How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$?

Question

Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$. How can I prove this sum? $$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$

My try:

Change the sums into an integrals: $\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+1)^2}=\int_0^1\frac{-\ln(x)}{1+x^5}dx$

Can somebody give a hint how on to integrate this integral.

Try substitution by letting $u=\ln(x)$ is not working and integration by part is making it more complicated than before. What kind of substitution should I be using?

Answer 1

Consider the function $f(x)=\frac12(3x^2-1)$ for $x\in(-1,1)$ and its periodic extension with period $2$. Since $f(x)=f(-x)$ we can write $$f(x)=\sum_{k=0}^{\infty}a_n\cos n\pi x$$ $$\int_{-1}^1\frac12(3x^2-1)dx=0=2a_0$$ $$\begin{align}\int_{-1}^1\frac12(3x^2-1)\cos n\pi x\,dx&=\left[\frac1{n\pi}\frac12(3x^2-1)\sin n\pi x+\frac{3x}{n^2\pi^2}\cos n\pi x-\frac3{n^3\pi^3}\sin n\pi x\right]_{-1}^1\\ &=\frac{6(-1)^n}{n^2\pi^2}=a_n\end{align}$$ So $$f(x)=\frac6{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos n\pi x$$ Note that $\cos\left(n\pi\left(1-\frac15\right)\right)=(-1)^n\cos\frac{n\pi}5$. So $$\begin{align}f\left(\frac45\right)&=\frac{23}{50}=\frac6{\pi^2}\sum_{n=1}^{\infty}\frac1{n^2}\cos\frac{n\pi}5\\ &=\frac6{\pi^2}\left(\frac{\sigma}2+\sum_{n=1}^{\infty}\frac1{25n^2}(-1)^n\right)\\ &=\frac6{\pi^2}\left(\frac{\sigma}2+\frac1{25}\left(-\frac12\right)\frac{\pi^2}6\right)\end{align}$$ Here $\sigma$ is the topical sum, where the coefficient of $1/n^2$ is $2\cos n\pi/5$ except where $n=5k$ where the $\cos 5k\pi/5=(-1)^k$ has been omitted. Thus $$\sigma=2\frac{\pi^2}6\left(\frac{23}{50}+\frac1{50}\right)=\frac{4\pi^2}{25}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int_{0}^{1}{\ln\pars{x} \over 1 + x^{5}}\,\dd x &= \int_{0}^{1}\ln\pars{x}\pars{{1 \over 1 + x^{5}} + {1 \over 1 - x^{5}}}\,\dd x - \int_{0}^{1}{\ln\pars{x} \over 1 - x^{5}}\,\dd x \\[3mm] & = 2\int_{0}^{1}{\ln\pars{x} \over 1 - x^{10}}\,\dd x - \int_{0}^{1}{\ln\pars{x} \over 1 - x^{5}}\,\dd x \\[3mm] & = {1 \over 50}\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,x^{-9/10}\,\dd x - {1 \over 25}\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,x^{-4/5}\,\dd x \\[3mm] & = -\,{1 \over 50}\lim_{\mu \to -9/10}\totald{}{\mu} \overbrace{\int_{0}^{1}{1 - t^{\mu} \over 1 - x}\,\dd x} ^{\ds{\Psi\pars{\mu + 1} + \gamma}}\ +\ {1 \over 25}\lim_{\mu \to -4/5}\totald{}{\mu} \overbrace{\int_{0}^{1}{1 - t^{\mu} \over 1 - x}\,\dd x}^{\ds{\Psi\pars{\mu + 1} + \gamma}} \\[3mm] & = \fbox{$\ds{{1 \over 50}\bracks{% 2\Psi\,'\pars{{1 \over 5}} - \Psi\,'\pars{{1 \over 10}}}}$} \approx -0.9780 \end{align}

$\Psi\pars{z}$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant. The final integral which involves the $\Psi$ function is a 'standard' identity and it appears in, for example, Abramowitz and Stegun table.

Answer 3

Let's define the sum we are looking for with $S$.

Furthermore, notice that ($k \in [1,4] $)

$$ S_k=\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+k)^2}=\sum_{n=0}^{\infty}\frac{1}{(5(2n)+k)^2}-\sum_{n=0}^{\infty}\frac{1}{(5(2n+1)+k)^2}=\\ \frac{1}{5^22^2}\left[\sum_{n=0}^{\infty}\frac{1}{(n+k/10)^2}-\sum_{n=0}^{\infty}\frac{1}{(n+(k+5)/10)^2}\right]=\\ \frac{1}{5^22^2}\left[\psi^{(1)}(k/10)-\psi^{(1)}((k+5)/10)\right] $$

where we used the definition of the Trigamma function

Taking now for example $\Delta_{14}=S_1-S_4$

$$ \Delta_{14} =\frac{1}{5^22^2}\left[\psi^{(1)}(1/10)-\psi^{(1)}(6/10)\right]-\frac{1}{5^22^2}\left[\psi^{(1)}(4/10)-\psi^{(1)}(9/10)\right]= \frac{1}{5^22^2}(\psi^{(1)}(1/10)+\psi^{(1)}(9/10))-\frac{1}{5^22^2}(\psi^{(1)}(4/10)+\psi^{(1)}(6/10)) $$

Trigamma reflection $\psi^{(1)}(1-z)+\psi^{(1)}(z)=\frac{\pi^2}{\sin^2(\pi z)}$ yields the massive simplification

$$ \Delta_{14}=\frac{\pi^2}{100}\left(\frac{1}{\sin^2(\pi/10)}-\frac{1}{\sin^2(4\pi/ 10)}\right)=\\ \frac{\pi^2}{125}(5+3\sqrt{5}) $$

Playing the same game with $\Delta_{23}=S_2-S_3$ we obtain

$$ \Delta_{23}=\frac{\pi^2}{125}(-5+3\sqrt{5}) $$

Doing the algebra yields

$$ S=\phi\Delta_{14}+\frac{\Delta_{23}}{\phi}=\frac{4 \pi^2}{25}\\ \bf{QED} $$

Answer 4

\begin{align*} \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi}{(5n+1)^{2}}-\frac{\phi}{(5n+4)^{2}}\right] & =\frac{2}{125}\left(5+2\sqrt{5}\right)\pi^{2}\\ \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi^{-1}}{(5n+2)^{2}}-\frac{\phi^{-1}}{(5n+3)^{2}}\right] & =\frac{2}{125}\left(5-2\sqrt{5}\right)\pi^{2} \end{align*}

Answer 5

Requested integral $$ \begin{align} &\int_0^1\frac{-\log(x)}{1+x^5}\,\mathrm{d}x\\ &=\int_0^\infty\frac{u}{1+e^{-5u}}\,e^{-u}\,\mathrm{d}u\\ &=\int_0^\infty\left(e^{-u}-e^{-6u}+e^{-11u}-e^{-16u}+\dots\right)u\,\mathrm{d}u\\ &=\Gamma(2)\left(1-\frac1{6^2}+\frac1{11^2}-\frac1{16^2}+\dots\right)\\ &=\sum_{k=0}^\infty\frac{e^{\pi i(k-1)/5}+e^{3\pi i(k-1)/5}+e^{5\pi i(k-1)/5}+e^{7\pi i(k-1)/5}+e^{9\pi i(k-1)/5}}{5k^2}\\ &=\tfrac{e^{-\pi i/5}}5\operatorname{Li}_2\left(e^{\pi i/5}\right) +\tfrac{e^{-3\pi i/5}}5\operatorname{Li}_2\left(e^{3\pi i/5}\right) +\tfrac{e^{-5\pi i/5}}5\operatorname{Li}_2\left(e^{5\pi i/5}\right) +\tfrac{e^{-7\pi i/5}}5\operatorname{Li}_2\left(e^{7\pi i/5}\right) +\tfrac{e^{-9\pi i/5}}5\operatorname{Li}_2\left(e^{9\pi i/5}\right)\\[3pt] &=\frac25\operatorname{Re}\left( e^{-\pi i/5}\operatorname{Li}_2\left(e^{\pi i/5}\right) +e^{-3\pi i/5}\operatorname{Li}_2\left(e^{3\pi i/5}\right) \right) +\frac{\pi^2}{60}\\[6pt] &=0.9779708947989040114122103535270704097171 \end{align} $$

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Full Sum

In this answer, it is shown that $$ \sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z) $$ Therefore, $$ \begin{align} \sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k} &=\overbrace{2\sum_{k=-\infty}^\infty\frac{1}{z+2k}}^{\text{twice the even terms}} -\overbrace{\sum_{k=-\infty}^\infty\frac{1}{z+k}}^{\text{all the terms}}\\ &=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt] &=\pi\csc(\pi z) \end{align} $$ Taking the derivative yields $$ \sum_{k=-\infty}^\infty\frac{(-1)^k}{(z+k)^2} =\pi^2\csc(\pi z)\cot(\pi z) $$ Therefore, $$ \begin{align} \sum_{n=0}^\infty(-1)^n\left[\frac1{(5n+1)^2}-\frac1{(5n+4)^2}\right] &=\frac1{25}\sum_{n=-\infty}^\infty\frac{(-1)^n}{\left(n+\frac15\right)^2}\\ &=\frac{\pi^2}{25}\csc\left(\frac\pi5\right)\cot\left(\frac\pi5\right)\\[6pt] &=\frac{\pi^2}{125}\sqrt{70+30\sqrt5} \end{align} $$ and $$ \begin{align} \sum_{n=0}^\infty(-1)^n\left[\frac1{(5n+2)^2}-\frac1{(5n+3)^2}\right] &=\frac1{25}\sum_{n=-\infty}^\infty\frac{(-1)^n}{\left(n+\frac25\right)^2}\\ &=\frac{\pi^2}{25}\csc\left(\frac{2\pi}5\right)\cot\left(\frac{2\pi}5\right)\\[6pt] &=\frac{\pi^2}{125}\sqrt{70-30\sqrt5} \end{align} $$ Thus, $$ \begin{align} &\sum_{n=0}^\infty(-1)^n\left[\frac\phi{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]\\ &=\frac{\pi^2}{25}\frac{\left(1+\sqrt5\right)\sqrt{7+3\sqrt5}+\left(-1+\sqrt5\right)\sqrt{7-3\sqrt5}}{\sqrt{10}}\\ &=\frac{2\pi^2}{25}\frac{\sqrt{9+4\sqrt5}+\sqrt{9-4\sqrt5}}{\sqrt5}\\ &=\frac{2\pi^2}{25}\frac{\left(2+\sqrt5\right)+\left(-2+\sqrt5\right)}{\sqrt5}\\ &=\frac{4\pi^2}{25} \end{align} $$