Prove that $\int\limits_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum\limits_{n=1}^\infty \frac{1}{(n+a)^2}$

Prove that $$\int_0^1 x^a(1-x)^{-1}\ln x \,dx = -\sum_{n=1}^\infty \frac{1}{(n+a)^2}$$

I know that we have a product of $x^a$, $\displaystyle\sum_{n=0}^\infty x^n$, and $\displaystyle\sum_{n=0}^\infty \frac{(1-x)^n}{n}$, but it hasn't helped me so far.

Any tips?

We are given that $a>-1$.

We first define the function $$ f(a,b)= \int_0^b \frac{x^a}{1-x}\,dx, \,\,\,b\in (0,1). $$ Using the fact that $$ \frac{1}{1-x}=\sum_{k\geq 0}x^k, $$ we obtain $$ f(a,b)= \int_0^b x^a\sum_{k\geq 0}x^k \,dx= \sum_{k\geq 0} \int_0^b x^{k+a}=\sum_{k\ge 0}\frac{b^{k+a+1}}{k+a+1}=\sum_{k\ge 1}\frac{b^{k+a}}{k+a}, $$ and thus $$ \frac{\partial}{\partial a}f(a,b)= \int_0^b \frac{x^a}{1-x}\ln x\,\,dx =-\sum_{k\ge 1}\frac{b^{k+a}}{(k+a)^2}. $$ We are allowed to differentiate the series and interchange $\partial/\partial a$ with the summation, since the differentiated series converges locally uniformly in $a$, when $a\in (0,\infty)$.

Now we are allowed to take the limits as $b\searrow 1$, and obtain the identity.

Justification of $ \,\,\,\displaystyle \lim_{b\searrow 1}\sum_{k=1}^\infty\frac{b^{k+a}}{(k+a)^2}=\sum_{k=1}^\infty\frac{1}{(k+a)^2}. $

It suffices to show that, for every $\varepsilon>0$, there exists a $b_0\in(0,1)$, such that $b\in (b_0,1)$ implies that $$ \sum_{k=1}^\infty\frac{1-b^{k+a}}{(k+a)^2}<\varepsilon. $$ Since $\,\,\displaystyle\sum_{k=0}^\infty\frac{1}{(k+a)^2}\,\,$ converges, there exists an $n_0$, such that $$\sum_{k=n_0+1}^\infty\frac{1}{(k+a)^2}<\frac{\varepsilon}{2}.$$ Clearly the function $$ f_{n_0}(b)=\sum_{k=1}^{n_0}\frac{b^{k+a}}{(k+a)^2}, $$ is continuous in $b\in\mathbb R$, as it is a polynomial, and hence there is a $b_0\in(0,1)$, such that $$ b\in (b_0,1) \quad\Longrightarrow\quad \frac{\varepsilon}{2}>\lvert f_{n_0}(b)-f_{n_0}(1)\rvert=\sum_{k=1}^{n_0}\frac{1-b^{k+a}}{(k+a)^2} $$ Altogether, if $b\in(b_0,1)$, then $$ \sum_{k=1}^\infty\frac{1-b^{k+a}}{(k+a)^2}=\sum_{k=1}^{n_0}\frac{1-b^{k+a}}{(k+a)^2}+\sum_{k=n_0+1}^\infty\frac{1-b^{k+a}}{(k+a)^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$

$$ \begin{align} &\int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x\\ &=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x\\ &=-\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\left[\left(1-x^a\right)+\left(x-x^{a+1}\right)+\left(x^2-x^{a+2}\right)+\dots\right]\,\mathrm{d}x\\ &=-\frac{\mathrm{d}}{\mathrm{d}a}\left[\left(1-\frac1{a+1}\right)+\left(\frac12-\frac1{a+2}\right)+\left(\frac13-\frac1{a+3}\right)+\dots\right]\\ &=-\left[\frac1{(a+1)^2}+\frac1{(a+2)^2}+\frac1{(a+3)^2}+\dots\right] \end{align} $$

Differentiation under the Integral $$ \begin{align} -\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x &=\lim_{h\to0}\int_0^1\frac{x^{a+h}-x^a}{h(1-x)}\,\mathrm{d}x\\ &=\lim_{h\to0}\int_0^1\frac{x^a}{1-x}\frac{x^h-1}{h}\,\mathrm{d}x \end{align} $$ Pointwise $\frac{x^h-1}{h}\to\log(x)$ and on $(0,1)$ we have $\left|\frac{x^h-1}{h}\right|\le|\log(x)|$. Dominated Convergence finishes things off.

Convergence of the Integral of the Sum

For $a\ge0$, note that on $[0,1]$, $\frac{1-x^a}{1-x}\le\max(a,1)$. Therefore, $$ \begin{align} \left|\int_0^1\frac{1-x^a}{1-x}\,\mathrm{d}x-\sum_{k=0}^{n-1}\int_0^1(1-x^a)x^k\,\mathrm{d}x\right| &=\left|\int_0^1\frac{1-x^a}{1-x}x^n\,\mathrm{d}x\right|\\ &\le\frac{\max(a,1)}{n+1} \end{align} $$ For $-1\lt a\lt0$, the term on the right becomes $$ \begin{align} \left|\int_0^1\frac{1-x^a}{1-x}x^n\,\mathrm{d}x\right| &=\left|\int_0^1\frac{x^{-a}-1}{1-x}x^{n+a}\,\mathrm{d}x\right|\\ &\le\frac1{n+a+1} \end{align} $$

An Alternate Approach

Note that integration by parts gives us $$ \begin{align} \int_0^1x^a\log(x)\,\mathrm{d}x &=\frac1{a+1}\int_0^1\log(x)\,\mathrm{d}x^{a+1}\\ &=-\frac1{a+1}\int_0^1x^a\,\mathrm{d}x\\ &=-\frac1{(a+1)^2} \end{align} $$ Then using the sum of a geometric series, we have $$ \begin{align} \int_0^1x^a(1-x)^{-1}\log(x)\,\mathrm{d}x &=\int_0^1\sum_{k=0}^\infty x^{a+k}\log(x)\,\mathrm{d}x\\ &=-\sum_{k=0}^\infty\frac1{(a+k+1)^2}\\ &=-\sum_{k=1}^\infty\frac1{(a+k)^2} \end{align} $$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}x^{a}\pars{1 - x}^{-1}\ln\pars{x}\,\dd x = -\sum_{j = 1}^{\infty}{1 \over \pars{\,j + a\,}^{2}}}$

With $\ds{\quad x \equiv \expo{-t}\quad\iff\quad t = -\ln\pars{x}}$: \begin{align} &\color{#00f}{\large\int_{0}^{1}x^{a}\pars{1 - x}^{-1}\ln\pars{x}\,\dd x}= \int_{\infty}^{0}\expo{-at}\bracks{1 - \expo{-t}}^{-1}\pars{-t} \pars{-\expo{-t}\,\dd t} \\[3mm]&= -\int_{0}^{\infty}t\expo{-\pars{a + 1}t}\sum_{j = 0}^{\infty}\expo{-jt}\,\dd t =-\sum_{j = 0}^{\infty}\int_{0}^{\infty}t\expo{-\pars{j + a + 1}t}\,\dd t \\[3mm]&=-\sum_{j = 0}^{\infty}{1 \over \pars{\,j + a + 1\,}^{2}}\ \overbrace{\int_{0}^{\infty}t\expo{-t}\,\dd t}^{\ds{1!\ = 1}} =\color{#00f}{\large -\sum_{j = 1}^{\infty}{1 \over \pars{\,j + a\,}^{2}}} \end{align}