β-function or Γ-function? $C\left( p,q\right) =\int _{0}^{1}x^{p}\left( \log x\right) ^{q}dx$ I want to solve this calculation.
$$C\left( p,q\right) =\int _{0}^{1}x^{p}\left( \log x\right) ^{q}dx$$
I thought of two different approaches.

The first one is way how I use Γ-function.
Put $y=\log x$,
$\begin{aligned}x=e^{y} ,dx=e^{q}dy\end{aligned}$
Then
$$C\left( p,q\right) =\int _{\infty }^{0}e^{(p+1)y}y^{q}dy$$
Γ-function is $\Gamma \left( s\right) =\int ^{\infty }_{0}e^{-x}x^{s-1}dx$.
But I can't make this form.

The second one is β-function.
$$\int _{0}^{1}x^{p}\left( \log x\right) ^{q}dx=\int _{0}^{1}x\log x\cdot x^{p-1}\left( \log x\right) ^{p-1}dx$$
β-function is $\beta \left( p,q\right) =\int ^{1}_{0}x^{p-1}\left( 1-x\right) ^{q-1}dx$.
But I can't make $(1-x)$ and erase $x\log x$

I tried to make it as official as possible, but I didn't understand. Please tell me what to do with this continuation.
 A: Let $x=e^y,  then $ $$C\left( p,q\right) =\int _{0}^{1}x^{p}\left( \log x\right) ^{q}dx= \int_{-\infty}^0e^{y(p+1)} \; y^qdy$$
Now, let $t=-(p+1)y$, then  $$C\left( p,q\right)=-\int_{+\infty}^{0} \dfrac{e^{-t}t^q}{(-(p+1))^q(p+1)}dt=\dfrac{(-1)^q}{(p+1)^{q+1}}\int_0^{+\infty} t^qe^{-t}dt =\dfrac{(-1)^q}{(p+1)^{q+1}}\Gamma(q+1) $$
N.B: you should have $p+1>0$ and $q+1>0$.
A: More elementary way
Via integration by part we observe that $$C(p,q) = \int_0^1 x^p\ln^q dx=\underbrace{\frac{x^{p+1}}{p+1}\ln^q }_{=0}-\frac{q}{p+1}\int_0^1 x^p\ln^{q-1} dx=-\frac{q}{p+1}C(p,q-1)$$ Now if we replace $q$ by $q-1$ in the last integral we have $$C(p,q-1)=-\frac{q-1}{p+1}C(p,q-2)$$ and replacing  by $, q-k$ for $q$ for all $0\leq k\leq q$ we obtained $\displaystyle  C(p,q-k)=-\frac{q-k+1}{p+1}C(p,q-k)$ Now since we need to obtained the general integral relation we shall use the idea of telescoping product that ie $$\frac{C(p,q)}{C(p,q-1)} \cdot  \frac{C(p,q-1)}{C(p,q-2)}\cdot  \frac{C(p,q-2)}{C(p,q-3)} \cdots \frac{C(p,1)}{C(p,q-q)}=(-1)^q\frac{q!}{(p+1)^q} C(p,q-q)=(-1)^q\frac{q!}{(p+1)^q}\int_0^1 x^pdx=(-1)^q\frac{q!}{(p+1)^{q+1}}$$
Therefore,  for all $ p,q\geq 0$ our required integral

$$\int_0^1 x^p \ln^q dx =(-1)^q \frac{q!}{(p+1)^{q+1}}=(-1)^q\frac{\Gamma(q+1)}{(p+1)^{q+1}}$$

