How to find the closed form of $\int _0^{\infty }\frac{\ln \left(x^n+1\right)}{x^n+1}\:\mathrm{d}x$

Question

Is there a closed form for $$\int _0^{\infty }\frac{\ln \left(x^n+1\right)}{x^n+1}\:\mathrm{d}x$$ I tried multiple techniques such as the elementary ones but none really work out which leads me to think that it can maybe be expressed in special functions. ¿Can you please help me find if this has a closed form or not?.

Answer 1

Consider the following identity, $$\int _0^{\infty }\frac{1}{\left(x^n+1\right)^m}\:dx=\frac{1}{n}\:\frac{\Gamma \left(\frac{1}{n}\right)\Gamma \left(m-\frac{1}{n}\right)}{\Gamma \left(m\right)}$$ If we differentiate both sides with respect to $m$ we get, $$\int _0^{\infty }\frac{\ln \left(x^n+1\right)}{\left(x^n+1\right)^m}\:dx=\frac{1}{n}\frac{\Gamma \left(\frac{1}{n}\right)\Gamma \left(m-\frac{1}{n}\right)\left(\psi \left(m\right)-\psi \left(m-\frac{1}{n}\right)\right)}{\Gamma \left(m\right)}$$ Now setting $m=1$ will get us the result of your integral, $$\boxed{\int _0^{\infty }\frac{\ln \left(x^n+1\right)}{x^n+1}\:dx=-\frac{1}{n}\Gamma \left(\frac{1}{n}\right)\Gamma \left(1-\frac{1}{n}\right)\left(\gamma +\psi \left(1-\frac{1}{n}\right)\right)}$$ Where $\gamma$ is the Euler–Mascheroni constant and $\psi $ the Digamma function.

Some interesting values can be obtained with this, $$\int _0^{\infty }\frac{\ln \left(x^2+1\right)}{x^2+1}\:dx=-\frac{1}{2}\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{1}{2}\right)\left(\gamma +\psi \left(\frac{1}{2}\right)\right)=-\frac{\pi }{2}\left(\gamma -\gamma -2\ln \left(2\right)\right)$$ $$=\pi \ln \left(2\right)$$

Answer 2

Denote $$I(n)=\int_0^{\infty} \frac{\ln(1+x^n)}{1+x^n}dx,\; \; \forall n>1$$ setting $1+x^n\to x$ and $x\to \frac{1}{x}=t$ we yield $$\begin{aligned}I(n)&= \frac{1}{n}\int_{1}^{\infty}\frac{\ln(x)}{x}\frac{\sqrt[n]{x-1}}{x(x-1)}dx\\& =-\frac{1}{n}\int_0^1{t^{-1/n}(1-t)^{1/n-1}}\ln tdt\\&=-\frac{1}{n}\frac{\partial}{\partial k}\int_0^1t^{k}(1-t)^{m}dt\end{aligned}$$ where $k=-\frac{1}{n}$ and $m=\frac{1}{n}-1$ and the last expression we have is nothing but the derivatives of beta function and hence $$\begin{aligned} I(n)& =-\frac{1}{n}\frac{\partial}{\partial k}\beta\left(1+k,m+1\right)\\&=-\frac{1}{n}\beta(1+k,m+1)\left(\psi^0\left(1+k\right)+\gamma\right)\\&=-\frac{1}{n}\underbrace{\Gamma\left(1-\frac{1}{n}\right)\Gamma\left(\frac{1}{n}\right)}_{\text{reflection formula}}\left(\psi^0(k+1)+\gamma\right)=-\frac{\pi}{n\sin(\frac{\pi}{n})}\left(\psi^0\left(1-\frac{1}{n}\right)+\gamma\right)\\&=-\frac{\pi}{n\sin(\frac{\pi}{n})}H_{-\frac{1}{n}}\end{aligned}$$

For $n=60$ it is interesting to derive the following closed form

$$\begin{aligned}I(60)=& \int_0^{\infty}\frac{\ln(1+x^{60})}{1+x^{60}}dx\\& =-\frac{\pi}{15}\frac{H_{-\frac{1}{60}}}{\sqrt{8-\sqrt{12-4\phi}-2\sqrt{3}\phi}}\end{aligned}$$ Notation: $\phi$ is Golden ratio.