A weighted normal-integral is doable?

Question

Question without words: $\int_{-\infty}^\infty \exp(-a^2x^2)/(1+x^2)=? \ $. Maple fails on this, but I saw this in a Russian example-book, so I am curious. I can do it for $a=0$ only.

Answer 1

Looking at

$$I(b)=\int_{-\infty}^\infty \exp(-bx^2)/(1+x^2)dx$$

with $b=a^2$, take the derivative w.r.t. $b$ and you can establish the following equation for $I$

$$\frac{dI(b)}{db}=I(b)-\sqrt{\frac{\pi}{b}}$$

you can solve this to arrive at the following representation for $I(b)$:

$$I(b)=\pi e^b - \sqrt{\pi}e^b\int_0^{b}t^{-\frac{1}{2}}e^{-t}dt$$

The last integral is an incomplete Gamma function which can be transformed into an erf function via the relation

$$\gamma(\frac{1}{2},b)=\sqrt{\pi}\text{erf}(\sqrt{b})$$

which will get you the expression Dr. Sonnhard Graubner suggested in the comments.