Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$.

Question

Inspired by my post, I go further to investigate the general integral and find a formula for $$ I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right] \tag*{} $$ Let’s start with its partner integral

$$ I(a)=\int_0^{\infty}\left(x^n+1\right)^a d x $$

and transform $I(a)$, by putting $y=\frac{1}{x^n+1}$, into a beta function $$ \begin{aligned} I(a) &=\frac{1}{n} \int_0^1 y^{-a-\frac{1}{n}-1}(1-y)^{-\frac{1}{n}-1} d y \\ &=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right) \end{aligned} $$ Differentiating $I(a)$ w.r.t. $a$ yields $$ I^{\prime}(a)=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)\left(\psi(-a)-\psi\left(-a-\frac{1}{n}\right)\right) $$ Then putting $a=-1$ gives our integral$$ \begin{aligned} I_n&=I^{\prime}(-1) \\&=\frac{1}{n} B\left(1-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(1)-\psi\left(1-\frac{1}{n}\right)\right] \\ &=-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right]\end{aligned} $$

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For examples,

$$ \begin{aligned}& I_2=-\frac{\pi}{2} \csc \frac{\pi}{2}\left[\gamma+\psi\left(1-\frac{1}{2}\right)\right]=\pi \ln 2,\\ & I_3=-\frac{\pi}{3} \csc \left(\frac{\pi}{3}\right)\left[\gamma+\psi\left(\frac{2}{3}\right)\right]=\frac{\pi \ln 3}{\sqrt{3}}-\frac{\pi^2}{9} ,\\ &I_4=-\frac{\pi}{4} \csc \left(\frac{\pi}{4}\right)\left[\gamma+\psi\left(\frac{3}{4}\right)\right]=\frac{3 \pi}{2 \sqrt{2}}\ln 2-\frac{\pi^2}{4 \sqrt{2}},\\ & I_5=-\frac{\pi}{5} \csc \left(\frac{\pi}{5}\right)\left[\gamma+\psi\left(\frac{4}{5}\right)\right]=-\frac{2 \sqrt{2} \pi}{5 \sqrt{5-\sqrt{5}}}\left[\gamma+\psi\left(\frac{4}{5}\right)\right], \\ & I_6=-\frac{\pi}{6} \csc \left(\frac{\pi}{6}\right)\left[\gamma+\psi\left(\frac{5}{6}\right)\right]=\frac{2 \pi}{3} \ln 2+\frac{\pi}{2} \ln 3-\frac{\pi^2}{2 \sqrt{3}}, \end{aligned} $$

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Furthermore, putting $a=-m$, gives $$\boxed{I(m,n)=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{(x^n+1)^m} dx = \frac{1}{n} B\left(m-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(m)-\psi\left(m-\frac{1}{n}\right)\right] }$$ For example,

$$ \begin{aligned} \int_0^{\infty} \frac{\ln \left(x^6+1\right)}{(x^6+1)^5} dx & =\frac{1}{6} B\left(\frac{29}{6}, \frac{1}{6}\right)\left[\psi(5)-\psi\left(\frac{29}{6}\right)\right] \\ & =\frac{1}{6} \cdot \frac{21505 \pi}{15552} \cdot\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right) \\ & =\frac{21505 \pi}{93312}\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right) \end{aligned} $$

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Are there any other methods?

Your comments and alternative methods are highly appreciated.

Answer 1

The integral admits elementary close-form as evaluated below

\begin{align} I_n=&\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} \overset{x\to 1/x}{dx}\\ =& \ \frac12 \int_0^{\infty} \frac{(1+x^{n-2})\ln \left(x^n+1\right)}{x^n+1}dx-\frac n2 \int_0^{\infty} \frac{x^{n-2}\ln x}{x^n+1}dx \end{align} where $\int_0^{\infty} \frac{x^{n-2}\ln x}{x^n+1}dx=\frac{\pi^2}{n^2}\csc\frac\pi n\cot\frac \pi n$ \begin{align} &\int_0^{\infty} \frac{(1+x^{n-2})\ln \left(x^n+1\right)}{x^n+1}dx\\ =& \int_0^{\infty}\int_0^1 \frac{nt^{n-1}(1+x^{n-2})}{(x^n+1)(t^n+x^n)}dt\ dx\\ =& \int_0^1 \frac{2t^{n-1}-t^{n-2}-1}{t^n-1}\int_0^\infty\frac n{x^n+1}dx \ dt \\ =&\ \frac\pi2\csc\frac\pi n \int_0^1 \frac{2t^{n-1}-t^{n-2}-1}{t^n-1}dt\\ =& \ \frac{4\pi}n \csc\frac\pi n \bigg(\frac n4\ln2 - \sum_{k=1}^{[\frac {n-1}2]}\frac{\ln\csc\frac{k\pi}n}{\csc^2\frac{k\pi}n}\bigg) \end{align} Substitute above results into $I_n$ to obtain $$\color{blue}{I_n= \frac{4\pi}n \csc\frac\pi n \bigg(\frac n4\ln2 -\frac\pi8 \cot\frac\pi n- \sum_{k=1}^{[\frac {n-1}2]}\frac{\ln\csc\frac{k\pi}n}{\csc^2\frac{k\pi}n}\bigg)} $$ In particular, the close-form yields \begin{align} I_2&=\pi \ln2\\ I_3 &=\frac\pi {\sqrt3}\ln3-\frac{\pi^2}9\\ I_4 &=\frac{3\pi}{2\sqrt2}\ln2-\frac{\pi^2}{4\sqrt2}\\ I_5 &= \frac{\pi\sqrt{\phi}}{\sqrt[4]5}\left(\frac12\ln5+\frac1{\sqrt5}\ln\phi\right)-\frac{\pi^2\phi^2}{5\sqrt5}\\ I_6&= \frac{2\pi}3\ln2+\frac\pi2\ln3 -\frac{\pi^2}{2\sqrt3}\\ I_7&= \frac{4\pi}7\csc\frac\pi7\bigg( \frac74\ln2-\frac\pi8\cot\frac\pi7 \\ &\hspace{3cm} - \frac{\ln\csc\frac{\pi}7}{\csc^2\frac{\pi}7}- \frac{\ln\csc\frac{2\pi}7}{\csc^2\frac{2\pi}7}-\frac{\ln\csc\frac{3\pi}7}{\csc^2\frac{3\pi}7}\bigg)\\ I_8&= \pi\sqrt{1+\frac1{\sqrt2}}\left(\ln2+\frac1{2\sqrt2}\ln(\sqrt2+1)-\frac\pi8(\sqrt2+1)\right)\\ I_9&= \frac{4\pi}9\csc\frac\pi9\bigg( \frac32\ln2+\frac38\ln3-\frac\pi8\cot\frac\pi9\\ &\hspace{3cm} - \frac{\ln\csc\frac{\pi}9}{\csc^2\frac{\pi}9}- \frac{\ln\csc\frac{2\pi}9}{\csc^2\frac{2\pi}9}-\frac{\ln\csc\frac{4\pi}9}{\csc^2\frac{4\pi}9}\bigg)\\ I_{10}&= \frac{\pi \phi }{10}\bigg(\frac5{2}\ln5+4\ln2 +3\sqrt5\ln\phi-\pi\sqrt[4]5 \ \phi^{3/2}\bigg)\\ \end{align}

Answer 2

You can obtain the antiderivative. Let $x^n=t$ to face $$I=\frac 1 n \int \frac{\log (t+1)}{t+1}\,t^{\frac{1}{n}-1}\,dt $$ $$I=n (t+1)^{\frac{1}{n}} \, _3F_2\left(-\frac{1}{n},-\frac{1}{n},-\frac{1}{n};1-\frac{1}{n},1 -\frac{1}{n};\frac{1}{t+1}\right)+t^{\frac{1}{n}} \log (t+1)-$$ $$\frac{n t^{\frac{1}{n}+1} }{n+1}\, _2F_1\left(1,1+\frac{1}{n};2+\frac{1}{n};-t\right) -(t+1)^{\frac{1}{n}} \log (t+1) \, _2F_1\left(-\frac{1}{n},-\frac{1}{n};\frac{n-1}{n};\frac{1}{t+1}\right)$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#44f}{{\tt I}_{n}} & \equiv \color{#44f}{\int_{0}^{\infty}{\ln\pars{x^{n} + 1} \over x^{n} + 1}\dd x} \sr{x^{n}\ \mapsto\ x}{=} \int_{0}^{\infty}{\ln\pars{1 + x} \over 1 + x}\ {1 \over n}\, x^{1/n - 1}\,\,\dd x \\[5mm] & = \left.{1 \over n}\partiald{}{\nu}\int_{0}^{\infty}x^{\color{red}{1/n} - 1}\,\, \pars{1 + x}^{\nu - 1}\,\dd x\right\vert_{\nu\ =\ 0}. \\[5mm] & \mbox{Note that}\ \pars{1 + x}^{\nu - 1} = \sum_{k = 0}^{\infty}{\nu - 1 \choose k}x^{k} = \sum_{k = 0}^{\infty}\bracks{{k - \nu \choose k}\pars{-1}^{k}}x^{k} \\[5mm] & = \sum_{k = 0}^{\infty}{\Gamma\pars{k - \nu + 1} \over \Gamma\pars{1 - \nu}}{\pars{-x}^{k} \over k!}\quad \mbox{such that} \\[5mm] \color{#44f}{{\tt I}_{n}} & \equiv \color{#44f}{\int_{0}^{\infty}{\ln\pars{x^{n} + 1} \over x^{n} + 1}\dd x} = {1 \over n}\partiald{}{\nu}\overbrace{\bracks{\Gamma\pars{\color{red}{1 \over n}} {\Gamma\pars{-\color{red}{1/n} - \nu + 1} \over \Gamma\pars{1 - \nu}}}}^{\ds{Ramanujan's\ Master\ Theorem}}\hspace{-1mm}_{\substack{\\[2mm]\nu\ =\ 0}} \\[5mm] & = \bbx{\color{#44f}{% -\,{\pi\csc\pars{\pi/n}H_{-1/n}\,\, \over n}}} \\ & \end{align}

Answer 4

For large $n$, a simple asymptotic behaviour of the integral can be deduced.

Let us first examine the structure of the integrand

$$y(x)=\frac{\ln \left(x^n+1\right)}{x^n+1} $$

Take the derivative of the function and set the result to zero to get the point $x=x_{m}$ where $y(x)$ reaches his maximum

$$1+(x_{m})^{n}=e$$ Putting this into $y(x)$ gives

$$y_{max}=y(x_{m})=\frac{\ln e}{e}=\frac{1}{e}$$

$y_{max}$ is independent of $n$

For integrals whose integrand, $y(x)$, has a sharp maximum a simple asymptotic formula exists.

$$I\approx \sqrt{2\pi\frac{y^{3}(x_{m})}{\left|y''(x_{m}) \right|}}$$

Here $y''(x_{m})$ is the second derivative of $y(x)$ at $x=x_{m}$.

I will skip elementary computations and write down final result $$I_{n}\approx \frac{\sqrt{2\pi}}{n(e-1)^{1-\frac{1}{n}}}$$

Below is a few numerical examples to show approximation errors produced by the last formula

$n=10$, the approximation error about $0.03$

$n=20$, the approximation error about $0.01$

$n=30$, the approximation error about $0.007$

The higher n is, the higher the accuracy of the last formula.