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$$\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(4x)\cdots\sin(2^kx)\over x^{k+1}}\mathrm dx=2^{0.5(k^2-k-2)}\pi\tag1$$

$k\ge0$

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

$$\int_{0}^{\infty}{\sin(x)\over x}\mathrm dx={\pi\over 2}$$

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  • By induction, integral by parts and differentiating under the integral sign. – Rab Oct 20 '17 at 21:08
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    The numerator is $2^{-k}(\sin x - \sin 3x + \cdots \pm \sin (2^{k+1} + 1)x)$ if $k$ is even and $2^{-k}(\cos x - \cos 3x + \cdots \pm \cos(2^{k + 1} + 1)x )$ if $k$ is odd, but dealing with the behavior at $x = 0$ is a bit tricky. You can consider the integral over $[\epsilon, \infty)$, expand it in terms of $\epsilon$, and discard all but the constant term, but that involves a bit of computation. – anomaly Oct 20 '17 at 21:23
  • I agree with anomaly. It is enough to consider the Fourier sine/cosine series of the numerator, then exploit well-known integrals. – Jack D'Aurizio Oct 20 '17 at 21:50
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    It's a classic Borwein integral. See Marco Cantarini's answer to a related question. The approach I posted also applies. – Random Variable Oct 20 '17 at 22:01
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    @anomaly With your equalities, just to make sure there is no misunderstanding. The sum is not altering. For example, with $k=2$, $$\sin x\sin 2x\sin 4x=\frac14\bigl(-\sin x+\sin 3x+\sin 5x-\sin 7x\bigr).$$ – mickep Oct 21 '17 at 08:08

1 Answers1

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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.

Argon
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