# Writing complicated integral in terms of the Gamma function

1. Apr 12, 2015

### davidbenari

1. The problem statement, all variables and given/known data
Write $\int_{0}^{1}x^2(ln\frac{1}{x})^3 dx$ in terms of the gamma function

2. Relevant equation
$\Gamma(p+1)=p\Gamma(p)$
3. The attempt at a solution
Say $x=e^{-u}$ one would eventually obtain the integral

$\int_{-\infty}^{0} u^3 e^{-u} du$

STEPS:
$x=e^{-u}$ $e^{u}=1/x$ $u=ln(1/x)$ $du=xdx$

$\int_{0}^{1}x^2(ln\frac{1}{x})^3 dx=\int_{0}^{-\infty} e^{-u} x u^3 \frac{du}{x}=\int_{-\infty}^{0} u^3 e^{-u} du$

Which wants to look like a gamma function but isnt because of the limits.

Last edited: Apr 12, 2015
2. Apr 12, 2015

### Staff: Mentor

I don't know what went wrong with the substitution (can you show the steps?), but the last integral diverges.

3. Apr 12, 2015

### davidbenari

4. Apr 12, 2015

### davidbenari

5. Apr 13, 2015

### Staff: Mentor

I don't think that is right.
And your integral limits look wrong, too.
x=1 corresponds to u=0, but x=0 corresponds to a different value for u.

6. Apr 13, 2015

### davidbenari

Yeah I've corrected it now. Thanks!