MHB Problem of the Week #86 - November 18th, 2013

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: For $x,y>0$, show that

\[\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}= 2\int_0^{\pi/2} \cos^{2x-1}\theta\sin^{2y-1}\theta\,d\theta\]

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Hint: [sp]1) Make the substitution $t=u^2$ in the definition of $\Gamma(x)$, where
\[\Gamma(x) = \int_0^{\infty} e^{-t}t^{x-1}\,dt\]

2) Use (1) to create a double integral for the expression $\Gamma(x)\Gamma(y)$

3) Change to polar coordinates in order to evaluate the double integral.[/sp]

 

[Chris L T521]

This week's problem was correctly answered by topsquark.

Here's my solution, which uses the approach outlined in the hint:

[sp]Let $t=u^2\implies 2u\,du = \,dt$. Therefore,
\[\Gamma(x) = \int_0^{\infty}e^{-t}t^{x-1}\,dt \xrightarrow{t=u^2}{} \Gamma(x) = 2\int_0^{\infty}e^{-u^2}(u^2)^{x-1}u\,du = 2\int_0^{\infty}e^{-u^2}u^{2x-1}\,du\]
Now let
\[\Gamma(y) = 2\int_0^{\infty}e^{-v^2}v^{2y-1}\,dv\]
and hence we see that
\[\Gamma(x)\Gamma(y) = 4\int_0^{\infty}\int_0^{\infty} e^{-(u^2+v^2)} u^{2x-1}v^{2y-1}\,du\,dv\]

Now, we make the conversion to polar. Let $u=r\cos\theta$, $v=r\sin\theta$ and thus $r^2=u^2+v^2$. We also see that the region defined by $0<u<\infty$ and $0<v<\infty$ is the first quadrant in the $uv$-system; thus, the appropriate region in the polar coordinate system would be $0< r< \infty$ and $0<\theta<\dfrac{\pi}{2}$. Thus, we see that

\[\begin{aligned} \Gamma(x)\Gamma(y) &= 4\int_0^{\pi/2}\int_0^{\infty} e^{-r^2} r^{2x+2y-2}\cos^{2x-1}\theta \sin^{2y-1}\theta r\,dr\,d\theta \\ &= \left(\color{purple}{ 2\int_0^{\infty} e^{-r^2}r^{2(x+y)-1}\,dr}\right) \left(2\int_0^{\pi/2} \cos^{2x-1}\theta\sin^{2y-1}\theta\,d\theta\right) \\ &= 2 \color{purple}{\Gamma(x+y)} \int_0^{\pi/2} \cos^{2x-1}\theta\sin^{2y-1}\theta\,d\theta.\end{aligned}\]

Therefore, we now see that

\[\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = 2\int_0^{\pi/2} \cos^{2x-1}\theta\sin^{2y-1}\theta\,d\theta.\][/sp]

Here's topsquark's solution, which nicely uses the Beta function:

[sp]I find it to be easier to deal with the Beta function, rather than the Gamma function directly:
[math]2 \int_0^{\pi/2} \cos^{2x - 1} \theta \cdot \sin^{2y - 1} \theta\,d \theta[/math]

[math]= \int_0^{\pi/2} \cos^{2x - 2} \theta \cdot \sin^{2y - 2} \theta \cdot 2\sin \theta \cos \theta \,d \theta[/math]

[math]= \int_0^{\pi/2} (1 - \sin^2 \theta )^{x - 1} \cdot (\sin^2 \theta )^{y - 1} \cdot 2\sin \theta\cos \theta\,d \theta[/math]

Let [math]t = \sin^2 \theta \implies dt = 2 \sin \theta \cos \theta \,d \theta[/math]

Thus
[math]2 \int_0^{\pi/2} \cos^{2x - 1} \theta \cdot \sin^{2y - 1} \theta \,d \theta = \int_0^1 (1 - t)^{x - 1} t^{y - 1}\,dt [/math]

[math]= B(y, x) = \frac{\Gamma(y) \Gamma(x)}{\Gamma(y + x)}[/math]

where $B(y,x)$ is the Beta function.[/sp]