The closed form representations of Integrals of logarithm functions

Question

I wish to find a closed form representations of the following integral $$\int\limits_{0}^1\frac{\log^p(x)\log^r\left(\frac{1-x}{1+x}\right)}{x}dx=?$$ Here $p\ge 1$ and $r\ge 0$ 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.)?

Answer 1

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

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

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

$$\begin{aligned} \int_{0}^{1}x^{-1+\epsilon}(1-x)^{\delta}\,\mathrm{d}x &= \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{\epsilon^{m}}{m!}\frac{\delta^{n}}{n!}\int_{0}^{1}\frac{\ln^{m}x\ln^{n}(1-x)}{x}\,\mathrm{d}x \\ &= \frac{\Gamma(\epsilon)\Gamma(1+\delta)}{\Gamma(1+\epsilon+\delta)} = \frac{1}{\epsilon}\frac{\Gamma(1+\epsilon)\Gamma(1+\delta)}{\Gamma(1+\epsilon+\delta)}.\end{aligned}$$ Using the Taylor series of $\ln\Gamma(1+\epsilon)$, 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:

$$\begin{aligned} \frac{\Gamma(1+\epsilon)\Gamma(1+\delta)}{\Gamma(1+\epsilon+\delta)} &\approx \exp\left(\sum_{k=2}^{p+2}\frac{(-1)^{k}\zeta(k)}{k}\left(\epsilon^{k} + \delta^{k} - (\epsilon+\delta)^{k}\right)\right) \\ &\approx 1 + \frac{\zeta(2)}{2}\left(\epsilon^{2} + \delta^{2} - (\epsilon + \delta)^{2}\right) + \cdots \\ &\,\quad+ \frac{(-1)^{p+2}\zeta(p+2)}{p+2}\left(\epsilon^{p+2} + \delta^{p+2} - (\epsilon+\delta)^{p+2}\right).\end{aligned}$$ 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

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

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

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

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

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

Answer 2

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 $$I_p = (-1)^{p + 1} p! \left (1 - \frac{1}{2^{p + 2}} \right ) \zeta (p + 2).$$