# Calculus-Infinite Integral

1. Apr 20, 2010

### TheForumLord

1. The problem statement, all variables and given/known data

Check whether the integral $$\int_{0}^{\infty}\frac{arctanx}{xln^{2}x}dx$$ converges.

2. Relevant equations
3. The attempt at a solution

The problematic points are: $$0, 1, \infty$$ . So I said:
$$\int_{0}^{\infty}\frac{arctanx}{xln^{2}x}dx = \int_{0}^{1}\frac{arctanx}{xln^{2}x}dx+ \int_{1}^{2}\frac{arctanx}{xln^{2}x}dx+ \int_{2}^{\infty}\frac{arctanx}{xln^{2}x}dx$$ .

The second integral converges [I've proved this by substition: $$x=1+t$$ and then comparison to the series $$g(x)=\frac{1}{x^{2}}$$... I did it by knowing that in 0:
$$ln(1+x)\approx x$$...
I have no idea how to deal with the two other integrals... The ln is my problem...

Hope you'll be able to help

2. Apr 20, 2010

### System

The first integral has two problems again
you should rewrite it as(you could choose any number in (0,1) not only half):

$$\int_0^{\frac{1}{2}} \frac{arctan x}{x ln^2x} \, dx + \int_{\frac{1}{2}}^1 \frac{arctan x}{x ln^2x} \, dx$$

you should know that

$$\frac{arctan x}{x ln^2x} \leq \frac{\pi}{2} \cdot \frac{1}{xl n^2 x}$$

and
$$\int \frac{1}{x ln^2 x} \, dx$$

is easy.

3. Apr 20, 2010

### elect_eng

System's idea is a good one, but I believe that it shows that the new integral diverges. If it had converged, then you would be ok, but since it diverges, then you still don't know if your original integral converges.

There is an interesting expansion for the logarithm.

$${\rm ln}(x)=\sum_{n=1}^{\infty} (-1)^{n+1}{{(x-1)^n}\over{n}}$$

valid for $$0<x\le 2$$

This may be useful for looking at the problem point x=1.

Last edited: Apr 20, 2010
4. Apr 20, 2010

### TheForumLord

Hmmmm... How did you prove that this integral convergs? (I mean the integral:
$$\int_{0}^{0.5} \frac{1}{xln^{2}x} dx$$ ) ...

I think this is what is missing for my proof...

5. Apr 20, 2010

### elect_eng

The antiderivative is -1/ln(x) I believe. This makes it clear that that integral converges. However, the next integral with the limit at 1 will diverge. This makes the situation a little more complicated. You now have to ask the question whether the divergence of the upper-bound integral proves the divergence of the actual integral. Maybe it does, but you need to make a proper arguement to prove it.

6. Apr 20, 2010

### TheForumLord

Thanks a lot!