How do I change one of my limit $-\infty$ to $\infty$ to evaluate this Integral? I'm trying to  evaluate:
$$\Large{\int_0^1 x^m \ln^n (x)\mathrm{dx}}$$
$$y=-\ln x \implies x=e^{-y} $$
$$dy/dx=1/x \implies dx=x dy= e^{-y} dy$$
$$ - \int_{-\infty}^{0} e^{-(m+1)y} (-1)^n y^n \mathrm{dy}$$
$$u=(m+1)y \implies du=dy$$
I Know how to solve this, If only I could changed the lower limit of $-\infty$ to $\infty$, how do I change that?
I hope you see I can't do the obvious $s=-u$ because I'm trying to use the Gamma function, so please don't suggest that or $u=-(m+1)y$
 A: Hint: Substitute $x=e^{-u}$:
$$
\begin{align}
\int_0^1x^m\log(x)^n\,\mathrm{d}x
&=(-1)^{n+1}\int_\infty^0u^ne^{-(m+1)u}\,\mathrm{d}u\\
&=(-1)^n\int_0^\infty u^ne^{-(m+1)u}\,\mathrm{d}u
\end{align}
$$
A: For a variety :
$$\begin{align}
\int_0^1 x^m \ln^n (x)\mathrm{dx} &=\int_0^1 \frac{\partial^n}{\partial m^n}x^m\mathrm{dx} \\~\\
&= \frac{\partial^n}{\partial m^n}\int_0^1x^m\mathrm{dx} \\~\\
&= \frac{\partial^n}{\partial m^n}\dfrac{1}{m+1} \\~\\
&= \dfrac{(-1)^nn!}{(m+1)^{n+1}} \\~\\
  \end{align}$$

This is actually a classic problem! For using Gamma function, $x = e^{\frac{-u}{m+1}}$ substitution works smoothly. See Sophomore's_dream
A: I just want to mention that ususal integration by parts also works, and you don't need anything fancy like $\Gamma$-Functions or Feynman's trick. Let's see:
$$
I(m,n)=\int_0^1x^m\ln^n(x)=\underbrace{\frac{n}{m+1}x^{m+1}\ln^n(x)\big|_0^1}_{=0}+\frac{n}{m+1}\int_0^1x^m\ln^{n-1}(x)
$$
Integrating again by parts, we see what happens
$$
I(m,n)=\underbrace{\frac{n}{(m+1)^2}x^{m+1}\ln^{n-1}(x)\big|_0^1}_{=0}-(-)^2\frac{n (n-1)}{(m+1)^2}\int_0^1x^m\ln^{n-2}(x)
$$
It's now obvious that this process repeats until the power of the logatrithm is brought down to one because than its derivative is just $1/x$, and the resulting integral will not vanish at the endpoints.  We get
$$
I(m,n)=\underbrace{\frac{(n-1)!}{(m+1)^{n}}x^{m+1}\ln(x)\big|_0^1}_{=0}-(-)^{n+1}\frac{n!}{(m+1)^n}\int_0^1x^m=(-1)^n\frac{n!}{(m+1)^{n+1}}
$$
