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 $$vdu$$
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