It may seem like a very basic question to some but I struggle at integrating $$\int_{-\infty} ^\infty \frac{1}{4\pi} e^{-\frac{1}{8}((u+v)^2+(u-v)^2)} dv$$
It gives me the error function $\mathrm{erf}(x)$ when I try to calculate it online which I am not familiar with.
Can anyone help me out please?
Start by looking at the exponent:$$(u+v)^2+(u-v)^2=u^2+2uv+v^2+u^2-2uv+v^2=2u^2+2v^2$$ Then the integral becomes: $$I=\int_{-\infty}^\infty \frac{1}{4\pi} e^{-\frac{1}{8}((u+v)^2+(u-v)^2)} dv=\frac1{4\pi}e^{-\frac{u^2}{4}}\int_{-\infty}^\infty e^{-\frac{v^2}{4}}dv$$ Change the variable $x=v/2$, so $dv=2dx$ and $$I=\frac 1{2\pi}e^{-\frac{u^2}{4}}\int_{-\infty}^\infty e^{-x^2}dx$$ Let's call $$G=\int_{-\infty}^\infty e^{-x^2}dx$$ Then $$G^2=\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}dxdy$$ Transforming to polar coordinates $$G^2=\int_0^\infty dr\int_0^{2\pi}d\theta re^{-r^2}=2\pi\int_0^\infty re^{-r^2}dr$$ Change the variable again $z=r^2$. Then $dz=2rdr$. I hope you can continue from here.
$$(u+v)^2+(u-v)^2=(u^2+2uv+v^2)+(u^2-2uv+v^2)=2(u^2+v^2)$$ now lets take a look at the exponential: $$e^{-\frac14(u^2+v^2)}=e^{-\frac{u^2}4}e^{-\frac{v^2}4}$$ we can clean the integral up a bit now: $$I=\frac{e^{-\frac{u^2}4}}{4\pi}\int_{-\infty}^\infty e^{-v^2/4}dv$$ if we make the substitution $x=v/2\Rightarrow dv=2dx$ and so: $$I=\frac{e^{-u^2/4}}{2\pi}\int_{-\infty}^\infty e^{-x^2}dx$$ and this is the famous gaussian integral, which the error function is the finite version of with a scaling factor. In my answer to this question I derived why this was true so you can take a look at that
$(u+v)^2+(u-v)^2=2u^2+2v^2$ so the integral is $\frac{e^{-\frac{u^2}{4}}}{4\pi}\int_{-\infty}^\infty e^{-\frac{v^2}{4}}dv=\frac{e^{-\frac{u^2}{4}}}{2\sqrt{\pi}}$.
Note: There is no $x$ in the expression! How did you get $erf(x)$? The upper limit would have been $x$, not $\infty$.
