gaussian-like integral $\int_{-\infty}^{\infty}\frac{1}{x^2+t^2}\exp^{-\frac{x^2}{2t}}dx$ I'm trying to figure out how to evaluate the integral
$\int_{-\infty}^{\infty}\frac{1}{x^2+t^2}\exp^{-\frac{x^2}{2t}}dx$
where $t\geq 0$.
I have tried a lot of methods but without success. Please help.
 A: Change variables in your integral, letting $x = ty$. Then your integral becomes
$${1 \over t} \int_{-\infty}^{\infty}\frac{1}{y^2 + 1}e^{-\frac{ty^2}{2}}dy$$
Thus you are looking for ${\displaystyle {1 \over t}f\bigg({t \over 2}\bigg)}$, where
$$f(u) = \int_{-\infty}^{\infty}\frac{1}{y^2 + 1}e^{-uy^2}dy$$
By differentiating under the integral sign, you have
$$f'(u) - f(u) = -\int_{-\infty}^{\infty}e^{-uy^2}dy$$
By changing variables $z = u^{1/2}y$ this is equal to
$$-u^{-{1 \over 2}}\int_{-\infty}^{\infty}e^{-z^2}dz$$
$$= -\sqrt{\pi}u^{-{1 \over 2}}$$
So $f(u)$ satisfies the differential equation
$$f'(u) - f(u) = -\sqrt{\pi}u^{-{1 \over 2}}$$
$$f(0) = \pi$$
Now we can use standard first order linear methods, using an integrating factor of $e^{-u}$ to get
$$(e^{-u} f(u))' = - \sqrt{\pi}u^{-{1 \over 2}}e^{-u}$$
Hence $f(u)$ is given by
$$-\sqrt{\pi}e^{u} (\int^u v^{-{1 \over 2}}e^{-v}\,dv + C)$$
Changing variables $v = w^2$ leads to
$$f(u) = -2\sqrt{\pi}e^{u} (\int^{u^{1 \over 2}} e^{-w^2}\,dw + C)$$
$$= -\pi e^{u}(erf(u^{1 \over 2}) + C)$$
Now it just becomes a question of plugging in $u = 0$ to get $C$, and then letting ${1 \over t}f(t/2)$ be your final answer. Plugging in $u = 0$ gives $C = -1$, so that
$$f(u) = -\pi e^{u}(\rm{erf}(u^{1 \over 2}) - 1)$$
$$= \pi e^u {\rm erfc} (u^{1 \over 2})$$
So your final answer is just
$$ {\pi \over t}e^{t \over 2}{\rm erfc}\bigg(\sqrt{t \over 2}\bigg)$$
