Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$? For $|x|<1,$ we have
$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$

$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k dx \\
& =-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2} \\
& =- \zeta(2) \\
& =-\frac{\pi^2}{6}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln xdx \\
& =\sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{(k+1)^2} \\
& =\zeta(3) \\
\end{aligned}
$$

$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^2 x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^2 xdx \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{2}{(k+1)^3} \\
& =-2 \zeta(4) \\
& =-\frac{\pi^4}{45}
\end{aligned}
$$

In a similar way, I dare guess that
$$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x  =(-1)^{n+1}\Gamma(n)\zeta(n+2),$$
where $n$ is a non-negative real number.
Proof:
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^n xdx \\
\end{aligned}
$$
Letting $y=-(k+1)\ln x $ transforms the last integral into a Gamma function as
$$
\begin{aligned}
\int_0^1 x^k \ln ^n x d x & =\int_{\infty}^0 e^{-\frac{k}{k+1}}\left(-\frac{y}{k+1}\right)^n\left(-\frac{1}{k+1} e^{-\frac{y}{k+1}} d y\right) \\
& =\frac{(-1)^n}{(k+1)^{n+1}} \int_0^{\infty} e^{-y} y^n d y \\
& =\frac{(-1)^n \Gamma(n+1)}{(k+1)^{n+1}}
\end{aligned}
$$
Now we can conclude that
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =(-1)^{n+1} \Gamma(n+1) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+2}} \\
& =(-1)^{n+1} \Gamma(n+1)\zeta(n+2)
\end{aligned}
$$
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln (1-x)$?
Your comments and alternative methods are highly appreciated?
 A: For natural number $n$, I can use differentiation of the integral
$$
\begin{aligned}
I(a) & =\int_0^1 x^a \ln (1-x) d x \\
& =-\int_0^1 x^a \sum_{k=0}^{\infty} \frac{1}{k+1}  x^{k+1} d x \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^{a+k+1} d x \\
& =-\sum_{k=0}^{\infty}\left(\frac{1}{k+1} \cdot \frac{1}{a+k+2}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ by $n$ times at $a=-1$ yields our integral.
$$
\begin{aligned}
I & =-\left.\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{(-1)^n n !}{(a+k+2)^{n+1}}\right|_{a=-1} \\
& =(-1)^{n+1} n ! \sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{1}{(k+1)^{n+1}} \\
& =(-1)^{n+1} n ! \zeta(n+2)
\end{aligned}
$$
A: Starting with the integral
$$ \int_{0}^{1} (1-x)^m \, x^{-t} \, dx = B(m+1, 1-t), $$
a case of the Beta function, then
\begin{align}
\partial_{m} \int_{0}^{1} (1-x)^m \, x^{-t} \, dx &= \partial_{m} \, B(m+1, 1-t) \\
\int_{0}^{1} \ln(1-x) \, (1-x)^m \, x^{-t} \, dx &= B(m+1, 1-t) \, \left(\psi(m+1) - \psi(m-t+2) \right),
\end{align}
where $\psi(x)$ is the digamma function. Setting $m=0$ leads to
$$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \frac{\psi(1) - \psi(1-t)}{1-t}. $$
Now using
$$ \psi(x+1) = - \gamma + \sum_{k=1}^{\infty} (-1)^{k+1} \, \zeta(k+1) \, x^{k} $$
then
$$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \sum_{k=1}^{\infty} (-1)^k \, \zeta(k+1) \, (1-t)^{k-1}. $$
Using the operator $(-1)^n \, \partial_{t}^{n}$ on both sides of this last expression the following is obtained.
\begin{align}
I_{n} &= \int_{0}^{1} \ln(1-x) \, \ln^{n}(x) \, x^{-t} \, dx \\
&= \sum_{k=1}^{\infty} (-1)^{k+n} \, \zeta(k+1) \, \partial_{t}^{n} \, (1-t)^{k-1} \\
&= \sum_{k=1}^{\infty} (-1)^{k} \, \zeta(k+1) \, \frac{(k-1)!}{(k-n-1)!} \,  (1-t)^{k-n-1} \\
&= \sum_{k=n}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n} \\
&= (-1)^{n+1} \, \zeta(n+2) \, n! + \sum_{k=n+1}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n}.
\end{align}
Setting $t = 1$ gives the desired result
$$ \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx = (-1)^{n+1} \, n! \, \zeta(n+2). $$
or
\begin{align}
(-1)^{n+1} \, n! \, \zeta(n+2) &= \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx \\
&= (-1)^n \, \partial_{t}^{n} \, \left. \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx \right|_{t=1} \\
&= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. \int_{0}^{1} (1-x)^{m} \, x^{-t} \, dx \right|_{t=1}^{m=0} \\
&= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. B(m+1, 1-t) \right|_{t=1}^{m=0}.
\end{align}
A: Without expanding using series at all (taking for granted any properties of special functions whose derivations require series manipulation):
$$\begin{align*}
I &= \int_0^1 \frac{\log(1-x) \log^n(x)}x \, dx \\[1ex]
&= \int_0^1 \frac{\log(x) \log^n(1-x)}{1-x} \, dx \tag{1} \\[1ex]
&= (-1)^n \int_0^\infty x^n \log\left(1-e^{-x}\right) \, dx \tag{2} \\[1ex]
&= (-1)^{n+1} n \int_0^\infty x^{n-1} \operatorname{Li}_2(e^{-x}) \, dx \tag{3} \\[1ex]
&= (-1)^{n+1} n (n-1) \int_0^\infty x^{n-2} \operatorname{Li}_3(e^{-x}) \, dx = \cdots \tag{4} \\
&\;\vdots \\
&= (-1)^{n+1} n! \int_0^\infty \operatorname{Li}_{n+1}(e^{-x}) \, dx \\[1ex]
&= (-1)^{n+1} n! \operatorname{Li}_{n+2}(1) \\[1ex]
&= \boxed{(-1)^{n+1} \Gamma(n+1) \zeta(n+2)} \tag{5}
\end{align*}$$


*

*$(1)$ : substitute $x\mapsto1-x$

*$(2)$ : substitute $x\mapsto 1-e^{-x}$

*$(3)$ : integrate by parts, recalling $\displaystyle \frac d{dx}\operatorname{Li}_2(x) = -\frac{\log(1-x)}x$ where $\operatorname{Li}_2$ is the dilogarithm

*$(4)$ : integrate by parts ad nauseam, using the recurrence $\displaystyle\frac d{dx}\operatorname{Li}_n(x)=\frac{\operatorname{Li}_{n-1}(x)}x$

*$(5)$ : $n!=\Gamma(n+1)$ and $\operatorname{Li}_n(1)=\zeta(n)$
