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Evaluate $$ \int_0^\infty\frac{x\log x}{(1+x^2)^2}dx $$

$$ \int\frac{x}{(1+x^2)^2}dx=\frac{-1}{2(1+x^2)}\\ \int_0^\infty\frac{x\log x}{(1+x^2)^2}dx= \bigg[\log x\frac{-1}{2(1+x^2)}\bigg]_0^\infty-\int_0^\infty\frac{1}{x}.\frac{-1}{2(1+x^2)}dx\\ =0+\int_0^\infty\frac{1}{2x(1+x^2)}dx=\frac{1}{2}\int\bigg[\frac{1}{x}-\frac{x}{1+x^2}\bigg]dx\\ =\frac{1}{2}[\log x]^\infty_0-\frac{1}{4}[\log|1+x^2|]_0^\infty=\frac{1}{4}\log\frac{x^2}{1+x^2}\\ =\bigg[\frac{1}{4}\log\frac{1}{\dfrac{1}{x^2}+1}\bigg]_0^\infty $$

Is there a better substitution that I can make to evaluate thi definite integral ?

Note: The solution given in my reference is $0$

Similar question is asked @Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$ but that is about the convergence of the given integral, here I am looking for a better and obvious substitution that will make the above definite integral easier to solve. But, hint about the solution is found there, Thanks @lab bhattacharjee

Sooraj S
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  • I'm rusty, but this feels like something suited to contour integration, if you're familiar with that/complex analysis. – Mark S. Jan 12 '20 at 19:53
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    Did you try to split the integral into $\int_0^1 + \int_1^\infty$ and use the substitution $u = 1/x$ in the last integral? – N3buchadnezzar Jan 12 '20 at 19:58
  • @N3buchadnezzar Nope, how do I even see if such a substitution helps ? – Sooraj S Jan 12 '20 at 19:59
  • @ss1729 The integral reminded me of $\int_{\mathbb{R}} \log(x)/(1+x^2) ,\mathrm{d}x$ and so I thought to use the same "trick". If you look at the plot it looks something like this https://imgur.com/a/BZgNKNP. Thus it is tempting to split the integral at 1 and see if the integral over 0 to 1 sort of cancels out the integral from 1 to $\infty$. Another point is that $\log x = - \log (1/x)$. In broader terms your integral satisfies my functional equation here: https://math.stackexchange.com/a/691064/18908 $R(x) = x/(1+x^2)^2$. – N3buchadnezzar Jan 12 '20 at 20:07

3 Answers3

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Note that, with $x=\frac1t$,

$$\int_1^\infty\frac{x\log x}{(1+x^2)^2}dx =- \int_0^1\frac{t\log t}{(1+t^2)^2}dt$$

Thus, the integral is evaluated to zero.

Quanto
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  • This is a nice substitution but how do I even imagine about such a substitution would help in the first place ? – Sooraj S Jan 12 '20 at 20:05
  • @ss1729 - There are numerous such integrals, with $\int_0^{\infty}\frac{\ln x}{1+x^2}=0$ being most known. I guess you have seen enough of them, you’d always have a watchful eye for it. – Quanto Jan 12 '20 at 20:10
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$$I=\int_{0}^{\infty}\dfrac{x\ln(x)}{(1+x^2)^2}dx$$ $$x=\dfrac{1}{t}$$

$$1=-\dfrac{1}{t^2}\dfrac{dt}{dx}$$ $$dx=-\dfrac{1}{t^2}dt$$

$$I=\int_{\infty}^{0}\dfrac{t\ln(t)}{(t^2+1)^2}dt$$ $$I=-\int_{0}^{\infty}\dfrac{t\ln(t)}{(t^2+1)^2}dt$$

As integration is independent of change of variable

$$I=-I$$ $$2I=0$$ $$I=0$$

So by this method we easily got answer as zero.

user3290550
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Thanks @lab bhattacharjee, for the hint.

Set $x=\tan t\implies dx=\sec^2t.dt$ $$ I=\int_0^{\pi/2}\frac{\tan t.\log(\tan t).\sec^2t.dt}{\sec^4t}=\int_0^{\pi/2}\sin t.\cos t\log(\tan t).dt\\ =\int_0^{\pi/4}\sin t.\cos t.\log(\tan t).dt+\int_0^{\pi/4}\sin t.\cos t.\log(\cot t).dt\\ =\int_0^{\pi/4}\sin t.\cos t.\log(\tan t).dt-\int_0^{\pi/4}\sin t.\cos t.\log(\tan t).dt=0 $$

Sooraj S
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