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Show that $$\int^1_0 x^{m-1}\ln^2(1-x)dx=\frac{2}{m}\sum_{k=1}^m\frac{H_{k}}{k}$$

We know: $$\ln(1-x)=-\sum^{\infty}_{n=1}\frac{x^n}{n}$$ and using cauchy product Therfore: $$\begin{align*} \ln^2(1-x)&=\sum_{n=1}^{\infty}x^{n+1}\sum_{k=1}^n\frac{1}{k(n-k+1)}\\ &=\sum^{\infty}_{n=1}\frac{x^{n+1}}{n+1}\sum_{k=1}^n\frac{1}{k}+\frac{1}{n-k+1}\\ &=2\sum_{n=1}^{\infty}\frac{x^{n+1}}{n+1}H_n \end{align*}$$

then: $$\begin{align*} \int^1_0 x^{m-1}\ln^2(1-x)dx&=2\int^1_0\sum_{n=1}^{\infty}\frac{x^{n+m}}{n+1}H_n\\ &=2\sum_{n=1}^{\infty}\frac{H_n}{(n+m+1)(n+1)}\\ &=\frac{2}{m}\sum_{n=1}^{\infty}\frac{H_n}{(n+1)}-\frac{H_n}{(n+m+1)} \end{align*}$$

How can I follow up on the solution? Thank you for your attention

user170231
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Delta
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  • You can also find this result in the book “(Almost) Impossible Integrals, Sums, and Series” by Cornel Ioan Vălean. – bob Jul 27 '24 at 09:07

2 Answers2

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Let $I_m=\int_0^1 x^{m-1}\ln^2(1-x)\,dx$ for $m\geqslant 0$.

Then, for $m>0$, integrating by parts twice, \begin{align} mI_m-(m-1)I_{m-1} &=-\int_0^1\big(x^{m-1}(1-x)\big)'\ln^2(1-x)\,dx \\&=-2\int_0^1 x^{m-1}\ln(1-x)\,dx \\&=\frac2m\int_0^1(1-x^m)'\ln(1-x)\,dx \\&=\frac2m\int_0^1\frac{1-x^m}{1-x}\,dx \\&=\color{gray}{\frac2m\int_0^1\sum_{k=1}^m x^{k-1}\,dx}=\frac2m H_m. \end{align}

metamorphy
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    The result can also be written as $$I_m=\frac{{H_m}^2+H_m^{(2)}}{m},\quad H_m^{(2)}:=\sum_{k=1}^m\frac1{k^2}$$ (proven directly or using this after $x=1-t$). – metamorphy Jul 16 '24 at 05:58
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OP wants to simplify $$\frac{2}{m}\sum_{n=1}^{\infty}\frac{H_n}{(n+1)}-\frac{H_n}{(n+m+1)} =\frac{2}{m}\sum_{n=1}^{\infty}\sum_{k=1}^n\frac{\tfrac1k}{(n+1)}-\frac{\tfrac1k}{(n+m+1)}$$ By $k=n$ diagonal trick $$\frac{2}{m}\sum_{k=1}^{\infty}\tfrac1k\sum_{n=k}^\infty\frac1{(n+1)}-\frac1{(n+m+1)} =\frac{2}{m}\sum_{k=1}^{\infty}\tfrac1k(\tfrac1{k+1}+\tfrac1{k+2}+...+\tfrac1{k+m}) $$ Can you complete by telescoping sums?

Bob Dobbs
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