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 $\theta$?


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

1 Answer 1


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 $a>0$.

  • 1
    $\begingroup$ Please, can you clarify why your starting integral is the same as the integral in OP? Tanks! $\endgroup$ Apr 22, 2017 at 13:47
  • $\begingroup$ I just used trivial substitutions, polar coordinates and ultimately the integral definitions of the $\Gamma$ and Beta functions. $\endgroup$ Apr 22, 2017 at 13:48
  • $\begingroup$ all too easy (+1) $\endgroup$
    – tired
    Apr 22, 2017 at 13:50
  • $\begingroup$ What I 'don't understand is the $1/\Lambda^2$ factor. $\endgroup$ Apr 22, 2017 at 14:03
  • $\begingroup$ @EmilioNovati: The very first step is to replace $x$ with $\frac{x}{\Lambda}$ and similarly $y$ with $\frac{y}{\Lambda}$. $\endgroup$ Apr 22, 2017 at 14:04

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