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### Can $\int_0^{\pi/2} \ln ( \sin(x)) \; dx$ be evaluated with “complex method”?

Can the following integral be evaluated using complex method by substituting $\sin(x) = {e^{ix}-e^{-ix} \over 2i}$? $$I=\int_0^{\pi/2} \ln ( \sin(x)) \; dx = - {\pi \ln(2) \over 2}$$
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### Find the following integral (most likely substitution)

$$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$$ I tried letting $x^2=\tan \theta$ but it didn't work. What should I do? Please don't give full solution, just a hint and I will continue.
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### How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$

How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
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### Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$

Prove that $$\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx=\frac{\pi\sqrt{2}}{2}\log\left(1+\frac{\sqrt{2}}{2}\right).$$ I managed to prove this result with some ...
### Improper integral of $\frac{\log\left(\sqrt{x^2+a^2}\right)}{x^2+b^2}$
Show that $$\int_{-\infty}^\infty \frac{\log\left(\sqrt{x^2+a^2}\right)}{x^2+b^2}\,dx = \frac{\pi}{b}\log\left(a+b\right)$$ for $a,b>0\in\mathbb{R}$. I stumbled on this answer empirically, but I'm ...