# Improper Integral Question: $\int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx$

I have to test the convergence of the integral :

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

Please suggest. Also, have to show that the value of the integral is zero ?

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Hint: Separate into integral $0$ to $1$, $1$ to $\infty$. For the second integral, let $u=1/x$ and simplify. You will get a very nice surprise. Actually, you don't need to break up the integral, but it may be clearer if you do. Existence of integral can be done by standard comparisons. – André Nicolas Apr 28 '12 at 20:58
This is just a special case of this math.stackexchange.com/questions/449070/… Where $R(x) = x/(1+x^2)^2$. – N3buchadnezzar Aug 27 '13 at 21:44

## 4 Answers

André's solution is very clever. Another way to solve it is exploiting the properties of odd functions. Let $x=e^u$, so that

$$\int\limits_0^\infty {\frac{{x\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx} = \int\limits_{ - \infty }^\infty {\frac{u}{{{{\left( {{e^{ - u}} + e^u} \right)}^2}}}du}$$

Note that $u$ is odd, and $e^u+e^{-u}$ is even, so that the integrand itself is odd. We also know any integral over $[a,\infty)$ or $(-\infty,b]$ exists because of exponential decay, and since $$\int\limits_0^\infty {\frac{u}{{{{\left( {{e^{ - u}} + e^{u}} \right)}^2}}}du}+\int\limits_{-\infty}^0 {\frac{u}{{{{\left( {{e^{ - u}} + e^{u}} \right)}^2}}}du} =0$$ we have $$\int\limits_{ - \infty }^\infty {\frac{u}{{{{\left( {{e^{ - u}} + e^{u}} \right)}^2}}}du} =0$$

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$$\int_0^{\infty} \dfrac{x \log x}{(1+x^2)^2}$$

A little observation would tell you that substitution $u=\dfrac{1}{1+x^2}$ would be of great help.

$du=\dfrac{-2x}{(1+x^2)^2}$. So now, the limits run from $1$ to $0$

$$\dfrac{-1}{2} \int_1^0\log \sqrt{\dfrac{1-u}{u}}=I=\dfrac{-1}{2} \int_1^0\log \sqrt{\dfrac{1-(1-u)}{1-u}}$$

$$2I= \int_0^1 \log \sqrt{ \dfrac{u}{1-u} \cdot \dfrac{1-u}{u}} du$$

Thus you have the desired result.

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There is a simple solution for this question. Note \begin{eqnarray} \int_0^\infty\frac{x\ln x}{(1+x^2)^2}dx&=&\int_0^1\frac{x\ln}{(1+x^2)^2}dx+\int_1^\infty\frac{x\ln x}{(1+x^2)^2}dx. \end{eqnarray} But for the second integral, we use the substitute $u=\frac{1}{x}$: \begin{eqnarray} \int_1^\infty\frac{x\ln x}{(1+x^2)^2}dx=-\int_0^1\frac{u\ln u}{(1+u^2)^2}du. \end{eqnarray} Thus $$\int_0^\infty\frac{x\ln x}{(1+x^2)^2}dx=0.$$

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You want to find

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

First, we find a primitive of the integrand, this is, we need

$$\int\frac{x\log x}{(1+x^2)^2} dx =F(x)$$

Use integration by parts with $u = \log x$ and $dv = \dfrac{x}{(1+x^2)^2}$.

This makes $du = dx/x$ and $v=-\dfrac{1}{2(1+x^2)}$, from where

$$\int\frac{x\log x}{(1+x^2)^2} dx = -\dfrac{\log x}{2(1+x^2)}+\int \dfrac{dx}{2x(1+x^2)}$$

To find the last integral, use partial fractions:

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

This means

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

So the primitive we're looking for is

$$\int\frac{x\log x}{(1+x^2)^2} dx = -\dfrac{\log x}{2(1+x^2)}+\frac{1}{2}\log x - \frac{1}{4}\log \left( {1 + {x^2}} \right)$$

Writing this more tidily,

$$\int {\frac{{x\log x}}{{{{(1 + {x^2})}^2}}}} dx = - \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$

Now we need to evaluate those expressions at $x \to 0$, and $x \to \infty$.

$$\int\limits_0^\infty {\frac{{x\log x}}{{{{(1 + {x^2})}^2}}}dx} = - \mathop {\lim }\limits_{x \to \infty } \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right] + \mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$

You can convince yourself pretty easily that

$$\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{{1 + {x^2}}} = \mathop {\lim }\limits_{x \to \infty } \log \frac{{\sqrt {1 + {x^2}} }}{x} = 0$$

so all it is needed is to find

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$

which is an indeterminate form $\infty -\infty$.

This might be really tedious to solve, so I will use something that might be frowned upon:

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]=$$

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \frac{1}{2}\log \left( {\frac{1}{{{x^2}}} + 1} \right)} \right] =$$

Here's the catch: for $x \to 0$

\eqalign{ & 1 + {x^2} \sim 1 \cr & \frac{1}{{{x^2}}} + 1 \sim \frac{1}{{{x^2}}} \cr}

so the limit can be simplified to

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\log x + \frac{1}{2}\log \left( {\frac{1}{{{x^2}}}} \right)} \right]=$$

$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left( {\log x - \log x} \right) = 0$$

So

$$\int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx =0$$

This is why maybe using subtitutions is more recommended.

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The answer you gave is wrong. Please review it. – Pedro Tamaroff Apr 29 '12 at 19:28
I cant seem to find that error, did I mess up the limits part? – Tomarinator Apr 30 '12 at 4:32
Not really. The primitive is not well calculated. Edit the post and tick the "community wiki" box, and I'll correct it for you to see it. – Pedro Tamaroff Apr 30 '12 at 4:37
I did it, checked the community wiki check box – Tomarinator Apr 30 '12 at 5:06
You can see it now. – Pedro Tamaroff Apr 30 '12 at 5:17