Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$?

Question

For $|x|<1,$ we have $$ \begin{aligned} & \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1} \end{aligned} $$

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$$ \begin{aligned} \int_0^1 \frac{\ln (1-x)}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k dx \\ & =-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2} \\ & =- \zeta(2) \\ & =-\frac{\pi^2}{6} \end{aligned} $$

$$ \begin{aligned} \int_0^1 \frac{\ln (1-x) \ln x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln xdx \\ & =\sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{(k+1)^2} \\ & =\zeta(3) \\ \end{aligned} $$

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$$ \begin{aligned} \int_0^1 \frac{\ln (1-x) \ln ^2 x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^2 xdx \\ & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{2}{(k+1)^3} \\ & =-2 \zeta(4) \\ & =-\frac{\pi^4}{45} \end{aligned} $$

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In a similar way, I dare guess that

$$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x =(-1)^{n+1}\Gamma(n)\zeta(n+2),$$

where $n$ is a non-negative real number.

Proof: $$ \begin{aligned} \int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^n xdx \\ \end{aligned} $$ Letting $y=-(k+1)\ln x $ transforms the last integral into a Gamma function as

$$ \begin{aligned} \int_0^1 x^k \ln ^n x d x & =\int_{\infty}^0 e^{-\frac{k}{k+1}}\left(-\frac{y}{k+1}\right)^n\left(-\frac{1}{k+1} e^{-\frac{y}{k+1}} d y\right) \\ & =\frac{(-1)^n}{(k+1)^{n+1}} \int_0^{\infty} e^{-y} y^n d y \\ & =\frac{(-1)^n \Gamma(n+1)}{(k+1)^{n+1}} \end{aligned} $$

Now we can conclude that $$ \begin{aligned} \int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =(-1)^{n+1} \Gamma(n+1) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+2}} \\ & =(-1)^{n+1} \Gamma(n+1)\zeta(n+2) \end{aligned} $$

Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln (1-x)$?

Your comments and alternative methods are highly appreciated?

Answer 1

For natural number $n$, I can use differentiation of the integral $$ \begin{aligned} I(a) & =\int_0^1 x^a \ln (1-x) d x \\ & =-\int_0^1 x^a \sum_{k=0}^{\infty} \frac{1}{k+1} x^{k+1} d x \\ & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^{a+k+1} d x \\ & =-\sum_{k=0}^{\infty}\left(\frac{1}{k+1} \cdot \frac{1}{a+k+2}\right) \end{aligned} $$ Differentiating $I(a)$ w.r.t. $a$ by $n$ times at $a=-1$ yields our integral.

$$ \begin{aligned} I & =-\left.\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{(-1)^n n !}{(a+k+2)^{n+1}}\right|_{a=-1} \\ & =(-1)^{n+1} n ! \sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{1}{(k+1)^{n+1}} \\ & =(-1)^{n+1} n ! \zeta(n+2) \end{aligned} $$

Answer 2

Starting with the integral $$ \int_{0}^{1} (1-x)^m \, x^{-t} \, dx = B(m+1, 1-t), $$ a case of the Beta function, then \begin{align} \partial_{m} \int_{0}^{1} (1-x)^m \, x^{-t} \, dx &= \partial_{m} \, B(m+1, 1-t) \\ \int_{0}^{1} \ln(1-x) \, (1-x)^m \, x^{-t} \, dx &= B(m+1, 1-t) \, \left(\psi(m+1) - \psi(m-t+2) \right), \end{align} where $\psi(x)$ is the digamma function. Setting $m=0$ leads to $$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \frac{\psi(1) - \psi(1-t)}{1-t}. $$
Now using $$ \psi(x+1) = - \gamma + \sum_{k=1}^{\infty} (-1)^{k+1} \, \zeta(k+1) \, x^{k} $$ then $$ \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx = \sum_{k=1}^{\infty} (-1)^k \, \zeta(k+1) \, (1-t)^{k-1}. $$ Using the operator $(-1)^n \, \partial_{t}^{n}$ on both sides of this last expression the following is obtained. \begin{align} I_{n} &= \int_{0}^{1} \ln(1-x) \, \ln^{n}(x) \, x^{-t} \, dx \\ &= \sum_{k=1}^{\infty} (-1)^{k+n} \, \zeta(k+1) \, \partial_{t}^{n} \, (1-t)^{k-1} \\ &= \sum_{k=1}^{\infty} (-1)^{k} \, \zeta(k+1) \, \frac{(k-1)!}{(k-n-1)!} \, (1-t)^{k-n-1} \\ &= \sum_{k=n}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n} \\ &= (-1)^{n+1} \, \zeta(n+2) \, n! + \sum_{k=n+1}^{\infty} (-1)^{k+1} \, \zeta(k+2) \, \frac{k!}{(k-n)!} \, (1-t)^{k-n}. \end{align} Setting $t = 1$ gives the desired result $$ \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx = (-1)^{n+1} \, n! \, \zeta(n+2). $$

or \begin{align} (-1)^{n+1} \, n! \, \zeta(n+2) &= \int_{0}^{1} \frac{\ln(1-x) \, \ln^{n}(x)}{x} \, dx \\ &= (-1)^n \, \partial_{t}^{n} \, \left. \int_{0}^{1} \ln(1-x) \, x^{-t} \, dx \right|_{t=1} \\ &= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. \int_{0}^{1} (1-x)^{m} \, x^{-t} \, dx \right|_{t=1}^{m=0} \\ &= (-1)^n \, \partial_{t}^{n} \, \partial_{m} \, \left. B(m+1, 1-t) \right|_{t=1}^{m=0}. \end{align}

Answer 3

Without expanding using series at all (taking for granted any properties of special functions whose derivations require series manipulation):

$$\begin{align*} I &= \int_0^1 \frac{\log(1-x) \log^n(x)}x \, dx \\[1ex] &= \int_0^1 \frac{\log(x) \log^n(1-x)}{1-x} \, dx \tag{1} \\[1ex] &= (-1)^n \int_0^\infty x^n \log\left(1-e^{-x}\right) \, dx \tag{2} \\[1ex] &= (-1)^{n+1} n \int_0^\infty x^{n-1} \operatorname{Li}_2(e^{-x}) \, dx \tag{3} \\[1ex] &= (-1)^{n+1} n (n-1) \int_0^\infty x^{n-2} \operatorname{Li}_3(e^{-x}) \, dx = \cdots \tag{4} \\ &\;\vdots \\ &= (-1)^{n+1} n! \int_0^\infty \operatorname{Li}_{n+1}(e^{-x}) \, dx \\[1ex] &= (-1)^{n+1} n! \operatorname{Li}_{n+2}(1) \\[1ex] &= \boxed{(-1)^{n+1} \Gamma(n+1) \zeta(n+2)} \tag{5} \end{align*}$$

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• $(1)$ : substitute $x\mapsto1-x$
• $(2)$ : substitute $x\mapsto 1-e^{-x}$
• $(3)$ : integrate by parts, recalling $\displaystyle \frac d{dx}\operatorname{Li}_2(x) = -\frac{\log(1-x)}x$ where $\operatorname{Li}_2$ is the dilogarithm
• $(4)$ : integrate by parts ad nauseam, using the recurrence $\displaystyle\frac d{dx}\operatorname{Li}_n(x)=\frac{\operatorname{Li}_{n-1}(x)}x$
• $(5)$ : $n!=\Gamma(n+1)$ and $\operatorname{Li}_n(1)=\zeta(n)$