Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n=3,4,5$.

share|improve this question

3 Answers 3

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.

share|improve this answer
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

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.

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

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.