Prove that $\int_0^\infty y^{\frac12} e^{-y^2} \int_0^\infty y^{-\frac12} e^{-y^2} =\frac{\pi}{2^{\frac32}}$

Question

My attempt:

$\int_0^\infty y^{\frac12} e^{-y^2}$

$=\frac12 e^{-z} z^ {-\frac14} dz$, using the transformation: $y^2=z$, i.e. $y=z^\frac12$

$=\frac12 e^{-z} z^ {1-\frac54} dz$

$=\Gamma\frac54$

$=\frac14\Gamma\frac14$

Similarly, $\int_0^\infty y^{-\frac12} e^{-y^2}= \Gamma \frac34$

I can't take it beyond this. Can someone please help? I know I'm supposed to get a $\Gamma \frac12$ but I didn't. Where did I go wrong?

Answer 1

You made a mistake in determining $\Gamma$ values. Let's use the Pi function, $\Pi(x)=\int_0^\infty t^x e^{-t}\,dt=\Gamma(x+1)$, which makes things clearer.

The first integral is $1/2\,\Pi(-1/4)$ and the second is $1/2\,\Pi(-3/4)$. We know $\Pi(1/4)=1/4\,\Pi(-3/4)$, so using the reflection formula $$1/4\,\Pi(-1/4)\,\Pi(-3/4)=\Pi(-1/4)\,\Pi(1/4)=\frac{\pi/4}{\sin(\pi/4)}=\frac{\pi}{2\sqrt 2}$$