How to analytically evaluate $\int_{0}^{\infty}\frac{x^r(-\log(1-\frac{x}{c}))^n}{c-x}dx$ How do you analytically evaluate $$\int_{0}^{\infty}\frac{x^r(-\log(1-\frac{x}{c}))^n}{c-x}dx$$ where $c,r$ are real constants, and $n$ is an integer. I've tried running it on Wolfram-alpha but the time limit exceeds. Is there a way to analytically approach it? How do I even argue that it exists?!
 A: First note the following:


*

*This integral does not exist if $r\geq 0$, since the decay at infinity is insufficient. 

*Similarly, $r>-1$ because otherwise the singularity in 0 is too large.

*Substituting $y=x/c$, you may assume that $c=1$ or $c=-1$. For $c=1$ the integral does not exist since $\frac{1}{x-1}\log(x-1)^n = \frac{d}{dx} \frac{\log(x-1)^{n+1}}{n+1}$ diverges as x tends to 1.


So instead of the integral you are probably searching for an anti-derivative. By the above considerations it suffices to consider (with $\pm 1$) 
$\int x^r \frac{\log(x+1)^n}{x+1} dx$
Integrating the second factor by parts this leads to 
$\int x^{r-1} \log(x+1)^n dx$
If you now additionally assume that $r$ is an integer you can substitute $x+1$ and reduce to considering 
$\int x^m \log(x)^n dx$   
Here, you substitute again by $x=e^y$ and obtain:
$\log ^{n+1}(x) (-(m+1) \log (x))^{-n-1} (-\Gamma (n+1,-(m+1) \log (x)))$  
TLDR: You want more assumptions and should use integration by parts and substitution.
A: Here is some notes. With substitution $x=c\dfrac{t}{t-1}$ we obtain 
$$I=\int_{0}^{\infty}\frac{x^r(-\ln(1-\frac{x}{c}))^n}{c-x}dx=\dfrac{c^r}{(-1)^{r+1}}\int_0^1t^r(1-t)^{-r-1}\ln^n(1-t)\,dt$$
and with substitution $1-t=e^{-u}$
\begin{align}
I
&= c^r(-1)^{n+r+1}\int_0^\infty(e^{u}-1)^ru^n\,du \\
&= c^r(-1)^{n+r+1}\int_0^\infty(1-e^{-u})^re^{ur}u^n\,du \\
&= c^r(-1)^{n+r+1}\int_0^\infty e^{ur}u^n\,du\sum_{k=0}^\infty\dfrac{\Gamma(-r+k)}{\Gamma(-r)}\dfrac{(e^{-u})^k}{k!} \\
&= c^r(-1)^{n+r+1}\sum_{k=0}^\infty\dfrac{\Gamma(-r+k)}{\Gamma(-r)k!}\int_0^\infty e^{-u(k-r)}u^n\,du 
\end{align}
for $r<0$ (only) we see
\begin{align}
I
&= c^r(-1)^{n+r+1}\sum_{k=0}^\infty\dfrac{\Gamma(k-r)}{\Gamma(-r)k!(k-r)^{n-1}}\Gamma(n+1) \\
&= c^r(-1)^{n+r+1}\dfrac{\Gamma(n+1)}{\Gamma(-r)}\sum_{k=0}^\infty\dfrac{\Gamma(k-r)}{k!(k-r)^{n-1}}
\end{align}
which is a closed form of integral.
