Trouble with Gamma function

Solve for $\displaystyle \Gamma(-3/2).$

$\displaystyle \Gamma(-3/2) = \frac{1}{2} \cdot \frac{3}{2} \Gamma(1/2)$

$\displaystyle = \frac{4}{3} \int_{0}^\infty t^{1/2} e^{-t} dt$ <-- With $\displaystyle t = u^2$,

$\displaystyle = \frac{8}{3} \int_{0}^\infty e^{-u^2} du$

My book says that I need to square the left side and turn the right side into a double integral of $\displaystyle e^{-(u^2 + v^2)}$ and then convert u into r cos theta, v into r sin theta, and go from there. But that is where I get lost. Obviously that makes u^2 + v^2 = r^2, but what the heck do I do with (a) du and dv, and (b) the limits of integration?

Re: Trouble with Gamma function

Are you familiar with polar coordinates? You express the function value in terms of r and theta. The transformation is x = rcos(theta), y = rsin(theta), so the Jacobian will be r. But even if you don't know this stuff, the integral should still be solvable by other means.

Gaussian integral - Wikipedia, the free encyclopedia

This will help.

Re: Trouble with Gamma function

Oh whoa. That's basically the normal curve. So really, I could just stick with *u* and integrate, getting $\displaystyle \frac {\sqrt{\pi}}{2}$? And then multiply times 8/3 to get the final answer?

Re: Trouble with Gamma function

Yes, that would be precisely correct. But I can't tell you whether or not the teacher wants more work. It would be the correct answer though.