I have to evaluate $\int_0^1 x^2 \ln(\frac1x)^3 $.I used gamma function and used substitution $t=\ln (\frac {1}{x})^3$. In this I get to integrate from $1$ to $-\infty$ with a minus sign outside.Because of this minus sign by interchanging upper and lower limit I get to integrate from $+\infty$ to $1$. Since $x>0$ I took this integration from 0 to 1. Can I make that change of $+\infty$ => 0.
The answer I got is $\Gamma(2)=1$.Is this correct.Any help is appreciated
On
According to WolframAlpha, the value of the integral is about $0.0740741$, which is certainly not $1$, so you've got an error somewhere. Babak shows you how to use the gamma function but, since I've already typed it up, here's an outline of another approach.
Note that $$\frac{d }{d x}\log ^n\left(\frac{1}{x}\right)=-\frac{n}{x} \log ^{n-1}\left(\frac{1}{x}\right).$$
This allows you to derive a reduction formula using integration by parts. As a result, the indefinite integral is $$ \frac{2 x^3}{27}+\frac{1}{3} x^3 \log ^3\left(\frac{1}{x}\right)+\frac{1}{3} x^3 \log ^2\left(\frac{1}{x}\right)+\frac{2}{9} x^3 \log \left(\frac{1}{x}\right). $$
The limit of this as $x$ goes down to zero is zero and the limit as $x$ goes up to $1$ is $2/27$. The the value of the integral is $2/27\approx 0.070741$.
Let me solve it from the first point. If I got correctly you are working on $$-\int_0^1x^2\ln^3(x)dx$$ Letting $x=\text{e}^{-y}$, the integral becomes $$(-1)^{4}\int_0^{\infty}y^3\text{e}^{-3y}dy$$ Now if we set $3y=u$, the latter integral becomes $$\int_0^{\infty}\frac{u^3}{3^3}\text{e}^{-u}\frac{du}{3}=\frac{1}{3^4}\Gamma(4)$$