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I will be grateful if you would write me a solution procedure for this integral

$$\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx. $$

I am sure that an antiderivative is

$$\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$

Now if I put $+\infty $ instead of $x$ I get

\begin{align*} \left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0} &= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right)-\frac{1}{4} \left( \frac{2 \log 1}{1}-\log 1 \right) \\ &= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right). \end{align*}

As you can see, it is useless. Can you help me please? Thanks

Can I use this solution below?

Let $$I=\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$

Now if I calculate the limit of I i get: $$\lim_{x\to\infty}I=0$$

So the final result is \begin{align*} \left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}=0 \end{align*}

Anakin
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  • Anakin, welcome to Math.SE! Hope you stay around and contribute to the site. To check the answer, Wolfram Alpha is helpful: http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=Integrate%5Bx*Log%5Bx%5D%2F(1%2Bx%5E2)%5E2%5D – gt6989b Jun 11 '13 at 15:47
  • Possible link issues, here is what WA gets:

    $$ \frac{1}{4}\left(\frac{2 x^2 \ln x}{1 + x^2} - \ln(1 + x^2)\right) $$

     – gt6989b Jun 11 '13 at 15:48
  • You can improve the readability of your question by utilizing various $\LaTeX$ commands and environments! Please let me know if this edited question is not what you intended. – Sangchul Lee Jun 11 '13 at 15:55

4 Answers4

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Putting $x=\tan\theta$ we get,

$$I=\int_0^{\frac\pi2}\frac{\tan\theta\ln \tan \theta}{\sec^2\theta}d\theta$$

$$=\frac12\int_0^{\frac\pi2}\sin2\theta\ln \tan \theta d\theta$$

Now as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

$$I=\frac12\int_0^{\frac\pi2}\sin2\left(\frac\pi2+0-\theta\right)\theta\ln \tan \left(\frac\pi2+0-\theta\right) d\theta$$

$$=-\int_0^{\frac\pi2}\sin2\theta \ln\tan \theta d\theta=-I$$

as $\sin(\pi-2\theta)=\sin2\theta$

and $\tan \left(\frac\pi2+0-\theta\right)=\cot\theta=(\tan\theta)^{-1}\implies \ln \tan\left(\frac\pi2+0-\theta\right)=-\ln\tan \theta$

$$\implies I=0$$

lab bhattacharjee
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Putting $x=\frac1y$

$$I=\int_0^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$$

$$=\int_{\infty}^0\frac{(-\ln y)}{y\left(1+\frac1{y^2}\right)^2}\cdot\frac{(-dy)}{y^2}$$

$$=\int_{\infty}^0\frac{\ln ydy}{(1+y^2)^2}$$

$$=-\int_0^{\infty}\frac{\ln dy}{(1+y^2)^2}\text{ as }\int_a^bf(x)dx=-\int_b^af(x)dx$$

$$=-I$$

lab bhattacharjee
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1

Related problems: (I). Recalling the Mellin transform

$$ F(s) = \int_{0}^{\infty} x^{s-1}f(x) dx \implies F'(s) = \int_{0}^{\infty} x^{s-1}\ln(x)f(x) dx$$

So, taking $f(x)=\frac{1}{(1+x^2)^2}$ and finding its Mellin transform

$$F(s) = \frac{1}{4}\,{\frac { \left( 2-s \right) \pi }{\sin \left( \frac{\pi s}{2} \right) }}.$$

Now, differentiating and taking the limit as $s\to 2$ gives the desired result

$$ \lim_{s\to 2} F'(s) = 0. $$

Community
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Mhenni Benghorbal
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As $x \to \infty$, $$ \frac{2x^2\log(x)}{1+x^2} - \log(1+x^2) \approx \frac{2x^2\log(x)}{x^2} - \log(x^2) = 0. $$

You can make the $\approx$ precise using explicit estimates.

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    I would actually put in the explicit estimates. Other than that, just saying the answer is not very helpful... – gt6989b Jun 11 '13 at 15:53