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If I may, I have a rather challenging integral. I am not so sure there is a closed form.

$$\int_0^1 \left(\frac{\ln x}{1-x}\right)^n \; dx$$

I have evaluated when $n=1$ and $2$. But, when $n=3, 4, 5,\ldots$, the solution involves varying zeta functions.

For instance, If $n=3$, we get $$\begin{align*} &-3\zeta(3)-\frac{\pi^2}{2}\\ n=4: &\;\ 12\zeta(3)+\frac{2\pi^2}{3}+\frac{4\pi^4}{45}\\ n=5: &\;\ -30\zeta(5)-30\zeta(3)-\frac{11\pi^4}{18}-\frac{5\pi^2}{6} \end{align*}$$

I have evaluated $$\int_0^1 \frac{\ln^n(x)}{1-x} \; dx=(-1)^n n!\zeta(n+1),$$ but that $n$ power in the denominator of the problem at hand changes things drastically.

Do you think it is possible to find a closed form? As I said, it may not have one but it would interesting to see some clever methods incorporating the zeta function.

By the way, Maple gave me those values for $n=3,4,5$.

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up vote 13 down vote accepted

In this answer I show that $$ \int_0^1\left(\frac{\log(t)}{1-t}\right)^n\mathrm{d}t=(-1)^nn\sum_{j=0}^{n-1}\genfrac{[}{]}{0}{0}{n-1}{j}\zeta(n-j+1) $$ where $\genfrac{[}{]}{0}{0}{n}{k}$ is a Stirling number of the first kind.

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Thanks much, RobJohn. Clever. I will research Stirling numbers – Cody Mar 6 '12 at 18:06
@Cody: The important fact for this problem is the definition: $$\sum_{k=0}^n\genfrac{[}{]}{0}{0}{n}{k}x^k=x(x+1)(x+2)\dots(x+n-1)$$ – robjohn Mar 6 '12 at 19:36
Thank you. Yes, I found that out. I even checked it with values of n and it is cool how it works. I have to admit, I do not understand every line of your proof. Notably, the 4th to 5th line where you change from combinatoric-type argument to the Stirlings. I will have to study more about it. – Cody Mar 7 '12 at 21:57
I think I may see. Did you use the relation to the binomial: $$(1-t)^{x}=\sum_{n=0}^{\infty}\sum_{k=1}^{n}S(n,k)x^{k}\frac{t^{n}}{n!}$$. I used S(n,k) instead of the brackets to indicate Stirling numbers of the 1st Kind. – Cody Mar 7 '12 at 22:44
@Cody: the only identity about the Stirling numbers I used is the one I quoted in the comment above, which can be rewritten as $$\frac{1}{n!}\sum_{k=0}^n\genfrac{[}{]}{0}{0}{n}{k}x^k=(-1)^n\binom{-x}{n}\tag{‌​1}$$ Using $(1)$ yields $$\sum_{n=0}^\infty\frac{t^n}{n!}\sum_{k=0}^n\genfrac{[}{]}{0}{0}{n}{k} x^k=\sum_{n=0}^\infty(-t)^n\binom{-x}{n}=(1-t)^{-x}$$ I think your exponent is missing a minus sign. – robjohn Mar 7 '12 at 23:56

I evaluated the integral you mentioned above and arrived at $$\int_{0}^{1}t^{m}ln^{n}(t)dt=(-1)^{n}\frac{n!}{(m+1)^{n+1}}$$.
May I ask how this is used to evaluate the problem at hand?. I tried using $(1-t)^{m}$ instead of $t^{n}$ letting $m=-n$. But it turned nasty when I used parts as I done with the first part. I am certain there is something I am not seeing. I did manage to get the $(-1)^{n}\cdot n$ portion as in RobJohn's solution, but not the Stirling/zeta portion

Here's how I managed the integral you mentioned. I just do not know how to relate it to the one I posted.


Use parts and get:


Using the limits gives:


Change n to n-1:


Sub this into [1]:


Now, continue repeating and generally we have:



Now, how can I relate to the Stirling and zeta to arrive at the closed form RobJohn showed?.

I tried the same sort of method and let $m=-n$. This kind of threw a wrench in the whole mess. I managed to see the $(-1)^{n}\cdot n$, but not the stirling/zeta portion.

Using parts: $$dv=(1-t)^{-n}dt, \;\ u=ln^{n}(t)dt, \;\ du=\frac{nln^{n-1}(t)}{t}dt$$


This leads to $$-\frac{n}{n-1}\int_{0}^{1}\frac{ln^{n-1}(t)}{t(1-t)^{n-1}}dt$$

That extra t in the denominator may be a culprit. Otherwise, it is $I_{n-1}$.

I could repeat as before, but sorry to say I got hung up.

Thanks for your input and help.

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Well, you can try $$ A_{n,m} = \int_0^1 (\log x)^n t^m\,dt $$ which has a nice closed form, then expand yours in terms of that.

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Thanks very much. I will give it a try. – Cody Mar 6 '12 at 18:05

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