Consider the following Gaussian Type Integral
$$I^2 = \frac{1}{4\pi kt}\iint_{\mathbb{R}^2} e^{-(x^2+y^2)/4kt} \ dydx$$
I know the standard method would use the polar coordinate transformation to calculate this integral, but I am kind of lost at here since there are several letter constant involving in the exponent. Could someone help me on what would be the actual transformation like in this cases? Many thanks.
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I think I had the answer by myself, my calculation is blow, would someone could tell am I get right thing? $$I^2 = \frac{1}{4\pi kt}\iint_{\mathbb{R}^2} e^{-(x^2+y^2)/4kt} \ dydx$$
take the coordinate transformation $x = cos\theta$, and $y = sin\theta$, then the integral become: $$I^2 = \frac{1}{4\pi kt}\int_{0}^{2\pi}\int_{0}^\infty e^{-r^2/4kt} r\ drd\theta$$
Take $ u = -\frac{r^2}{4kt}$, then $\ du =-\frac{r}{2kt}dr $ $$ = \frac{1}{2\pi}\cdot-\int_{0}^{2\pi}d\theta\int_{0}^\infty e^{u} \frac{r}{2kt}\ dr$$ $$ = \frac{1}{2\pi}\int_{0}^{2\pi}d\theta\int_{-\infty}^0 e^{u} du$$ $$ = \frac{1}{2\pi}\cdot2\pi\cdot1=1$$
Step 1: Switch to polar coordinates and integrate out the angle: $I^2=\frac{1}{2kt}\int_0^\infty e^{-\frac{r^2}{4kt}}rdr$.
Step 2: Let $u=\frac{r^2}{4kt}$ Then $rdr=2ktdu$ so integral becomes $I^2=\int_0^\infty e^{-u}du=1$.