How to show that $\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(4x)\cdots\sin(2^kx)\over x^{k+1}}\mathrm dx=2^{0.5(k^2-k-2)}\pi?$ $$\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(4x)\cdots\sin(2^kx)\over x^{k+1}}\mathrm dx=2^{0.5(k^2-k-2)}\pi\tag1$$
$k\ge0$
Experimental using wolfram integrator, let me to conclude the closed form for $(1)$ is as follow above.
But I don't know how shows it.
How can we prove $(1)$?
We have 
$$\prod_{j=0}^{k-1}\cos(2^jx)={\sin(2^kx)\over 2^k\sin(x)}\tag2$$
I am sure how is this product will relate to $(1)$
Setting $k=0$ we have the well known
$$\int_{0}^{\infty}{\sin(x)\over x}\mathrm dx={\pi\over 2}$$
 A: Using the same technique as here.
Let $$f(z) = \frac{\sin(z)\cdots\sin(2^{k-1}z)e^{iz2^k}}{z^{k+1}}$$
Then
$$0 = \oint_C f(z)\,dz$$
where $C$ is the classic semicircular contour contour indented around $z=0$. As the radius increases, the integrand goes to zero around that arc since as in the linked post, the exponential order $2^k$ is greater than the exponential order of all the sinusoids $\sum_{n=0}^{k-1} 2^n = 2^k-1$ and so the exponential term dominates all the sinusoids and causes exponential decay; moreover, the denominator grows as $R^{k+1} e^{(k+1)i\theta}$.
On the other hand, making the indentation smaller just adds a contribution of $\pi i\operatorname{Res}_{z=0}f(z)$. Then:
$$P.V. \int_\mathbb R f(z)\,dz=\pi i\operatorname{Res}_{z=0}f(z)$$
We have a simple pole since $\sin z\sim z$ at $z=0$.
$$\operatorname{Res}_{z=0}f(z) = \lim_{z\to 0} zf(z) = \\
\lim_{z\to 0}\frac{\sin(z)}{z}\frac{\sin(2z)}{z}\cdots\frac{\sin(2^{k-1}z)}{z}e^{2^kix} =\prod_{n=0}^{k-1}2^n = 2^{k(k-1)/2}$$
Now we can take the imaginary part of the integral above:
$$I=\int_0^\infty \frac{\sin(x)\cdots\sin(2^{k-1}x)\sin(2^k x)}{x^{k+1}}\,dx=\frac{1}{2}\Im\left(\pi i\operatorname{Res}_{z=0}f(z)\right) \\
=\pi 2^{(k^2-k-2)/2}$$
and confirm the hypothesis.
