We have some good integrators here...well...let see this one done. Prove
Good luck...
Next, I see trying to find
by the substitution . Thus gives , and thus we see:
.
Next, try the substitution . This gives:
.
.
This is the Eulerian integral of the first kind, and so we have:
Where is the Beta function.
Now,
so
A little more to follow.
--Kevin C.
I'm getting a different result on the last steps:
The gamma function has a relationship known as Euler's reflection formula:
.
So:
and so:
This combines with the earlier results to give:
.
I'm not sure why I obtained a different value for the arguement of that gamma.
--Kevin C.