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I am trying to solve this integral

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

where

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

I think the best way to solve our integral $\int_{0}^{+\infty }\frac{x\ln(x)}{(1+x^2)^2}$ is to use this formula:

$$[F(x)]_{a}^{b}=\lim_{x\to b-}F(x)-\lim_{x\to a+}F(x)$$

so $$\int_{0}^{+\infty }\frac{x\cdot \ln(x)}{(1+x^2)^2}=[F(x)]_{0}^{\infty}=\lim_{x\to \infty -}F(x)-\lim_{x\to 0+}F(x)$$ $$\int_{0}^{+\infty }\frac{x\cdot \ln(x)}{(1+x^2)^2}=\lim_{x\to \infty -}\frac{1}{4}\left( \frac{-2\ln(x)}{1+x^2}+\ln\frac{x^2}{1+x^2} \right)-\lim_{x\to 0+}\frac{1}{4}\left( \frac{-2\ln(x)}{1+x^2}+\ln\frac{x^2}{1+x^2} \right)$$

Would you please guide me through how to solve the following limits? $$\lim_{x\to \infty -}F(x)=\lim_{x\to \infty -}\frac{1}{4}\left( \frac{-2\ln(x)}{1+x^2}+\ln\frac{x^2}{1+x^2} \right)=\quad ?$$ $$\lim_{x\to 0+}F(x)=\lim_{x\to 0+}\frac{1}{4}\left( \frac{-2\ln(x)}{1+x^2}+\ln\frac{x^2}{1+x^2} \right)=\quad ?$$

Thank you

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up vote 2 down vote accepted

We answer your question, and at the end make a suggestion that makes the integration work, and to a fair degree the limit computation, unnecessary.

The behaviour as $x\to\infty$ is the more straightforward one. Your expression as two components, which we can treat separately. The least problematic is $$\lim_{x\to\infty}\frac{x^2}{1+x^2}.$$ For as $x\to\infty$, $\frac{x^2}{1+x^2}\to 1$, and so by continuity the logarithm approaches $\ln 1$, which is $0$.

Finding $$\lim_{x\to\infty} -\frac{2\ln x}{1+x^2}$$ is not difficult, particularly if we operate informally, since $x^2$ grows much faster than $\ln x$. There are various formal calculations we can make to confirm. For instance, we can use L'Hospital's Rule.

Dealing with the limit as $x\to 0^+$ is more complicated. This is because each of the two components "blows up." So to find the limit we will somehow have to combine them.

Our expression is equal to $$\frac{1}{4}\left(-2\frac{\ln x}{1+x^2} +2\ln x-\ln(1+x^2)\right).$$ The $\ln(1+x^2)$ part is totally harmless. Getting rid of irrelevant constants, we want to find $$\lim_{x\to 0^+} \left(-\ln x +\frac{\ln x}{1+x^2}\right),$$ which simplifies to $$\lim_{x\to 0^+} -\frac{x^2\ln x}{1+x^2}.$$ The denominator $1+x^2$ is harmless. We want $$\lim_{x\to 0^+} -x^2\ln x.$$ This limit is $0$. There are many ways to show this. For example, we can rewrite our expression as $\dfrac{-\ln x}{1/x^2}$ and use L'Hospital's Rule. Or else we can write $x=e^{-w}$, and study $\frac{w}{e^{2w}}$ as $w\to\infty$.

All this work for nothing!

Remark: To do the problem with a lot less work, break up the integral into $0$ to $1$ and $1$ to $\infty$. For the $1$ to $\infty$ part, make the change of variable $x=1/t$. Very nice things will happen!

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