Closed-form for rational log integral: $\int_0^1\left(\frac{\ln x}{1-x}\right)^{n}dx$

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.