Show that $\int_1^x \log^kt\log^l\frac{x}{t} dt = xQ(\log x),$ where $Q(t)$ is a polynomial with leading term $l!t^k$ I need to show that $$\int_1^x  \log^kt\log^l\frac{x}{t} dt = xQ(\log x),$$ where $Q(t)$ is a polynomial with leading term $l!t^k$.
Here $k,l$ are nonnegative integers.
I wrote $\int_1^x  \log^kt\log^l\frac{x}{t} dt = \int_1^x  \log^kt(\log x -\log t)^l dt = \sum_{j=0}^{l} (-1)^j {l \choose j} \log^{l-j}x \int_1^x \log^{k+j}t dt$.
However I calculated that $\int_1^x \log^{k+j}t dt = xq_j(\log x)$ where $q_j(t)$ is a polynomial of degree $k+j$. Therefore $\log^{l-j}x \int_1^x \log^{k+j}t dt$ should be of the form $xR(\log x)$ where $R(t)$ is a polynomial of degree $k+l$.
So its seems like I am getting a polynomial of the wrong degree and I'm not sure where the $l!$ coefficient would come from either.
 A: In its current form the result is indeed not correct (for $k = l = 0$ the integral is $x-1$, which cannot be written as $x$ times a polynomial in $\log(x)$). You can fix this by either modifying the right-hand side or replacing the lower limit by $0$.
For $k,l \in \mathbb{N}_0$ and $x>0$ let
\begin{align}
I_{k,l} (x) &= \int \limits_1^x \log^k (t) \log^l \left(\frac{x}{t}\right) \, \mathrm{d} t \, , \\
J_{k,l} (x) &= \int \limits_0^x \log^k (t) \log^l \left(\frac{x}{t}\right) \, \mathrm{d} t \, .
\end{align}
Then
\begin{equation}
I_{k,l} (\mathrm{e}^y) \overset{t = \mathrm{e}^{y u}}{=} y^{k+l+1} \int \limits_0^1 u^k (1-u)^l \mathrm{e}^{y u} \, \mathrm{d} u = y^{k+l+1} (-1)^l \frac{\mathrm{d}^k}{\mathrm{d}y^k} \left[\mathrm{e}^y \frac{\mathrm{d}^l}{\mathrm{d}y^l} \int \limits_0^1 \mathrm{e}^{-y(1-u)} \, \mathrm{d} u\right] \, .
\end{equation}
Evaluating the integral and using the general Leibniz rule to compute the derivatives, we find
\begin{equation}
I_{k,l} (\mathrm{e}^y) = (-1)^k \left[\mathrm{e}^y \sum \limits_{m=0}^k \binom{k}{m} (-1)^m (l+k-m)! y^m - \sum \limits_{n=0}^l \binom{l}{n} (k+l-n)! y^n\right] \, ,
\end{equation}
so
\begin{equation}
I_{k,l} (x) = x P_{k,l} (\log(x)) + \tilde{P}_{k,l} (\log(x))
\end{equation}
with polynomials $P_{k,l}$ and $\tilde{P}_{k,l}$.
Similarly, we compute
\begin{equation}
J_{k,l}(\mathrm{e}^y) \overset{t = \mathrm{e}^{y-u}}{=} \mathrm{e}^y \int \limits_0^\infty (y-u)^k u^l \mathrm{e}^{-u} \, \mathrm{d} u = \mathrm{e}^y \sum \limits_{m=0}^k \binom{k}{m} (l+k-m)! (-1)^{k-m} y^m\, .
\end{equation}
Therefore,
\begin{equation}
J_{k,l} (x) = x Q_{k,l} (\log(x))
\end{equation}
with a polynomial $Q_{k,l}$, which has the desired leading term.
