Convolution of $1/(1+x^2)$ and $\exp(-x^2/(4t))$ Is there a closed form formula for the convolution of $1/(1+x^2)$ and $\exp(-x^2/(4t))$, where $t>0$, i.e. the integral
$$\int_{-\infty}^\infty \frac{1}{1+u^2}\exp\left(-\frac{(x-u)^2}{4t}\right)\,du?$$
I doubt it, but someone may be able to use special functions.
 A: The integral may be evaluated using the convolution theorem.  The FT of this convolution is the product of the individual FTs; we then take the inverse FT.
The individual functions are $f(x)=1/(1+x^2)$ and $g(x)=e^{-x^2/(4 t)}$.  The FT of the above convolution is then
$$\hat{f}(k) \hat{g}(k) = 2 \sqrt{\pi^3 t} e^{-t k^2} e^{-|k|}$$
The integral is then the inverse FT of the above:
$$\begin{align}\int_{-\infty}^{\infty} \frac{du}{1+u^2} e^{-(u-x)^2/(4 t)} &= \sqrt{\pi t} \int_{-\infty}^{\infty} dk \, e^{-t k^2} e^{-|k|} e^{-i k x}\\ &=\sqrt{\pi t} \int_{-\infty}^{0} dk \, e^{-t k^2} e^{k} e^{-i k x} + \sqrt{\pi t} \int_{0}^{\infty} dk \, e^{-t k^2} e^{-k} e^{i k x}\\ &=2 \sqrt{\pi t} \Re{\left [\int_0^{\infty} dk\, e^{-t k^2} e^{(1-i x)k}  \right ]}\\ &= \pi e^{-(x^2-1)/(4 t)} \Re{\left [e^{-i x/(2 t)} \text{erfc}\left(\frac{-1+i x}{2 \sqrt{t}} \right ) \right]} \end{align}$$
A: An equivalent integral,
$$\int_{-\infty}^{\infty}\frac{e^{-x^2}dx}{(x-a)^2+b^2},$$
for $a,b>0$ can be expressed in terms of error function:
$$\frac{\pi}{b}e^{b^2-a^2}\left\{\cos2 ab-\mathrm{Re}\left[e^{2iab}\mathrm{erf}(b+ia)\right]\right\}.\tag{1}$$ 

Added derivation: Write
$$\frac{1}{(x-a)^2+b^2}=\frac{1}{2ib}\left(\frac{1}{x-a-ib}-\frac{1}{x-a+ib}\right).
\tag{2}$$
Henc our problem reduces to calculating $\displaystyle\int_{-\infty}^{\infty}\frac{e^{-x^2}dx}{x-\alpha}$ with $\alpha\notin\mathbb{R}$. Now suppose e.g. that $\mathrm{Im}\,\alpha>0$ and then deform the integration contour by translating it upwards. The final contour will consist of 3 parts: two horizontal segments $C_-=(\alpha-\infty,\alpha-\epsilon]$ and $C_+=[\alpha+\epsilon,\alpha+\infty)$ and a demi-circle $C_{0}$ of radius $\epsilon$ centered at $\alpha$. 


*

*The integral over $C_0$ in the limit $\epsilon\rightarrow0$ gives simply $i\pi \, e^{-\alpha^2}$.

*The integrals over $C_+$ and $C_-$ give
\begin{align}
\int^{\infty}_{\epsilon}\frac{e^{-(\alpha+s)^2}ds}{s}+\int^{-\epsilon}_{-\infty}\frac{e^{-(\alpha+s)^2}ds}{s}&=-2e^{-\alpha^2}\int_{\epsilon}^{\infty}\frac{\sinh2\alpha s}{s}e^{-s^2}ds\rightarrow\\&\rightarrow -2e^{-\alpha^2}\int_{0}^{\infty}\frac{\sinh2\alpha s}{s}e^{-s^2}ds\quad \mathrm{as}\;\epsilon\rightarrow0.
\end{align}

*The last integral can be rewritten as
\begin{align}
\int_0^{\infty}\frac{\sinh2\alpha s}{s}e^{-s^2}ds&=2\int_0^{\infty}e^{-s^2}\left(\int_0^{\alpha}\cosh 2\nu s \,d\nu\right)ds=\\ &=\int_0^{\alpha}\left(\int_{0}^{\infty}2\cosh2\nu s\,e^{-s^2}\right)d\nu=\\ &=\sqrt{\pi}
\int_0^{\alpha}e^{\nu^2}d\nu=-\frac{i\pi}{2}\,\mathrm{erf}(i\alpha).
\end{align}


Summarizing all of this, we get
$$ \int_{-\infty}^{\infty}\frac{e^{-x^2}dx}{x-\alpha}=i\pi e^{-\alpha^2}\left[1+\mathrm{erf}(i\alpha)\right].\tag{3}$$
Combining (2) and (3), we then get (1). $\blacksquare$ 
