Xrfgtjtk

I wish to find a closed form representations of the following integral
$$intlimits_{0}^1frac{log^p(x)log^rleft(frac{1-x}{1+x}right)}{x}dx=?$$
Here $pge 1$ and $rge 0$ are nonnegative integers. It can be expressed in terms of a linear combination of well known constants (such as: Riemann zeta values,$pi$ et. al.)?

I've asked a similar question in the recent past. If $r = 0$, it seems like the integral does not converge. If $r = 1$, then yes, one may express the integral in terms of $pi$ and odd values of $zeta(s)$. If $r > 1$, then the problem is much harder, and most likely requires values of the polylogarithm, e.g. $operatorname{Li}_{4}(1/2)$. See, for example, 1 and 2 for the difficulties encountered.

I give a partial answer to your question; I focus on the case $r = 1$. The integral becomes

$$ I = int_{0}^{1}frac{ln^{p}xln(1-x)}{x},mathrm{d}x - int_{0}^{1}frac{ln^{p}xln(1+x)}{x},mathrm{d}x = I_{1} - I_{2}.$$

For any fixed $p$, $I_{1}$ can be evaluated using the beta function. We have

$$begin{aligned} int_{0}^{1}x^{-1+epsilon}(1-x)^{delta},mathrm{d}x &= sum_{m=0}^{infty}sum_{n=0}^{infty}frac{epsilon^{m}}{m!}frac{delta^{n}}{n!}int_{0}^{1}frac{ln^{m}xln^{n}(1-x)}{x},mathrm{d}x \
&= frac{Gamma(epsilon)Gamma(1+delta)}{Gamma(1+epsilon+delta)} = frac{1}{epsilon}frac{Gamma(1+epsilon)Gamma(1+delta)}{Gamma(1+epsilon+delta)}.end{aligned}$$
Using the Taylor series of $lnGamma(1+epsilon)$, we need the coefficient of $epsilon^{p+1}delta$ of the ratio of gamma functions. The form of the Taylor expansion is such that we only need to expand the exponential to first order, because the power in $delta$ is $1$. Furthermore, we'll need to keep terms up to $p+2$ because only the cross terms contribute:

$$begin{aligned} frac{Gamma(1+epsilon)Gamma(1+delta)}{Gamma(1+epsilon+delta)} &approx expleft(sum_{k=2}^{p+2}frac{(-1)^{k}zeta(k)}{k}left(epsilon^{k} + delta^{k} - (epsilon+delta)^{k}right)right) \
&approx 1 + frac{zeta(2)}{2}left(epsilon^{2} + delta^{2} - (epsilon + delta)^{2}right) + cdots \ &,quad+ frac{(-1)^{p+2}zeta(p+2)}{p+2}left(epsilon^{p+2} + delta^{p+2} - (epsilon+delta)^{p+2}right).end{aligned}$$
The binomial coefficient is $p+2$, so we have that the entire coefficient is $(-1)^{p+1},zeta(p+2)$. Therefore $I_{1} = (-1)^{p+1},p!,zeta(p+2)$.

In my link, I give a way to evaluate $I_{2}$ with any allowed $p$. It starts with the integral

$$begin{aligned} int_{0}^{1}x^{-1+epsilon}(1+x)^{delta},mathrm{d}x &= sum_{m=0}^{infty}sum_{n=0}^{infty}frac{epsilon^{m}}{m!}frac{delta^{n}}{n!}int_{0}^{1}frac{ln^{m}xln^{n}(1+x)}{x},mathrm{d}x \ &= frac{1}{epsilon}{}_{2}F_{1}(-delta, epsilon; 1+epsilon; -1) = sum_{k=0}^{infty}frac{(-1)^{k}}{k+epsilon}frac{(-delta)_{k}}{k!}. end{aligned}$$

We need the $epsilon^{p}delta$ coefficient. The $epsilon^{p}$ is found using power series. To simplify the $(-delta)_{k}$, we separate the $k=0$ term with the rest of the series. This zeroth order term does not contribute to the integral. The rest of the sum is

$$begin{aligned} &,quad-sum_{k=1}^{infty}frac{(-1)^{k}delta}{k+epsilon}frac{(1-delta)(2-delta)cdots(k-1-delta)}{(1)(2)cdots(k-1)(k)} \ &= -sum_{k=1}^{infty}frac{(-1)^{k}delta}{k(k+epsilon)}(1-delta)left(1-frac{delta}{2}right)cdotsleft(1 - frac{delta}{k-1}right) end{aligned}$$

and because we just need $delta$, the product $prod_{j=1}^{k-1}(1-delta/j)$ does not contribute. The $epsilon^{p}delta$ coefficient is then

$$ (-1)^{p+1}sum_{k=1}^{infty}frac{(-1)^{k}}{k^{p+2}} = (-1)^{p},eta(p+2)$$

and the value of the integral is $I_{2} = (-1)^{p},p!,eta(p+2)$. The closed form of $I$ is therefore

$$begin{aligned} I &= p!,zeta(p+2)left((-1)^{p+1} - (-1)^{p}(1-2^{-p-1})right) \
&= boxed{2(-1)^{p}(2^{-p-2} - 1),p!,zeta(p+2).} end{aligned}$$

Here is an alternative approach for the case $r = 1, p in mathbb{N}$ that has already been given by @Ininterrompue.

Let
$$I_p = int_0^1 ln^p x ln left (frac{1 - x}{1 + x} right ) frac{dx}{x}.$$
Since
$$ln left (frac{1 - x}{1 + x} right ) = - 2 sum_{n = 0}^infty frac{x^{2n+1}}{2n + 1}, qquad |x| < 1,$$
the integral may be written as
$$I_p = -2 sum_{n = 0}^infty frac{1}{2n + 1} int_0^1 x^{2n} ln^p x , dx.$$
Integrating by parts $p$-times gives
begin{align}
I_p &= - 2 (-1)^p p! sum_{n = 0}^infty frac{1}{(2n + 1)^{p + 2}}\
&= -frac{(-1)^p p!}{2^{p + 1}} sum_{n = 0}^infty frac{1}{(n + 1/2)^{p + 2}}\
&= -frac{(-1)^p p!}{2^{p + 1}} cdot frac{(-1)^{p + 2}}{(p + 1)!} psi^{(p + 1)} left (frac{1}{2} right )\
&= -frac{1}{(p + 1) 2^{p + 2}} psi^{(p + 1)} left (frac{1}{2} right ). tag1
end{align}
Here $psi^{(m)} (x)$ is the polygamma function with the following series representation of
$$psi^{(m)} (z) = (-1)^{m + 1} m! sum_{n = 0}^infty frac{1}{(n + z)^{m + 1}},$$
having been used.

Finally, as (see here)
$$psi^{(m)} left (frac{1}{2} right ) = (-1)^{m + 1} m! (2^{m + 1} - 1) zeta (m + 1),$$
in terms of the Riemann zeta function, $zeta (z)$, one can rewrite (1) as
$$I_p = (-1)^{p + 1} p! left (1 - frac{1}{2^{p + 2}} right ) zeta (p + 2).$$