Evaluating $\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$ I have been working on the following integral:
$$\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$$
where $t$ is any nonnegative real number. Would anyone be able to provide a hint or provide a solution on how such an integral should be approached? I have a hunch it involves differentiation under the integral sign.

Using differentiation under the integral sign makes it quite easy: let
$f(t) = \int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$, then
$$f'(t) = \int_0^\infty \frac{\partial}{\partial t}\left(\frac{e^{-tx^2}-e^{-x^2}}{x}\right)dx = \int_0^\infty \frac{-x^2 e^{-tx^2}}{x} dx = -\int_0^\infty x e^{-tx^2} dx = \left[\sqrt{t}x = s, dx = \frac{ds}{\sqrt{t}}\right] = 
-\frac{1}{t}\int_0^\infty s e^{-s^2} ds = -\frac{1}{2t} \int_0^\infty e^{-s^2} d(s^2) = 
-\frac{1}{2t} e^{-s^2} \vert_0^{+\infty} = -\frac{1}{2t}
$$
So, $f'(t) = -\frac{1}{2t}$, $f(1) = 0$ (easy to see), so $f(t) = -\frac{1}{2} \ln t$.
