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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.

FD_bfa
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Elizabeth Huffman
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  • https://math.stackexchange.com/questions/808678/evaluate-int-0-infty-frac-ln-x1x2dx/808693#808693 If not clear, tell me – Claude Leibovici Mar 25 '23 at 07:47
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    $$\int_{1}^{\infty}\frac{\log x}{x^2+1}dx=\int_{1}^{0}\frac{-\log t}{\frac{1}{t^2}+1}:\frac{-dt}{t^2}=-\int_{0}^{1}\frac{\log t}{t^2+1}dt\quad\text{with}\quad x=\frac{1}{t}$$

    $$\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=\int_{0}^{1}\frac{\log x}{x^2+1}dx+\int_{1}^{\infty}\frac{\log x}{x^2+1}dx=\int_{0}^{1}\frac{\log x}{x^2+1}dx-\int_{0}^{1}\frac{\log t}{t^2+1}dt=0$$

     – JJacquelin Mar 25 '23 at 10:34
  • You've linked to an answer that links to three other sources; (III) uses the residue theorem, which can evaluate the integral you need to differentiate with respect to $s$. The others really on a formula for the Beta function in terms of the Gamma function, but also on the latter's reflection formula, which can be proven with the residue theorem or other advanced techniques. In terms of understanding the mandated solution technique rather than @JJacquelin's option, I'm not sure what the least painful path is at your current level. – J.G. Mar 25 '23 at 10:52
  • I’ve worked through a similar problem here https://math.stackexchange.com/a/4663462/1141581 using residue calculus. I don’t know how much help this is though if you are a beginner in calculus. – bob Mar 25 '23 at 11:38
  • This maybe irrelevant but since you are a "beginner to calculus" , perhaps the most suitable approach for you would be the one used by JJacquelin provided you do show that $\int_{0}^{1}\frac{\ln(x)}{1+x^{2}},dx$ is finite. But even that point maybe ignored for a beginner. – Mr.Gandalf Sauron Mar 25 '23 at 12:18
  • @Mr.GandalfSauron The integral over $[1,\infty)$ is convergent, for example by applying the substitution $x=e^y.$ and the inequality $e^y\ge y$ for $y\ge 0.$ The remaining integral over $(0,1]$ can be treated by substitution $x=v^{-1},$ as observed by JJacquelin. – Ryszard Szwarc Mar 25 '23 at 14:33

1 Answers1

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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 $$\lim_{s\to 1^+}F'(s)=\lim_{s\to 1^+}\left[\frac{\pi}{2}\operatorname{csc}(s\pi/2)\right] \\ =\lim_{s\to 1^+}\left[\frac{\pi}{2}\cdot\left(-\operatorname{csc}(s\pi/2)\operatorname{cot}(s\pi/2)\frac{\pi}{2}\right) \right] \\ =-\frac{\pi^2}{4}\lim_{s\to 1^+}\left[\operatorname{csc}(s\pi/2)\operatorname{cot}(s\pi/2)\right] \\ =-\frac{\pi^2}{4}\cdot 1\cdot 0 \\ =0.$$

K.defaoite
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