Let $F(x)= \int_{0}^{\infty} \frac{\sin^{2} (x\alpha)}{\alpha^{2}} d\alpha$. Is it true $F, F^{-1}\in L^{1}(\mathbb R)$?

Question

Define $$F(x)= \int_{0}^{\infty} \frac{\sin^{2} (x\alpha)}{\alpha^{2}} d\alpha, \ (x\in \mathbb R).$$ It is clear to me that, the integral converges for every real $x$ (as near origin integrand is continuous, and at infinity we can handle it by $\alpha^{-2}$ ).

But here I am interested to learn some method (if exists) to compute the integral of this kind and behaviour of the function $F(x)$.

My Questions: (1) What can we say about $F(x)$ ? (2) Is it true that $\int_{\mathbb R} F(x) dx < \infty. $ (3) Is it true that $\int_{\mathbb R} \frac{1}{F(x)} dx < \infty ? $

Thanks,

Answer 1

The antiderivative of the integrand is $$\frac{2 a x \text{Si}(2 a x)+\cos (2 a x)-1}{2 a}$$ in which appears the famous sine integral. Going to bounds, you obtain the result already given by Samrat Mukhopadhyay $$\frac{\pi |x|}{2}$$ provided that $x\in \mathbb{R}$

Answer 2

Hint: Note that, $$F(x)=|x|I$$ where $$I=\int_{0}^{\infty}\frac{\sin^2{\alpha}}{\alpha^2}d\alpha=\frac{\pi}{2}$$