Why $\int_{0}^{+\infty} \frac{\sin(x)}{x} \: dx = \int_{0}^{+\infty} \frac{\sin^{2}(x)}{x^{2}} \: dx$? In an exercise, I saw the following equality :
$$ \int_{0}^{+\infty} \frac{\sin(x)}{x} \: dx = \int_{0}^{+\infty} \frac{\sin^{2}(x)}{x^{2}} \: dx $$
At first, I was surprised by this equality. It is very easy to prove using integration by parts. Still, I don't find this explanation "intuitive". Is there a more intuitive explanation ?
 A: Well, in general, 
$$\int_{-\infty}^{\infty} dx \, f(x)$$
is equal to the Fourier transform of $f$, $\hat{f}(k)$, at $k=0$.  In this case, the FT of $\sin{x}/x$ is 
$$\hat{f}(k) = \begin{cases} \pi & |k| \le 1\\0&|k| \gt 1 \end{cases}$$
so that the integral on the left is $(1/2) \pi$.
To consider the integral on the right, we use the fact that the Fourier transform of a product of two functions is equal to $1/(2 \pi)$ times the convolution of the transforms of the functions.  In this case, the transform would be the above rectangle function convolved with itself.  But because we are evaluating the integral on the RHS, which is the FT at $k=0$, we just need the convolution at $k=0$, which is just the integral of  square of the above rectangle function, or $2 \pi^2/(2 \pi)$, so that the integral on the RHS is, again $\pi/2$.
The salient fact here is that the transforms of the above integrands are based on rectangle functions, which exhibit a form of invariance in their product.
A: You can generalize this result using the residue theorem of complex analysis and the theorem of the upper half plane, then you don't need partial integration, but this is the way, calculate as much as you can and see at the ending, that it is the same.
Don't think too much about this problem, because it's just coincidence.
A: If you use integration by parts with $ u=\sin(x)^2 $, then we have 
$$ \int_{0}^{+\infty} \frac{\sin^{2}(x)}{x^{2}} \: dx =  \int_{0}^{+\infty}\frac{2\sin(x)\cos(x)}{x}dx = \int_{0}^{+\infty}\frac{\sin(2x)}{x}dx =I. $$
Now, use the change of variables $2x=t$, you get the equality of the two integrals
$$ \int_{0}^{+\infty}\frac{\sin(t)}{t}dt. $$
