What is the integral $\int_0^\infty \log \left (1+\left (\frac A {\sqrt t}\right)^n\right) \, dt$?

Question

Let $A>0$ be a positive real number and $n \geq 4$ be an integer. What is the integral $$ \int_0^\infty \log \left (1+\left (\frac A {\sqrt t}\right)^n\right) \, dt \ ? $$ I achived a calculation using $n=4$ but for more general $n$ I'm failing. Any help would be appreciated.

Answer 1

Note \begin{eqnarray} &&\int_0^\infty \log \left (1+\left (\frac A {\sqrt t}\right)^n\right) \, dt \\ &\overset{\frac A {\sqrt t}\to t}=&2A^2\int_0^\infty t^{-3}\log \left (1+t^n\right) \, dt \overset{IBP}=2A^2n\int_0^\infty \frac{t^{n-3}}{1+t^n} \, dt \end{eqnarray} for $n>2$. Now you can use this result (Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$) to get the integral

Answer 2

$$I=\int \log \left (1+\left (\frac A {\sqrt t}\right)^n\right) \, dt$$ $$t=A^2 x^{-\frac 2 n}\quad \implies \quad I=-\frac {2A^2 }{n }\int \frac {\log(1+x) }{ x^{1+\frac{2}{n}}}\,dx$$ $$J=\int \frac {\log(1+x) }{ x^{1+\frac{2}{n}}}\,dx=-\frac{n}{2} x^{-\frac 2 n} \log (x+1)+\frac{n}{2} \int \frac{dx}{(1+x)\, x^{\frac 2 n}}$$ $$K=\int \frac{dx}{(1+x)\, x^{\frac 2 n}}=\frac {n}{n-2}x^{\frac{n-2}{n}}\, _2F_1\left(1,\frac{n-2}{n};\frac{2(n-1)}{n};-x\right)$$ Back to $t$ $$I=\frac{n}{2}t-\frac{n}{2} t \, _2F_1\left(1,-\frac{2}{n};1-\frac{2}{n};-\left(\frac{A}{\sqrt{ t}}\right)^n\right)+t \log \left(1+\left(\frac{A}{\sqrt{t}}\right)^n\right)$$ No problem to compute $$L_k=\int_0^{kA^2} \log \left (1+\left (\frac A {\sqrt t}\right)^n\right) \, dt$$ $$L_k=\frac {kA^2}2\left(n+2 \log \left(1+k^{-n/2}\right)-\frac{2 n }{n+2}k^{n/2} \, _2F_1\left(1,\frac{n+2}{n};\frac{2(n+1)}{n};-k^{n/2}\right)\right)$$ and, when $k \to \infty$ the result already given by @Caesar.tcl in comments

$$\int_0^\infty\log\left(1+\left(\frac{A}{\sqrt{t}}\right)^n\right)\mathrm{d}t=\pi A^2\csc\left(\frac{2\pi}{n}\right)$$

Answer 3

$$I=\int_0^\infty \ln \left (1+\left (\frac a {\sqrt t}\right)^n\right) \, dt$$

set $x=\frac a {\sqrt t}$, so $dx=-\frac12\frac a {\left(\sqrt{t}\right)^3}dt$, but $\sqrt t = \frac ax$, so $dx=-\frac12\frac{x^3}{a^2}dt$ $$dt=-\frac{a^2}{x^3}dx$$ $$I=\int_0^\infty \ln \left (1+\left (\frac a {\sqrt t}\right)^n\right) \, {dt}=\int_\infty^0 \ln\left(1+x^n \right)\cdot\left(-\frac{a^2}{x^3}\right)\,{dx} = a^2\int_0^\infty \frac{\ln\left(1+x^n \right)}{x^3}\,{dx}$$ $$\frac I {a^2} = \int_0^1 \frac{\ln\left(1+x^n \right)}{x^3}\,{dx} + \int_1^\infty \frac{\ln\left(1+x^n \right)}{x^3}\,{dx}\overset{u=\frac1x}=\int_0^1 \frac{\sum_{k=1}^{\infty}{\frac{x^{nk}}{k}}}{x^3}\,{dx} - \int_1^0 \frac{\ln\left(1+u^{-n} \right)}{u}\,{du} {=} \int_0^1 \sum_{k=1}^{\infty}{\frac{x^{k-3}}{k}}\,{dx} + \int_0^1 \frac{\ln\left(1+u^{-n} \right)}{u}\,{du} = \sum_{k=1}^{\infty}{\int_0^1 \frac{x^{k-3}}{k}\,{dx}} + + \int_0^1 \frac{\ln\left(1+u^{-n} \right)}{u}\,{du} = \sum_{k=1}^{\infty}{\left. {\frac{x^{k-2}}{k(k-2)}} \right\rvert_0^1} + \int_0^1 \frac{\ln\left(1+u^{-n} \right)}{u}\,{du} = \sum_{k=1}^{\infty}{{\frac{1}{k(k-2)}}} + \int_0^1 \frac{\ln\left(1+u^{-n} \right)}{u}\,{du} $$