I'm trying to solve
$\int_0^∞\int_0^∞ e^{C/2(x^2+y^2)}dxdy$
by Laplace transform, where C is a constant.
I know that I need to let $x=r cosθ$ and $y=r sinθ$. But then when I diffrentiating x and y respecting to r and substitute into the equation. It's getting complicated. I tried to use trigonometric identities to solve the equation but the results come out is complex. Any solution that makes it simple?
In $\mathbb{R}^+\times\mathbb{R}^+$ the length of the curve $x^2+y^2=\rho^2$, for any $\rho>0$, is given by $\frac{\pi}{2}\rho$.
It follows that $$ \iint_{(0,+\infty)^2}\exp\left(\frac{C}{2(x^2+y^2)}\right)\,dx\,dy=\int_{0}^{+\infty}\frac{\pi}{2}\rho\exp\left(\frac{C}{2\rho^2}\right)\,d\rho=\int_{0}^{+\infty}\frac{\pi}{2u^3}e^{\frac{C}{2}u^2}\,du $$ is divergent for any value of $C$, and $$ \iint_{(0,+\infty)^2}\exp\left(\frac{C}{2}(x^2+y^2)\right)\,dx\,dy=\int_{0}^{+\infty}\frac{\pi}{2}\rho\exp\left(\frac{C\rho^2}{2}\right)\,d\rho=\int_{0}^{+\infty}\frac{\pi}{4}e^{\frac{C}{2}u}\,du $$ is convergent to $\frac{\pi}{2|C|}$ for any $C<0$. You may regard the last integral as a value of the Laplace transform of $1$, but that is pretty artificial and not really needed.