# Expressing double integrals in polar coordinates

I have the double integrals

$\int_{x}^\infty \int_0^\infty e^{-\Lambda^a(x^a+y^a)}dxdy$

where $a$ and $\Lambda$ are real positive constants.

I would like to express it in polar coordinates. What are my $r$ and $\theta$?

Thanks.

• Are you sure about the limits of integration in this order? Apr 22, 2017 at 13:36

The given integral equals $$\frac{1}{\Lambda^2}\iint_{0\leq x \leq y}\exp\left[-(x^a+y^a)\right]\,dx\,dy =\frac{4}{a^2\Lambda^2}\int_{0\leq u\leq v} (uv)^{2/a-1}e^{-(u^2+v^2)}\,du\,dv$$ or $$\frac{4}{a^2\Lambda^2}\int_{0}^{+\infty}\int_{\pi/4}^{\pi/2}\rho(\rho^2\sin\theta\cos\theta)^{2/a-1} e^{-\rho^2}\,d\theta\,d\rho$$ or $$\frac{2}{a^2\Lambda^2}\,\Gamma\left(\frac{1}{2}+\frac{2}{a}\right)\,2^{-2/a}\int_{0}^{\pi/2}\left(\sin t\right)^{2/a-1}\,dt = \color{red}{\frac{2^{-2/a} \sqrt{\pi }\; \Gamma\left(\frac{1}{2}+\frac{2}{a}\right)\, \Gamma\left(\frac{1}{a}\right)}{a^2 \Lambda^2\, \Gamma\left(\frac{1}{2}+\frac{1}{a}\right)}}$$ for any $a>0$.
• I just used trivial substitutions, polar coordinates and ultimately the integral definitions of the $\Gamma$ and Beta functions. Apr 22, 2017 at 13:48
• What I 'don't understand is the $1/\Lambda^2$ factor. Apr 22, 2017 at 14:03
• @EmilioNovati: The very first step is to replace $x$ with $\frac{x}{\Lambda}$ and similarly $y$ with $\frac{y}{\Lambda}$. Apr 22, 2017 at 14:04