How can one evaluate this monster $\int_0^\infty \int_0^1 \frac{\ln(1+t^x)}{(1+t)(1+x^2)}\ dt\ dx$

Question

When I encountered this integral I couldn't imagine it having such an elegant closed form, and yet here it is:

$$\int_0^\infty \int_0^1 \frac{\ln(1+t^x)}{(1+t)(1+x^2)}\ \mathrm{d}t\ \mathrm{d}x=\frac{\pi}{4}\ln^2(2)$$

Or at least this is what I conjectured looking at the numerical value of the integral. Wolfram wasn't able to catch the closed form, and neither was I.

My attempts in solving the integral included exchanging the order of integration, ending up with

$$\int_0^1 \frac{1}{1+t} \int_0^\infty \frac{\ln(1+t^x)}{1+x^2}\ \mathrm{d}x\ dt$$ but this doesn't look simpler. However, the inner integral can be reduced to $$\int_0^\infty \frac{\ln(1+t^x)}{1+x^2}\ \mathrm{d}x= $$ $$=\int_0^1\frac{\ln(1+t^x)}{1+x^2}\ \mathrm{d}x\ +\int_1^\infty\frac{\ln(1+t^x)}{1+x^2}\ \mathrm{d}x=$$ $$=\int_0^1\frac{\ln(1+t^x)}{1+x^2}\ \mathrm{d}x\ +\int_0^1\frac{\ln(1+t^{\frac{1}{x}})}{1+x^2}\ \mathrm{d}x$$

and here I tried to use Taylor series, but couldn't make progress.

Suggestions would be awesome.

Answer 1

This is the solution based in the amazing change of variable given by @eyeballfrog in the comments. Indeed it can be shown easily that the map $$ \varphi :(0,1)\times (0,\infty )\to (0,1)\times (0,\infty ),\, (t,x)\mapsto (t^x,1/x) $$ is a diffeomorphism, with $|\det \partial \varphi (t,x)|=\frac{t^{x-1}}{x}$, therefore $$ \int_{(0,\infty )\times (0,1)}\frac{\log (1+t^x)}{(1+t)(1+x^2)}\,d (x,t)=\int_{(0,\infty )\times (0,1)}\frac{\log (1+t) xt^{x-1}}{1+t^x}\cdot \frac{1}{1+x^2}\,d (x,t)\\ =\int_{0}^{\infty }\frac1{1+x^2}\left((\log 2)^2-\int_{0}^1 \frac{\log (1+t^x)}{1+t}\,d t\right)\,d x\\ \therefore\quad \int_{(0,\infty )\times (0,1)}\frac{\log (1+t^x)}{(1+t)(1+x^2)}\,d (x,t)=\frac{(\log 2)^2}{2}\cdot \frac{\pi}{2}=\frac{(\log 2)^2\pi}{4} $$ ∎

Answer 2

We can start by interchanging the order of integration. To do so, we need to justify that the integral is absolutely convergent, so we will bound the integrand from above by an integrable function. Let $f(x,t)=\frac{\ln(1+t^x)}{(1+t)(1+x^2)}$. Then, for $x\geq 0$ and $t\in [0,1]$, we have $$|f(x,t)|\leq \frac{|\ln(1+t^x)|}{(1+t)(1+x^2)}\leq \frac{t^x}{(1+t)(1+x^2)}.$$ The last inequality follows from the fact that $\ln(1+u)\leq u$ for all $u>-1$, so $\ln(1+t^x)\leq t^x$. Now, $$\int_0^\infty \int_0^1 \frac{t^x}{(1+t)(1+x^2)}\ dt\ dx=\int_0^\infty \frac{dx}{1+x^2}\int_0^1 \frac{t^x}{1+t}\ dt.$$ The inner integral can be evaluated using the substitution $u=t^x$, $du=xt^{x-1}dt$, to get $$\int_0^1 \frac{t^x}{1+t}\ dt=\int_0^1 \frac{du}{u^{1/x}(1+u^{1/x})}=\frac{1}{x}\int_0^{\infty} \frac{du}{u^{1/x}(1+u^{1/x})}.$$ Now, let $v=u^{1/x}$, $dv=\frac{du}{x}u^{-1/x}$, to get $$\int_0^{\infty} \frac{du}{u^{1/x}(1+u^{1/x})}=\int_0^\infty \frac{dv}{v(1+v)}=\ln 2.$$ Therefore, $$\int_0^\infty \int_0^1 \frac{\ln(1+t^x)}{(1+t)(1+x^2)}\ dt\ dx=\int_0^\infty \frac{\ln(2)}{(1+x^2)}\ dx=\frac{\pi}{4}\ln^2(2),$$ where the last equality follows from the integral representation of the dilogarithm function.