# Integrating a x^k ln(x) Function with Gamma Function

• clandarkfire
In summary, the conversation discusses how to solve the integral -\int^1_0 x^k\ln{x}\,dx using the gamma function. The hint is to rewrite the integral in terms of the gamma function. The conversation explores different substitutions and methods, ultimately leading to the solution using the substitution x=e^{-u/k}. The conversation also clarifies the limits of integration for this substitution.

## Homework Statement

"Show that $$- \int^1_0 x^k\ln{x}\,dx = \frac{1}{(k+1)^2} ; k > -1$$.

Hint: rewrite as a gamma function.

## Homework Equations

Well, I know that $$\Gamma \left( x \right) = \int\limits_0^\infty {t^{x - 1} e^{ - t} dt}$$.

## The Attempt at a Solution

I've tried various substitutions, beginning with u=k*ln(x), but I'm not getting very far. To write it as a gamma function, I'd have to change the limits from (0 to 1) to (0 to infinity) and I can't find a way to do that.

Can someone point me in the right direction?

I wonder why you should use the $\Gamma$ function for this elementary integral. Perhaps I'm missing something, but if I'm not wrong, it should be pretty easily solvable by integration by parts.

If you are forced to use the $\Gamma$ function, I'd indeed try the substitution
$$x=\exp (-u/k),$$
which leads to an integral you can evaluate immediately in terms of the $\Gamma$ function.

• 1 person
It does seem pretty straightforward with integration by parts, but since I'm told to use the gamma function, I'd at least like to know how to do that.
If I use the substitution $$x=e^{-u/k}$$, I get $$dx = -k*e^{-u/k}\,du$$ The integral then becomes $$\int^1_0 e^{-u}*-\frac{u}{k}*-k*e^{-u/k}\,du = \int^1_0 e^{-(u + \frac{u}{k})}u\,du$$,
but the problem remains that it's between 0 and 1, not 0 and infinity. How do I get to a gamma function from there?
Thanks!

clandarkfire said:
It does seem pretty straightforward with integration by parts, but since I'm told to use the gamma function, I'd at least like to know how to do that.
If I use the substitution $$x=e^{-u/k}$$, I get $$dx = -k*e^{-u/k}\,du$$

Isn't that a ##\frac {-1} k## instead of ##-k## out in front?

The integral then becomes $$\int^1_0 e^{-u}*-\frac{u}{k}*-k*e^{-u/k}\,du = \int^1_0 e^{-(u + \frac{u}{k})}u\,du$$,
but the problem remains that it's between 0 and 1, not 0 and infinity. How do I get to a gamma function from there?
Thanks!

If ##x=e^{-u/k}## and ##x## goes from ##0## to ##1##, how do you get ##u## going from ##0## to ##1##?

Ah! That clears it up. Thanks!