Perhaps a number of you are familiar with the solution to this integral. if not, see if you can work it out.

If you do them individually, you just get

But that is certainly not the solution.

It has the same solution as:

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- Jul 6th 2008, 07:15 AMgalactuscool integral
Perhaps a number of you are familiar with the solution to this integral. if not, see if you can work it out.

If you do them individually, you just get

But that is certainly not the solution.

It has the same solution as:

- Oct 25th 2008, 09:33 PMMathstud28
I have worked on this integral for about two weeks and have finally come up with a solution

First split it into

Then make the sub on the second integral turning it into

Integration by parts on both gives

Now consider the gamma function

Well since the Gamma function is uniformly convergent for all values of x that make it converge we have that

The value of Gamma' evaluated at one...I cannot prove..haha - Oct 26th 2008, 06:34 AMgalactus
Wow, I had long forgotten about this. You done fine.

Here is a tidbit you may find nice:

The DiGamma function:

- Oct 26th 2008, 10:54 AMMathstud28
- Dec 28th 2008, 10:27 AMPaulRS
**Lemma 1.**Let be a function with continuous derivative in its domain.

Then where

__Proof__

By the Fundamental Theorem of Calculus:

So:

Now integrating by parts: and*Lemma 1*follows.

**Lemma 2.**We have:

__Proof__

First take in*Lemma 1:*We get

Thus: Now note that the integral in the RHS converges as - this in fact shows the existance of the gamma constant- then by definition of the gamma constant:

Now take with in*Lemma 1*:

Thus: and then

Therefore:

**Lemma 3.**:

*Proof:*now since we are done

**Proof :**

Let be the integral then letting we get

Now note that: ( )

Thus:

Then:

Applying*Lemma 2*: - Dec 29th 2008, 01:07 PMPaulRS
__Lemma__.

**Proof**By definition:

Note that: thus:

We get: Consider that: then

Exchange the summation order ( this can be justified working with the remainder, see here):

**Proof.**

Letting we have:

Now use the Power series expansion of :

So by Lemma 3of the previous post: And finally, by the Lemma we've just seen**:**