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.
 A: 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 $f(t) = -\frac{1}{2} \ln t$.
A: If $f'(x)$ is continuous and the integral exists then
$$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(0)-f(\infty)] \ln(b/a);~ a, b>0~~~~(1),$$
it is called Frullani's integral, see
https://en.wikipedia.org/wiki/Frullani_integral
Proof of Frullani's theorem
$$I=\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx=\int_{0}^{\infty} \frac{e^{(\sqrt{t}x)^2}-e^{(x)^2}}{x}dx=-\frac{1}{2}\ln(t)~~~~(2)$$
as $a=\sqrt{t}$ and $b=1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{\expo{-tx^{2}} - \expo{-x^{2}} \over x}\,\dd x} =
\\[5mm] = &\
-\int_{0}^{\infty}\ln\pars{x}\expo{-tx^{2}}\pars{-2tx}\dd x +
\int_{0}^{\infty}\ln\pars{x}\expo{-x^{2}}\pars{-2x}\dd x
\\[5mm] = &\
\int_{0}^{\infty}\ln\pars{x^{1/2} \over t^{1/2}}
\expo{-x}\,\dd x -
{1 \over 2}\int_{0}^{\infty}\ln\pars{x}\expo{-x}\dd x =
\bbx{-\,{1 \over 2}\,\ln\pars{t}} \\ &
\end{align}
A: Without using Frullani integral's, we have
$$\int \frac{e^{-kt^2}}x \,dx=\frac 12 {\text{Ei}\left(-t x^2\right)}$$ where appears the exponential integral function (we could also use the gamma function).
If $a>0$, then
$$\int_a^\infty \frac{e^{-kt^2}}x \,dx=\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)+\log
   \left(a^2\right)\right)$$
$$\int_a^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx=\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)-\Gamma
   \left(0,a^2\right)\right)$$ Using Taylor series
$$\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)-\Gamma
   \left(0,a^2\right)\right)=-\frac{\log (t)}{2}+\frac{1}{2} a^2 (t-1)+\frac{1}{8} a^4
   \left(1-t^2\right)+O\left(a^6\right)$$
A: Firstly, if you want to prove Frullani's integral:
$$I(a,b)=\int_0^\infty\frac{f(ax)-f(bx)}{x}dx$$
$$\frac{\partial I}{\partial a}=\int_0^\infty f'(ax)dx=\frac1a\int_0^\infty f'(x)dx$$
$$\frac{\partial I}{\partial b}=\int_0^\infty f'(bx)dx=\frac1b\int_0^\infty f'(x)dx$$
also:
$$dI=\frac{\partial I}{\partial a}da+\frac{\partial I}{\partial b}db$$
and so:
$$\int dI=\int_0^\infty f'(x)dx\left(\int\frac {da}a-\int\frac{db}b\right)$$
and so:
$$\int_0^\infty\frac{f(ax)-f(bx)}{x}=\left[\lim_{x\to\infty }f(x)-\lim_{x\to 0^+}f(x)\right]\left(\ln a-\ln b\right)$$

now that we have this we want to define our function, we can say:
$$f(x)=e^{-x^2}$$
and so this gives us: $a=\sqrt{t},b=1$ and so finally our integral is equal to:
$$I=\left[f(\infty)-f(0)\right]\left(\ln\sqrt{2}-\ln 1\right)$$
$$I=(0-1)(\ln\sqrt{t}-0)=-\ln\sqrt{t}=-\frac12\ln t$$
A: take $u=x^2$ to get $$ \int_{0}^{\infty}\dfrac{e^{-tu}-e^{-u}}{2u}du = \dfrac{-\ln(u)}{2} $$
