Calculate integral of Gaussian product by parts Have an issue calculating convolution of a Gaussian.
$$\int_{-\infty}^{\infty}e^{-\pi (t-x)^2}e^{-\pi x^2}dx.$$
How is it possible to calculate this integral by parts?
EDITED:
I got to a point where
$$e^{-\pi t^2}\int_{-\infty}^{\infty}e^{-2\pi x^2 }e^{2\pi xt}dx.$$
$$u=e^{-2\pi x^2}, du=e^{-2\pi x^2}(-4\pi x)$$
$$dv=e^{2\pi xt}dx, v=\frac{1}{2\pi t}e^{2\pi xt}$$
then
$$uv=\frac{1}{2\pi t}\frac{e^{2\pi xt}}{e^{2\pi x^2}}\Biggr|_{-\infty}^{\infty}=0$$
$$vdu = \frac{-4\pi}{2\pi t} \int_{-\infty}^{\infty}xe^{2\pi xt}e^{-2\pi x^2 }dx$$
and here I don't know what to do with $$vdu$$
 A: I am not sure it is easy to do by parts. Picking any one of the exponents as $dv$ would require integrating that, which does not have a closed form.
Here is a different way instead. When you multiply both exponents together, and factor $-\pi$ from the resulting power, you get
$$
(t-x)^2+x^2
 = 2x^2-2tx+t^2
 = \frac{4x^2-4tx+2t^2}{2}
 = \frac12 (2x-t)^2+\frac{t^2}{2},
$$
so now you have to integrate
$$
\begin{split}
\int_{-\infty}^\infty \exp\left(-\frac{\pi}2 (2x-t)^2-\frac{\pi t^2}{2}\right) dx
 &= \int_{-\infty}^\infty \exp\left(-\frac{\pi}2 (2x-t)^2\right)
                          \exp\left(-\frac{\pi t^2}{2}\right) dx \\
 &= \exp\left(-\frac{\pi t^2}{2}\right)
    \int_{-\infty}^\infty \exp\left(-\frac{\pi}2 (2x-t)^2\right) dx
\end{split}
$$
and now change variables to $u=\frac{2x-t}{\sqrt{\pi/2}}$ to reduce this to a standard Gaussian.

UPDATE
Perhaps a similar approach can lead to an integration by parts. You have:
$$
\begin{split}
I &= \int_{-\infty}^\infty \exp\left(-\pi(x-t)^2-\pi t^2\right) dx \\
  &= \int_{-\infty}^\infty \exp\left(-\pi x^2 +2\pi xt -2\pi t^2\right) dx \\
  &= e^{-2\pi t^2}
     \int_{-\infty}^\infty e^{-\pi x^2}e^{2\pi xt} dx
\end{split}
$$
and now let $u = e^{-\pi x^2}$ and $dv = e^{-2\pi tx} dx$. You can differentiate $u$ and $dv$ integrates well since there is only one power of $x$ in the exponent. Can you now finish this?

UPDATE 2
So you have continued this line; I got
$$
I = -\int_{-\infty}^\infty vdu
  = \frac{1}{t} \int_{-\infty}^\infty xe^{2\pi xt}e^{-\pi x^2 }dx
$$
and we can complete the square in the exponent:
$$
2\pi xt - \pi x^2= -\pi\left(x^2-2xt+t^2\right)+\pi t^2 = -\pi(x-t)^2+\pi t^2,
$$
so now we use the extra linear factor we got from integration by parts to substitute $u = -\pi(x-t)^2$ and then $du = -2\pi(x-t)dx$, and so
$$
\begin{split}
I &= \frac{e^{\pi t^2}}{t} \int_{-\infty}^\infty xe^{-\pi(x-t)^2}dx \\
  &= \frac{e^{\pi t^2}}{t}
     \left[
        \int_{-\infty}^\infty (x-t)e^{-\pi(x-t)^2}dx
      + \int_{-\infty}^\infty te^{-\pi(x-t)^2}dx
     \right] 
\end{split}
$$
which you can now finish using the above substitution
