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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 $n=3,4,5$.

3 Answers 3

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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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    @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.2012-03-07
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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.

$\int_{0}^{1}t^{m}ln^{n}(t)dt$

Use parts and get:

$\frac{t^{m+1}}{m+1}ln^{n}(t)-\frac{n}{m+1}\int_{0}^{1}t^{m}ln^{n}(t)dt$

Using the limits gives:

$\frac{-n}{m+1}\int_{0}^{1}t^{m}ln^{n-1}(t)dt$........[1]

Change n to n-1:

$\int_{0}^{1}t^{m}ln^{n-1}(t)dt=-\frac{n-1}{m+1}\int_{0}^{1}t^{m}ln^{n-2}(t)dt$

Sub this into [1]:

$(-1)^{2}\frac{n(n-1)}{(m+1)^{2}}\int_{0}^{1}t^{m}ln^{n-2}(t)dt$

Now, continue repeating and generally we have:

$\int_{0}^{1}t^{m}ln^{n}(t)dt=(-1)^{n}\frac{n!}{(m+1)^{n}}\int_{0}^{1}x^{m}dx$

$=(-1)^{n}\frac{n!}{(m+1)^{n+1}}$

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

$v=\frac{1}{(1-t)^{n-1}(n-1)}$

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.2012-03-06