How to show $\int_0^{\infty} \frac{\sin(bx)\sin(x)}{x^2} \prod_{k=1}^n \cos^{p_k}(a_kx) \, \text{d} x= \frac{\pi}{2}$? I have found the following complicated integral in Table of Integrals, Series and Products (page 469; No. 37); the interesting thing about this integral is that for arbitrary parameters $p_k,a_k>0$ and arbitrary natural number $n$ it has the value $\frac{\pi}{2}$:
$$\int_0^{\infty} \frac{\sin(bx)\sin(x)}{x^2} \prod_{k=1}^n \cos^{p_k}(a_kx) \text{d} x= \frac{\pi}{2}$$
A large integral that has only one value when integrated over the interval $[0, \infty]$, but how I can prove this interesting fact? Series expansion in the trigonometric functions does not make sense, I think. Can I use induction for proving this identity?

The previous link to a PDF of the book no longer works.
 A: EDIT: My answer was more complicated than it needed to be, so I made it simpler.

This is basically what is called a Borwein integral.
As was mentioned by user37238 in the comments, there is an additional condition stated in the table, namely $b> \sum_{k=1}^n a_{k}p_{k} $.
But this condition is not quite correct. It should be $$b \ge 1 +  \sum_{k=1}^n a_{k}p_{k}.$$
Presumably we also want the $p_{k}$'s to be positive integers; otherwise, the integral is not well-defined.

Consider the complex function $$f(z) =   \frac{e^{ibz}\sin(z)}{z^2} \prod_{k=1}^n \cos^{p_k}(a_{k}z).$$
What we want to show is that if $b \ge 1 +  \sum_{k=1}^n a_{k}p_{k}$, then  $$ \text{PV} \int_{-\infty}^{\infty} \frac{e^{ibx}\sin(x)}{x^2} \prod_{k=1}^n \cos^{p_k}(a_kx) \, d x= i \pi \operatorname{Res}[f(z), 0] = i \pi (1) =i \pi.  $$
(The result then follows if we equate the imaginary parts on both sides of the equation.)
To show this, all we need to do is argue that the magnitude of $$g(z) = e^{ibz} \sin(z) \prod_{k=1}^n \cos^{p_k}(a_kx)$$ is bounded in the upper half-plane if $b \ge 1 + \sum_{k=1}^n a_{k}p_{k}$.
It will then follow immediately from the estimation lemma that $\int f(z) \, dz$ vanishes along the upper half of the circle $|z|=R$ as $R \to \infty$.

If the upper half-plane, the magnitude of $e^{i \alpha   z}, \, \alpha \ge 0$, is less than or equal to $1$.
The function $g(z)$ can be expressed as $$e^{ibz}\left(\frac{e^{iz}-e^{-iz}}{2i} \right) \prod_{k=1}^{n} \left(\frac{e^{i a_{n}z}+e^{-ia_{n}}z}{2} \right)^{p_{n}}.$$
If we expand, we get a linear combination of $2^{n+1} $ exponential functions of the form $e^{i \beta_{k} z}$, where $ \beta_{k} \in \mathbb{R}$.
The $\beta_{k}$'s range from $b+1+\sum_{k=1}^{n}a_{k}p_{k}$ to $b-1-\sum_{k=1}^{n}a_{k}p_{k} $.
If $b \ge  1+ \sum_{k=1}^{n} a_{n}p_{n}$, then the magnitude of $g(z)$ is bounded in the upper half-plane since all of the exponential functions are of the form $e^{i \alpha z}, \, \alpha \ge 0$.

A slight modification of the above argument shows that $$\int_{0}^{\infty} \frac{\sin (bx) \sin(rx)}{x^2} \prod_{k=1}^n \cos^{p_k}(a_kx) \, d x= \frac{\pi r}{2}, \quad b,r,a_{k} >0, \ p_{k} \in\mathbb{Z_{>0}}, \ b \ge r+ \sum_{k=1}^{n} a_{k}p_{k}.  $$
