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I need to prove the following in the most simple way:

$$\int_{0}^{\infty} (\frac{\sin(x)}{x})^{2} = \int_{0}^{\infty} \frac{\sin(x)}{x}$$

I got an hint to use integration by parts, so this is my try:

$$\int_{0}^{\infty} (\frac{\sin(x)}{x})^{2} = \left [ \frac{\sin^{2}(x)\cdot (-1)}{x} \right ]_{0}^{\infty}+\int_{0}^{\infty}\frac{2\sin(x)\cdot \cos(x)}{x}$$

But I got stuck here and I don't know how to continue to get the right result.

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    $\begingroup$ Use $2\sin x \cos x=\sin 2x.$ $\endgroup$ May 14 at 14:30

1 Answer 1

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The first term is equal to $0$, this is easy. As for the second term:

$\int\limits_0^{\infty} \frac{2\sin(x)\cos(x)}{x}dx=2\int\limits_0^{\infty}\frac{\sin(2x)}{2x}dx=2\int\limits_0^{\infty}\frac{\sin(t)}{t}\frac{1}{2}dt=\int\limits_0^{\infty}\frac{\sin(t)}{t}dt$

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