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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?

Addendum. Following i707107's advice, I obtained the following formula

\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.

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When I clicked the question I said: surely sos440 will answer this... – Pedro Tamaroff Mar 25 '13 at 15:28
@PeterTamaroff, I also wished I could have answered it. :) – Sangchul Lee Mar 25 '13 at 15:37
I know for $m=1$ it's an integral representation of Euler's constant. But how do you evaluate the integral for $m=2$? – Random Variable Mar 25 '13 at 22:10
@RandomVariable, You can refer to my blog posting for the case $m=2$. Indeed, the first method is what I began with for the case $m\geq 3$. – Sangchul Lee Mar 25 '13 at 23:57
Could you prove positivity of $q$ as in your guess? It seems hard to prove it because $\zeta$ values alternates sign. – i707107 Mar 28 '13 at 15:31
up vote 5 down vote accepted

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 $m\geq 3$.

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I just grasped what you are trying to say. It seems illuminating! Thank you. I will try it. – Sangchul Lee Mar 27 '13 at 1:48
Thanks! I obtained an analytic formula for this integral, which coincided with my previous calculation for other values of $m$ as well. Now it remains to identify the limit of the integral as $s \to 1$. – Sangchul Lee Mar 27 '13 at 2:52

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