Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
6  
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

3 Answers 3

up vote 4 down vote accepted

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 $$\int\limits_{ - \infty }^\infty {\frac{u}{{{{\left( {{e^{ - u}} + e^{u}} \right)}^2}}}du} =0$$

share|improve this answer

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.

share|improve this answer
    
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
1  
You can see it now. –  Pedro Tamaroff Apr 30 '12 at 5:17

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

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.