# Generalise logarithmic integral $\int_0^1\ln^n (\frac{1-x}{1+x})\mathrm dx$ related to Zeta function

Context:

I've been recently putting integrals of some random functions in online calculators for fun. An interesting kind of logarithmic integrals that I observed are as follows :

\begin{align} &\int_0^1\ln^2\left(\frac{1-x}{1+x}\right)\mathrm dx= \frac{\pi^2}{3}=2\zeta(2)\\ &\int_0^1\ln^3\left(\frac{1-x}{1+x}\right)\mathrm dx=-9\zeta(3)\\ &\int_0^1\ln^4\left(\frac{1-x}{1+x}\right)\mathrm dx = \frac{7\pi^4}{15}=42\zeta(4)\\ &\int_0^1\ln^5\left(\frac{1-x}{1+x}\right)\mathrm dx=-225\zeta(5)\\ &\int_0^1\ln^6\left(\frac{1-x}{1+x}\right)\mathrm dx=\frac{31\pi^6}{21}= 1395\zeta(6) \end{align}

Question:

How can we generalise this for :

$$I_n = \int_0^1\ln^n\left(\frac{1-x}{1+x}\right)\mathrm dx$$

Notation: The superscript at logarithm means for exponentiation.

• Also, why did you write $\frac{4185\, \zeta(6)}{3}$ instead of $1395\, \zeta(6)$? Jul 24 at 15:05
• @StevenClark Sorry, I thought it wouldn't be divisible. Jul 24 at 15:29

\begin{align} &\int_0^1\ln^n\left(\frac{1-x}{1+x}\right)\overset{x\to \frac{1-x}{1+x} }{dx}\\ =&\ 2\int_0^1\frac{\ln^n x}{(1+x)^2}dx= 2\int_0^1\ln^n x \ d\left(\frac{x}{1+x}\right)\\ \overset{ibp}=& -2n\int_0^1\frac{\ln^{n-1} x}{1+x}dx =-2n\left(\int_0^1 \frac{\ln^{n-1} x}{1-x} {d}x-\int_0^1 \frac{2x\ln^{n-1} x}{1-x^2} \overset{x^2\to x}{dx}\right)\\ =&-2n \left(1- \frac1{2^{n-1}}\right) \int_0^1 \frac{\ln^{n-1} x}{1-x} {d}x = 2n!(-1)^{n}\left(1-\frac1{2^{n-1}}\right) \zeta(n) \end{align}
• Amazing how changing the denominator from $1+x$, to $1-x$ and $1-x^2$ works out so nicely. Great solution, thank you for the added steps! Jul 24 at 14:37
• @An_Elephant - The substitution $x\to \frac{1-x}{1+x}$ is well known for integrals over $(0,1)$. The 1st term of IBP $\frac{x\ln^n x}{1+x}$ vanishes at both $0$ and $1$. Jul 24 at 16:14