Let $a_k\gt 0$ and $a_0\gt \sum_{k=1}^n a_k$ . Show that $\int_0^{\infty} \prod_{k=0}^n \frac {\sin (a_k x)}{x} dx=\frac {\pi}{2}\prod_{k=1}^n a_k$ 
Let $a_k\gt 0$ and $a_0\gt \sum_{k=1}^n a_k$ . Show that $$\int_0^{\infty} \prod_{k=0}^n \frac {\sin (a_k x)}{x} dx=\frac {\pi}{2}\prod_{k=1}^n a_k$$

I saw this question on the internet somewhere a few days back and thought to give it a try. The question looks so absurd due to the arbitrary inputs in the sines and their products. 
I tried giving it a shot using Laplace and Mellin transforms but realized they were turning to dead end. i also tried using a little complex analysis by writing sines in terms of $e$ but to no avail.  
One method which seemed quite promising was the Feynman's technique because one peculiar thing about RHS is that it is independent of $a_0$ and also a point to note is the constraint on $a_0$( It is greater than sum of other $a_k$'s). Differentiating both sides w.r.t $a_0$ would give a $0$ on RHS while some integral on LHS which we need to prove is $0$. But couldn't much continue with this thought. 
The place where I saw this question also had an answer but it used principle of induction which I pretty don't like much so it would be very much better if I could get methods without involving any type of induction. 
Thanks!!!! 
 A: Note from here that
$$\int_{-\infty}^{+\infty}\,\int_{a}^b\exp(\text{i}kx)\,f(k)\,\text{d}k\,\text{d}x=2\pi\,f(0)\,,\tag{*}$$
where $f:[a,b]\to\mathbb{C}$ is a sufficiently well behaved function (e.g., $f$ is continuous), and $a,b\in\mathbb{R}$ are such that $a<0<b$.  Basically, this is what physicists tend to write: $$\int_{-\infty}^{+\infty}\,\exp(\text{i}kx)\,\text{d}x=2\pi\,\delta(k)\,,$$
where $\delta$ is the Dirac delta-distribution.  
We also have
$$\frac{\sin(a_jx)}{x}=\frac{a_j}{2}\,\int_{-1}^{+1}\,\exp(\text{i}a_jxk_j)\,\text{d}k_j$$
for $j=0,1,2,\ldots,n$.  The required integral is
$$I:=\int_{0}^{\infty}\,\prod_{j=0}^n\,\frac{\sin(a_jx)}{x}\,\text{d}x=\frac{1}{2}\,\int_{-\infty}^{+\infty}\,\prod_{j=0}^n\,\frac{\sin(a_jx)}{x}\,\text{d}x\,.$$
That is,
$$I=\frac{\prod\limits_{r=0}^n\,a_r}{2^{n+2}}\,\prod_{j=1}^n\,\int_{-1}^{+1}\,\text{d}k_j\,\int_{-\infty}^{+\infty}\,\text{d}x\,\int_{-1}^{+1}\,\text{d}k_0\,\exp\Biggl(\text{i}\,x\,\left(a_0k_0+\sum_{j=1}^n\,a_jk_j\right)\Biggr)\,.$$
Thus,
$$I=\frac{\prod\limits_{r=0}^n\,a_r}{2^{n+2}}\,\prod_{j=1}^n\,\int_{-1}^{+1}\,\text{d}k_j\,\int_{-\infty}^{+\infty}\,\text{d}x\,\frac{1}{a_0}\,\int_{A}^{B}\,\text{d}\kappa\,\exp(\text{i}\,x\,\kappa)$$
where $$A:=-1+\frac{1}{a_0}\,\sum_{j=1}^n\,a_jk_j<0\,,$$ $$B:=+1+\frac{1}{a_0}\,\sum_{j=1}^n\,a_jk_j>0\,,$$
and $\kappa:=a_0k_0$.  (The results $A<0$ and $B>0$ follow from the hypothesis that $a_0>\sum\limits_{j=1}^n\,a_j$.)  Using (*), we get
$$I=\frac{\prod\limits_{r=1}^n\,a_r}{2^{n+2}}\,\,\prod_{j=1}^n\,\int_{-1}^{+1}\,\text{d}k_j\,(2\pi)=\frac{\pi}{2}\,\prod_{r=1}^n\,a_r\,.$$

Apparently, from the Wikipedia page, in general case (where $a_0,a_1,a_2,\ldots,a_n$ are positive real numbers with no extra conditions), we have
$$\text{vol}_n(V)=\frac{1}{n!\,\prod\limits_{r=1}^n\,a_r}\,\sum_{\gamma\in \{-1,+1\}^n}\,\epsilon_\gamma\,\beta_\gamma^n\,\text{sign}(\beta_\gamma)\,,\tag{#}$$
where
$\text{vol}_n$ is the $n$-dimensional volume, 
$$V:=\Biggl\{(k_1,k_2,\ldots,k_n)\in[-1,+1]^n\,\Big|\,-a_0<\sum_{j=1}^n\,a_jk_j<+a_0\Biggr\}\,,$$
$$\epsilon_\gamma:=\gamma_1\gamma_2\cdots\gamma_n\,,$$
and $$\beta_\gamma:=a_0+\sum_{j=1}^n\,a_j\gamma_j\,.$$  Here, $\gamma_j$ is the $j$-coordinate of each $\gamma\in\{-1,+1\}^n$.  It would be interesting to prove (#), from which we have
$$\int_0^\infty\,\prod_{j=0}^n\,\frac{\sin(a_jx)}{x}\,\text{d}x=\frac{\pi}{2}\,\left(\frac{1}{2^nn!}\,\sum_{\gamma\in \{-1,+1\}^n}\,\epsilon_\gamma\,\beta_\gamma^n\,\text{sign}(\beta_\gamma)\right)\,.$$
Note also that (#) is equivalent to $$\text{vol}_n(W)=\frac{1}{n!}\,\sum_{\gamma\in \{-1,+1\}^n}\,\epsilon_\gamma\,\beta_\gamma^n\,\text{sign}(\beta_\gamma)\,,$$
where $$W:=\Biggl\{(\kappa_1,\kappa_2,\ldots,\kappa_n)\in\prod_{j=1}^n\,\left[-a_j,+a_j\right]\,\Big|\,-a_0<\sum_{j=1}^n\,\kappa_j<+a_0\Biggr\}\,.$$
