Question: Is there a closed form for the integral $$ I_n=\int _0^{\infty }\frac{\ln \left(x^n+1\right)}{x^n+1}\:\mathrm{d}x$$ I tried multiple techniques, including elementary ones, but none really worked out. This led me to think that it it maybe be expressed in special functions.
Can you please help me find if this has a closed form or not?.
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, $$\int _0^{\infty }\frac{\ln \left(x^2+1\right)}{x^2+1}\:dx=-\frac{1}{2}\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{1}{2}\right)\left(\gamma +\psi \left(\frac{1}{2}\right)\right)=-\frac{\pi }{2}\left(\gamma -\gamma -2\ln \left(2\right)\right)$$ $$=\pi \ln \left(2\right)$$
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.
Herein assume $n > 1$, so that the integral converges. Define $$J_n(m) := \int_0^\infty \frac{dx}{(1 + x^n)^m},$$ so that our integral is $I_n = -J_n'(0)$. Substituting $x = \tan^{\frac 2n} \theta$ transforms the integral to $$J_n(m) = \frac2n \int_0^{\frac\pi2} \sin^{\frac{2 - n}n} \theta \cos^{\frac{(2 m - 1) n - 2}n} \theta \,d\theta = \frac1n \mathrm{B}\left(\frac1n, m - \frac1n\right) = n \frac{\Gamma\left(\frac1n\right) \Gamma(m - \frac1n)}{\Gamma(m)} ,$$ where $\mathrm B$ and $\Gamma$ respectively denote the beta and gamma functions. Now, differentiating with respect to $m$ and setting $m = 1$ recovers $$I_n = -J_n'(1) = -\frac1n \Gamma\left(\frac1n\right) \Gamma\left(1 - \frac1n\right) \left(\gamma + \psi\left(1 - \frac1n\right)\right) ,$$ where $\psi$ denotes the digamma function, i.e., $\psi := (\log \Gamma)'$. Invoking Euler's Reflection Formula ($\Gamma(x) \Gamma(1 - x) = \pi \csc \pi x$) and the writing the digamma function in terms of the harmonic numbers $H_s$ (generalized to real index $s$) as $\psi(x) = H_{x - 1} - \gamma$ gives $$\boxed{\int_0^\infty \frac{\log(1 + x^n) \,dx}{1 + x^n} = -\frac{\pi}{n} H_{-\frac1n} \csc \frac\pi n}.$$
The special case $n = 2$ gives the well-known value $I_2 = \pi \log 2$, which can also be derived by other means.
Expanding in a series as $n \searrow 1$ gives $$\int_0^\infty \frac{\log(1 + x^n) \,dx}{1 + x^n} \sim \frac{1}{(n - 1)^2} + \frac{1}{(n - 1)} + \zeta(3) (n - 1) + \cdots ,$$ where $\cdots$ denotes a remainder in $O((n - 1)^2)$, where $\zeta$ denotes the Riemann zeta function.
Expanding in a series as $n \to \infty$ gives $$\int_0^\infty \frac{\log(1 + x^n) \,dx}{1 + x^n} \sim \frac{\zeta(2)}n + \frac{\zeta(3)}{n^2} + \frac{7 \zeta(4)}{2 n^3} + \cdots ,$$ where now $\cdots$ denotes a remainder in $O\left(\frac1{n^4}\right)$.
