How to calculate this definite integral $\int_0^1\frac{{\ln^4 x }}{1+x^2}dx=\frac{5\pi^5}{64}$

Question

$$\displaystyle\int_0^1\dfrac{{\ln^4 x}}{1+x^2}\text{d}x=\dfrac{5\pi^5}{64}$$

let $x=e^{-t}$， $$ \displaystyle\int_0^1\dfrac{({\ln x})^4}{1+x^2}\text{d}x=\displaystyle\int_0^{+\infty}\dfrac{t^4\text{e}^{-t}}{1+\text{e}^{-2t}}\text{d}t=\displaystyle\sum_{k=0}^{\infty}(-1)^k\displaystyle\int_0^{+\infty}t^4\text{e}^{-(2k+1)t}\text{d}t. $$ let $u=(2k+1)t$， $$ \displaystyle\sum_{k=0}^{\infty}(-1)^k\displaystyle\int_0^{+\infty}t^4\text{e}^{-(2k+1)t}\text{d}t=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^5}\Gamma(5)=24\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^5}. $$ I don't know how to solve this series.

So how can I solve this series? And is there any other ways to solve this definite integral. Thank you.

Answer 1

I'd like to add a sel-contained answer. We may consider that $$ f(x) = \sum_{n\geq 1}\frac{\sin(nx)}{n} $$ is the $2\pi$-periodic extension of a linear function, which equals $\frac{\pi-x}{2}$ on $(0,2\pi)$. The convergence is uniform over any compact subset of $\mathbb{R}\setminus 2\pi\mathbb{Z}$. If we integrate both sides four times, we get that $$ g(x)=\sum_{n\geq 1}\frac{\sin(nx)}{n^5} $$ is the $2\pi$-periodic extension of a polynomial with degree five, $p(x)=-\frac{x^5}{240}+\frac{\pi x^4}{48}-\frac{\pi ^2 x^3}{36}+\frac{\pi ^4 x}{90}$.
The convergence is uniform over $\mathbb{R}$, hence by evaluating $g$ and $p$ at $\pi/2$ we get $$ \sum_{n\geq 1}\frac{\sin(n\pi/2)}{n^5} = \sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^5} = p\left(\frac{\pi}{2}\right)=\frac{5\pi^5}{1536}.$$ Not by chance, this is related to the value of the wanted integral. By termwise integration

$$ \int_{0}^{1}\log^4(x)\sum_{k\geq 0}(-1)^k x^{2k}\,dx = \sum_{k\geq 0}(-1)^k\int_{0}^{1}x^{2k}\log^4(x)\,dx =\sum_{k\geq 0}\frac{24(-1)^k}{(2k+1)^5}$$ so $$ \int_{0}^{1}\frac{\log^4(x)\,dx}{1+x^2} = 24 p\left(\frac{\pi}{2}\right)=\color{red}{\frac{5\pi^5}{64}}.$$ Summarizing, it is enough to exploit the Fourier series of Bernoulli polynomials.

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As an alternative, we may use Feynman's trick. By Euler's Beta function and the reflection formula for the $\Gamma$ function we have that $$ \int_{0}^{+\infty}\frac{x^s\,dx}{1+x^2} = \frac{\pi}{2}\sec\left(\frac{\pi s}{2}\right)$$ holds for any $s\in(-1,1)$. If we differentiate (with respect to $s$) both sides four times, then perform an evaluation at $s=0$, we get $$ \int_{0}^{1}\frac{\log^4(x)\,dx}{1+x^2}=\frac{3\pi^5}{8}\cdot[z^4]\sec(z)=\frac{3\pi^5}{8}\cdot[z^4]\frac{1}{1-\frac{z^2}{2}+\frac{z^4}{24}}=\frac{3\pi^5}{8}\left(\frac{1}{2^2}-\frac{1}{24}\right) $$ and the conclusion is just the same.

Answer 2

Your approach works perfectly fine! What you have there, in the last line, is the so-called Dirichlet Beta Function $\beta(s)$. What you are looking for in particular is the value of $\beta(5)$ which is in fact expressable in terms of $\pi^5$ yet alone (as linked by Zacky). Similiar to the Riemann Zeta Function, where we have a formula for $\zeta(2n)$, there is a formula for computing $\beta(2n+1)$ for $n\in\Bbb N_0$. In order to be precise we have that

$$\beta(2n+1)~=~(-1)^n\frac{\pi^{2n+1}}{4^{n+1}(2n)!}\operatorname{E}_{2n}\tag1$$

Here $E_n$ denotes a Euler Number. Using $(1)$ you will obtain the value you are looking for. To give some more context. Dr. Sonnhard Graubner gave the value in terms of the Hurwitz Zeta Function $\zeta(s,a)$, which has a quite simple relation to the Dirichlet Beta Function (similiar with the mentioned Lerch Transcendent). Allawonder essentially gave you the integral representation of $\beta(s)$.

Currently I am not aware of a simpler derivation, not relying on $(1)$. However, as with $\beta(3)$ there might exist an elementary way of evaluating the integral or the sum, respectively.

Answer 3

Starting from the indefinite $$\int\frac{t^4e^{-t}}{1+e^{-2t}}\mathrm d t,$$ put this into the form $$\frac12\int\frac{2t^4}{e^t+e^{-t}}\mathrm d t=\frac12\int\frac{t^4}{\cosh t}\mathrm d t.$$ If we integrate by parts, this will definitely reduce the problem since the $$\int\frac{1}{\cosh t}\mathrm d t$$ may be easily done, for example by substituting $x=\tanh(t/2),$ or here since you'll eventually compute with it, write $1/\cosh t$ as $$\frac{2e^t}{1+(e^t)^2}.$$

Answer 4

\begin{align} J&=\int_0^1 \frac{\ln^4 x}{1+x^2}\,dx\\ &=\frac{1}{2}\int_0^\infty \frac{\ln^4 x}{1+x^2}\,dx\\ J_n&=\int_0^\infty \frac{\ln^{2n} x}{1+x^2}\,dx\\ J&=\frac{1}{2}J_2\\ K_n&=\int_0^\infty \int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\ &\overset{u=yx}=\int_0^\infty\left(\int_0^\infty\frac{y\ln^{2n} u}{(u^2+y^2)(1+y^2)}\,du\right)\,dy\\ &=\frac{1}{2}\int_0^\infty\left[\ln\left(\frac{u^2+y^2}{1+y^2}\right)\right]_{y=0}^{y=\infty}\frac{\ln^{2n} u}{1-u^2}\,du\\ &=\int_0^\infty \frac{\ln^{2n+1}u}{u^2-1}\,du\\ &=2\int_0^1 \frac{\ln^{2n+1}u}{u^2-1}\,du\\ &=2\int_0^1 \frac{\ln^{2n+1}u}{u-1}\,du-2\int_0^1 \frac{u\ln^{2n+1}u}{u^2-1}\,du\\ &=\left(2-\frac{1}{2^{2n+1}}\right)\int_0^1 \frac{\ln^{2n+1}u}{u-1}\,du\\ &=(2n+1)!\left(2-\frac{1}{2^{2n+1}}\right)\zeta(2n+2) \end{align}On the other hand, \begin{align} K_1&=2\int_0^\infty \frac{\ln^2 x}{1+x^2}\,dx\int_0^\infty \frac{1}{1+y^2}\,dy\\ &=\pi J_1\\ K_2&=6J_1^2+\pi J_2\\ &=\frac{6K_1^2}{\pi^2}+\pi J_2\\ J_2&=\frac{K_2}{\pi}-\frac{6K_1^2}{\pi^3}\\ &=\frac{945\zeta(6)}{4\pi}-\frac{6075\zeta^2(4)}{8\pi^3}\\ \end{align}Moreover,\begin{align}\zeta(4)&=\frac{\pi^4}{90}\\ \zeta(6)&=\frac{\pi^6}{945} \end{align}Therefore,\begin{align}\boxed{J=\frac{5}{64}\pi^5}\end{align}

NB: for $n\geq 0$ integer \begin{align}\int_0^\infty \frac{\ln^{2n+1} x}{1+x^2}\,dx=0\end{align}

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[#ffd,15px]{\int_{0}^{1}{\ln^{4}\pars{x} \over 1 + x^{2}}\,\dd x = {5\pi^{5} \over 64}}:\ {\large ?}}$

\begin{align} \int_{0}^{1}{\ln^{4}\pars{x} \over 1 + x^{2}}\,\dd x & = \left.\partiald[4]{}{\mu}\int_{0}^{1}{x^{\mu} \over 1 + x^{2}}\,\dd x \,\right\vert_{\ \mu\ =\ 0} = \left.\partiald[4]{}{\mu}\int_{0}^{1}{x^{\mu} - x^{\mu + 2} \over 1 - x^{4}}\,\dd x\,\right\vert_{\ \mu\ =\ 0} \\[5mm] & \stackrel{x^{\large 4}\ \mapsto\ x}{=}\,\,\, \left.{1 \over 4}\,\partiald[4]{}{\mu}\int_{0}^{1}{x^{\mu/4 - 3/4} - x^{\mu/4 - 1/4} \over 1 - x}\,\dd x\,\right\vert_{\ \mu\ =\ 0} \\[5mm] & = {1 \over 4}\,\partiald[4]{}{\mu}\bracks{% \int_{0}^{1}{1 - x^{\mu/4 - 1/4} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{\mu/4 - 3/4} \over 1 - x}\,\dd x}_{\ \mu\ =\ 0} \\[5mm] & = {1 \over 4}\,\partiald[4]{}{\mu}\bracks{% \Psi\pars{{\mu \over 4} + {3 \over 4}} - \Psi\pars{{\mu \over 4} + {1 \over 4}}}_{\ \mu\ =\ 0}\label{1}\tag{1} \\[5mm] & = {1 \over 4}\,{1 \over 4^{4}}\bracks{% \Psi^{\pars{\tt IV}}\pars{3 \over 4} - \Psi^{\pars{\tt IV}}\pars{1 \over 4}} \\[5mm] & = \left.{1 \over 1024}\,\totald[4]{\bracks{\pi\cot\pars{\pi z}}}{z} \,\right\vert_{\ z\ =\ 1/4}\label{2}\tag{2} \\[5mm] & = \left.{8\pi^{5}\cot\pars{\pi z}\csc^{2}\pars{\pi z} \bracks{\cot^{2}\pars{\pi z} + 2\csc^{2}\pars{\pi z}} \over 1024} \right\vert_{\ z\ =\ 1/4} \\[5mm] & = \bbox[15px,#ffd,border:1px solid navy]{5\pi^{5} \over 64}\ \approx\ 23.9078 \end{align}

$\ds{\Psi}$ is the Digamma Function.

\ref{1}: See ${\bf\color{black}{6.3.22}}$ in this link.

\ref{2}: Euler Reflection Formula ${\bf\color{black}{6.3.7}}$