Does $\int_0^{+\infty}\frac{\ln(1+x^n)-\ln(1+x^{n+1})}{(1+x^2)\ln(x)}\, dx$ converge? How?

Question

I found this improper integral and I have been trying it for a long time. Try to determine whether it converges or not, and if it converges to what

$$\int_0^{+\infty}\frac{\ln(1+x^n)-\ln(1+x^{n+1})}{(1+x^2)\ln(x)}\, dx$$

I tried to compare the integral with $$\int_0^{+\infty}\frac{\ln(x^n)-\ln(x^{n+1})}{(1+x^2)\ln(x)}\, dx$$ but i couldn't get a result.

Any help will be greatly appreciated.

Answer 1

Define $I$ as the integral that we need to compute and $I_1 = \int_0^{1}\frac{\ln(1+x^n)-\ln(1+x^{n+1})}{(1+x^2)\ln(x)}\, dx$. Then, $$\begin{align*} I &= I_1+\int_{1}^{\infty}\frac{\ln(1+x^n)-\ln(1+x^{n+1})}{(1+x^2)\ln(x)}\, dx\\ &\overset{x = 1/t}{=} I_1-\int_{0}^{1}\frac{\ln((1+t^n)/t^n)-\ln((1+t^{n+1})/t^{n+1})}{(1+t^2)\ln(t)}\, dt \\ &= \int_{0}^{1}\frac{\ln(t^n)-\ln(t^{n+1})}{(1+t^2)\ln(t)}\, dt = -\int_0^1 \frac{1}{1+t^2}dt = -\frac{\pi}{4}. \end{align*}$$

Answer 2

Mathematica suggests, that the integral does not depend on $n$

        (NIntegrate[ 
          Log[1 + x^(# + 1)] - Log[1 + x^#]/((1 + x^2) Log[x]), 
          {x, 0, \[Infinity]}] &) /@ Range[7] 

       {0.785398,0.785398,0.785398,0.785398,0.785398,0.785398,0.785398}

$\pi$/4 - 0.7853981592949361` = 4 *10^-9