?$ – saulspatz Feb 5 '18 at 17:08
• 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 – ancientmathematician Feb 5 '18 at 17:08
• @saulspatz if $\phi_1,\cdots,\phi_n$, $a_1,\cdots,a_m$ be real and $a$ be positive and $a > |\phi_1| + \cdots |\phi_n|+|a_1|+\cdots+|a_m|$. – achille hui Feb 5 '18 at 17:18
• @ancientmathematician I have a hard copy ;-) BTW, the integral is $\frac{\pi}{2} \phi_1\cdots \phi_n$ assume above condition is satisfied. – achille hui Feb 5 '18 at 17:20
• @ancientmathematician I found a copy that was easier to read at books.google.com/… – saulspatz Feb 5 '18 at 17:25

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 $$a > \sum_{p=1}^n |\phi_p| + \sum_{q=1}^m | a_q|$$

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

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

• Perfect:) Thanks – Alex Feb 5 '18 at 18:25
• What a masterful demonstration – Biggs Feb 5 '18 at 20:17
• Excellent, @achille hui. But have you any idea why Whittaker and Watson chose to call the $\sin$ parameters $\phi_i$ but the $\cos$ ones $a_i$? – ancientmathematician Feb 6 '18 at 9:00
• @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 – achille hui Feb 6 '18 at 9:32
• \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. – achille hui Feb 6 '18 at 9:32