Calculate an improper integral and find an expression for computing it using the gamma function

[Like Tony Stark]

[MATH]I=\int_0^1 Ln(\frac{1}{x})^{a-1} dx [/MATH]
With [MATH]a=2[/MATH] and [MATH]a=3[/MATH]
I differentiated both sides (because it converges uniformly), so I got:

[MATH]I'=\int_0^1 \frac{-a(-Ln(x)^{a-2})+(-Ln(x)^{a-2})}{x} dx [/MATH]
With that, I used u-substitution ([MATH]u=lnx[/MATH]) to get:

[MATH]I'=\int_\infty ^0 -a(-u^{a-2})+(-u^{a-2}) du [/MATH]
This is easy to integrate. But once it is solved, what should I do?

 

[Prof]

I am not familiar with your method of solving it. I would start by rewriting the integrand in terms of ln(x) using exponent and logarithm properties.
Then use a change of variable to t where t = -ln x. The integral in terms of t will start to resemble the gamma function. See if you can go from there.