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Good evening, have problems with the next integral $$\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx$$ I tried to do it using Faymann's trick but it does not work out ... they could help me with the procedure or an idea

Making the change of variable with $x=\tan(y)$, what I got is, $$\int\ln(\tan(y))dy=\int\ln(\sin(y))dy-\int\ln(\cos(y))dy$$

how can you continue?

Thank you

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  • $\begingroup$ Try a change of variables $x = 1/u$. What do you get? $\endgroup$ – John Barber Jan 22 '18 at 4:39
  • $\begingroup$ this integral is very similar to what you have here, probably some answer can help you. $\endgroup$ – Masacroso Jan 22 '18 at 4:40
  • $\begingroup$ Hi, how about, with the change of variable gives me the same integral, es decir, [\int_{0}^{\infty}\frac{\ln(u)}{u^{2}+1}du] $\endgroup$ – user431277 Jan 22 '18 at 4:47
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    $\begingroup$ Possible duplicate of Evaluate $\int_0^\infty\frac{\ln x}{1+x^2}dx$ $\endgroup$ – projectilemotion Jan 22 '18 at 17:07
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Enforcing $x=1/u$ gives $dx=-\frac{1}{u^2}du$ and $$I=\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}} dx = \int_{0}^{\infty}\frac{\ln(1/u)}{1+1/u^{2}}\frac{1}{u^2}du =-\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx=-I$$

Hence $$ I=\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}} dx =0$$

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  • $\begingroup$ Hi, how could you explain to me that I = 0, it is assumed that the change of variable and the derivative will be negative, and because of the logarithms, you have another negative and it becomes positive, that is, the whole integral it remains the same only with the letter u $$\int_{0}^{\infty}\frac{\ln(1)-\ln(u)}{\frac{u^{2}+1}{u^{2}}}\left(-\frac{1}{u^{2}}du\right)$$ $\endgroup$ – user431277 Jan 22 '18 at 5:19
  • $\begingroup$ @SantiagoSeeker apply the change of variable properly . zero becomes infinity vice versa and inverting the born of integration you recover the negative $\endgroup$ – Guy Fsone Jan 22 '18 at 5:21
  • $\begingroup$ Ahhh .... interesting ... Thank you very much!, Excuse one more question, Faymann's trick in what circumstances does it apply? $\endgroup$ – user431277 Jan 22 '18 at 5:27
  • $\begingroup$ it is possible that the series doesnt converges, so the answer is correct whenever the integral converges $\endgroup$ – Masacroso Jan 22 '18 at 6:13
  • $\begingroup$ @Masacroso which series $\endgroup$ – Guy Fsone Jan 22 '18 at 6:14
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It is trivially zero by symmetry, it is enough to enforce the substitution $x\mapsto \frac{1}{z}$ on the interval $(1,+\infty)$. You may also notice that $$ \int_{0}^{1}\frac{\log x}{1+x^2}\,dx = \sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{2n}\log(x)\,dx = \sum_{n\geq 0}\frac{(-1)^{n+1}}{(2n+1)^2}$$ equals the opposite Catalan's constant. Through Feyman's trick, $$ \int_{0}^{1}\frac{x^\alpha}{1+x^2}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{2n+1+\alpha} $$ leads to the very same conclusion by differentiating both sides with respect to $\alpha$.

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Note $$I=\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx=\int_{0}^{1}\frac{\ln(x)}{1+x^{2}}dx+\int_{1}^{\infty}\frac{\ln(x)}{1+x^{2}}dx.$$ Under $x=\frac1u$, the second part of the above becomes $$ \int_{1}^{\infty}\frac{\ln(x)}{1+x^{2}}dx=-\int_{1}^{0}\frac{\ln(\frac1u)}{1+\frac{1}{u^{2}}}\frac{du}{u^2}=-\int_{0}^{1}\frac{\ln(u)}{1+u^{2}}du$$ and hence $$ I=0. $$

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