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Im aware this integral has been evaluated here before, i started with $$\int _0^{\infty}\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx=2\int _0^1\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx+2G$$ Solving the integral on the left is very easy, then solving for the other one gets its value but the main point of my question comes from the substitution $ x=\frac{1-t}{1+t}$ $$I=\int _0^1\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx=\int _0^1\frac{\ln \left(1+t^2\right)}{1+t^2}\:dt+\ln \left(2\right)\int _0^1\frac{1}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt$$ And $I$ cancels out, can anyone tell me whats happening here?

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    What are the bounds for $t$ ? – Claude Leibovici Aug 06 '20 at 10:25
  • 0 to 1 as well, see $\frac{1-x}{1+x}\rightarrow \frac{1-1}{1+1}=0,\frac{1-x}{1+x}\rightarrow :\frac{1-0}{1+0}=1$ –  Aug 06 '20 at 10:32
  • then $dx=-\frac{2}{\left(1+t\right)^2}:dt$ where the- sign is used to flip the bounds –  Aug 06 '20 at 10:33
  • You have $t=\frac {1-x}{1+x}$ . So, if $x\to 0$ then $t \to1$ and if $x\to \infty$ then $t \to -1$ (at least for me) – Claude Leibovici Aug 06 '20 at 10:35
  • You are misunderstanding, i am not seeking to solve the one from 0 to infinity but the one from 0 to 1 with said substitution –  Aug 06 '20 at 10:37
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    You just found that $\int_0^1 \frac{\ln 2}{1+t^2}dt=2\int_0^1 \frac{\ln(1+t)}{1+t^2}dt$ and also discovered that that substitution isn't useful for that integral. – Zacky Aug 06 '20 at 10:53
  • If you want performing a change of variable try $x=\tan t$ and consider $\displaystyle A=\int_0^{\frac{\pi}{4}}\ln(\cos t)dt,B=\int_0^{\frac{\pi}{4}}\ln(\sin t)dt$ – FDP Aug 06 '20 at 15:37
  • That means your subbing doesn't work for $I$. – Ali Shadhar Aug 06 '20 at 18:25

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Letting $x=\tan \theta$ yields

$$ \begin{aligned} \int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x &=\int_0^{\frac{\pi}{4}} \ln \left(\sec ^2 \theta\right) d \theta \\ &=-2 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta \\ &=-2\left(-\frac{\pi}{4} \ln 2+\frac{1}{2} G\right) \\ &=\frac{\pi}{2} \ln 2-G \end{aligned} $$

By my post, $$\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x =\frac{\pi}{2}\ln2-G$$ where $G$ is the Catalan’s constant.

Lai
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