Question: Find a fast calculation for $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$
I saw the above question in an another thread (linked below) but I'm just a beginner to calculus.
I would appreciate it if someone could explain in detail how to solve the particular question asked in this thread using the method presented below (taken as a direct quote from one of the answers in the linked thread).
Answer: Here is a general approach. Consider the integral
$$ F(s) = \int_{0}^{\infty}\frac{x^{s-1}}{1+x^2}=\frac{\pi}{2\sin(s\pi/2)},\quad 0<Re(s)<2. $$
which is the Mellin transform of the function $\frac{1}{1+x^2}$. Now, our integral can be evaluated as
$$ I = \lim_{s\to 1} F'(s) = 0.$$
Note:
1) To evaluate the above integral, you can use the technique.
2)
$$ F(s)= \int_{0}^{\infty} x^{s-1}f(x)dx \implies F'(s)= \int_{0}^{\infty} x^{s-1}\ln(x)f(x)dx .$$
https://math.stackexchange.com/a/652734/1149900
I have read it several times but I am unable to understand the explanation.
The answer starts by taking integral $$F(s)=\int_0^\infty\frac{x^{s-1}}{1+x^2}\mathrm dx= \frac{\pi}{2}\operatorname{csc}(s\pi/2) \\ 0<\Re s<2$$ As known. So evaluating this integral is a whole different problem on its own. Given that we know this result, you can calculate straightforwardly using the Leibniz integral rule, $$F'(s)=\frac{\partial}{\partial s}\int_0^\infty\frac{x^{s-1}}{1+x^2}\mathrm dx$$ Now, since $$\frac{\partial}{\partial s}x^{s-1}=\frac{\partial}{\partial s}\exp((s-1)\log x) \\ =\exp((s-1)\log x)\cdot \log x \\ =x^{s-1}\log x$$ We get $$F'(s)=\int_0^\infty\frac{x^{s-1}\log(x)}{1+x^2}\mathrm dx$$ Therefore, $$\lim_{s\to 1^+}F'(s)=\lim_{s\to 1^+}\int_0^\infty\frac{x^{s-1}\log(x)}{1+x^2}\mathrm dx=\int_0^\infty\frac{\log(x)}{1+x^2}\mathrm dx$$ On the other hand $$\lim_{s\to 1^+}F'(s)=\lim_{s\to 1^+}\left[\frac{\pi}{2}\operatorname{csc}(s\pi/2)\right] \\ =\lim_{s\to 1^+}\left[\frac{\pi}{2}\cdot\left(-\operatorname{csc}(s\pi/2)\operatorname{cot}(s\pi/2)\frac{\pi}{2}\right) \right] \\ =-\frac{\pi^2}{4}\lim_{s\to 1^+}\left[\operatorname{csc}(s\pi/2)\operatorname{cot}(s\pi/2)\right] \\ =-\frac{\pi^2}{4}\cdot 1\cdot 0 \\ =0.$$