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.
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$\begingroup$ Are you sure about the limits of integration in this order? $\endgroup$– Emilio NovatiApr 22, 2017 at 13:36
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1 Answer
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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 $a>0$.
answered Apr 22, 2017 at 13:41
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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
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$\begingroup$ 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
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$\begingroup$ What I 'don't understand is the $1/\Lambda^2$ factor. $\endgroup$ Apr 22, 2017 at 14:03
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$\begingroup$ @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