We have some good integrators here...well...let see this one done. Prove

Good luck...

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- June 15th 2008, 07:54 PMMathstud28Challenge
We have some good integrators here...well...let see this one done. Prove

Good luck... - June 15th 2008, 07:58 PMtopsquark
- June 15th 2008, 08:32 PMTwistedOne151First step
Well, right away I see you can make the substitution . Then we have

Which takes care of the term containing*a*. Proving that

will require a bit more thought. I'm thinking a beta function integral may be involved at some point.

--Kevin C. - June 15th 2008, 09:43 PMMathstud28
- June 16th 2008, 08:52 AMTwistedOne151Next steps
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. - June 16th 2008, 09:13 AMTwistedOne151I'm not getting the same result
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.