Solve integral using Plancherel's formula

Question

This is from a test in Fourier analysis:

Define $$ f(\xi)=\int_0^1 \sqrt{x}\rm{sin}(\xi x) \rm{d}x $$ Calculate $$ \int_{-\infty}^\infty |f(x)|^2 \rm{d}x $$

I started with Plancherel's formula, i.e. $$ \int_{-\infty}^\infty |f(x)|^2 \rm{d}x = \frac{1}{2\pi}\int_{-\infty}^\infty |f(\xi)|^2 \rm{d}\xi $$

but what now? Am I supposed to compute the integral of $f(\xi)$ somehow or what should I do?

Answer 1

Added the "idea flow" to the solution.. I don't know maybe it's useful.

The idea is that the definition of $f(\xi)$ looks remarkably similar to the definition of the Fourier transform. We want to calculate the $2$-norm of $f(\xi)$, and we want to use Plancherel's theorem; So we need to make Fourier transform come into play somehow. You note that if the function you're integrating is odd, then the $\cos x$ part in the Fourier transform vanishes; so we

Define $\sqrt x$ in $(-1, 1)$ so that it is an odd function. It's fourier transform is

$$\mathcal F (t) = \int_{-1}^1 \sqrt x e^{ixt} dx = \int_{-1}^1 \sqrt x \cos(tx) dx + i \int_{-1}^1 \sqrt x \sin(tx) dx = 2i \int_0^1 \sqrt x \sin(tx) dx = 2if(t)$$

Now $\mathcal F$ and our modified $\sqrt x$ have the same $||\cdot||_2$ norm (plancherel's theorem!) , and you basically want to find $||\mathcal F||_2$.

You should be good to go now! :-)