# References to integrals of the form $\int_{0}^{1} \left( \frac{1}{\log x}+\frac{1}{1-x} \right)^{m} \, dx$

While extending my calculation techniques, with aid of Mathematica, I found that

\begin{align*}
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{3} \, dx
&= -6 \zeta ‘(-1) -\frac{19}{24}, \\
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{4} \, dx
&= -10 \zeta ‘(-2)-2 \zeta ‘(-1)-\frac{37}{72}, \\
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{5} \, dx
&= -\frac{35}{3} \zeta ‘(-3)-\frac{15}{2} \zeta ‘(-2)-\frac{5}{3} \zeta ‘(-1)-\frac{3167}{8640}, \\
\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{6} \, dx
&=-\frac{21}{2} \zeta ‘(-4)-14 \zeta ‘(-3)-\frac{31}{4} \zeta ‘(-2)-\frac{3}{2} \zeta ‘(-1)-\frac{1001}{3600}.
\end{align*}

I conjectures that these relations extends also to higher degrees:

My Guess. For $m \geq 3$, we can write
$$\int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{m} \, dx = – \Bigg( q + \sum_{k=1}^{m-2} q_k \zeta'(-k) \Bigg)$$
for some positive rational numbers $q$ and $q_k$.

Is there any reference regarding this problem?

\begin{align*}
& \int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{m} \, dx \\
&= -\frac{H_{m-1}}{(m-1)!} + \frac{1}{(m-1)!} \sum_{k=1}^{m-1} \left[{{m-1}\atop{k}}\right] \zeta(1-k) \\
&\quad – \frac{1}{(m-2)!}\sum_{j=1}^{m-1}\sum_{l=m-j}^{m-1} \binom{m}{j} \binom{m-2}{j-1} \left[{{j-1}\atop{l+j-m}}\right] \{ \zeta'(1-l) + H_{m-j-1} \zeta(1-l) \}
\end{align*}

valid for $m \geq 2$, where $\left[{{n}\atop{k}}\right]$ denotes the unsigned Stirling’s number of the first kind.

#### Solutions Collecting From Web of "References to integrals of the form $\int_{0}^{1} \left( \frac{1}{\log x}+\frac{1}{1-x} \right)^{m} \, dx$"

You can try
$$\int_0^1\left(\frac{1}{\log x}+\frac{1}{1-x}\right)^m (-\log x)^{s-1}dt$$
for $s$ with sufficiently large real part.

This will give you an expression involving $\Gamma$ and $\zeta$ functions.
Then use the analytic continuation to $\sigma>0$ and plug in $s=1$.

I checked this method for $m=2$ and got the right answer. I am sure that this will work in $m\geq 3$.