Solving $\int_0^{+\infty}\frac{e^{-\alpha x^2} - \cos{\beta x}}{x^2}dx$ I need to find solution of $$\int_0^{+\infty}\frac{e^{-\alpha x^2} - \cos{\beta x}}{x^2}dx$$
I know that Leibniz rule can help but I don't know how to use it.
Could you help me please?
Thank you.

Assume that $\alpha>0$ and $\beta>0$. You may write
$$
\int_0^{+\infty}\frac{e^{-\alpha x^2} - \cos{\beta x}}{x^2}dx=\int_0^{+\infty}\frac{e^{-\alpha x^2} - 1}{x^2}dx+\int_0^{+\infty}\frac{1 - \cos{\beta x}}{x^2}dx
$$ then use two standard results. By an integration by parts, we have
$$
\begin{align}
\int_0^{+\infty}\frac{e^{-\alpha x^2} - 1}{x^2}dx &=\left. -\frac{e^{-\alpha x^2} - 1}{x}\right|_0^{+\infty}-2\alpha\int_0^{+\infty}e^{-\alpha x^2}dx=-\sqrt{\pi\alpha}.
\end{align}
$$ On the other hand, using a standard result, we have
$$
\begin{align}
\int_0^{+\infty}\frac{1 - \cos{\beta x}}{x^2}dx &=2\int_0^{+\infty}\frac{\sin^2{(\beta x/2)}}{x^2}dx=\frac{\pi}2\beta.
\end{align}
$$ Then

$$
\int_0^{+\infty}\frac{e^{-\alpha x^2} - \cos{\beta x}}{x^2}dx=\frac{\pi}2\beta-\sqrt{\pi\alpha}.
$$

