# 0->+inf ln integration

• Nov 16th 2013, 07:15 AM
0->+inf ln integration
Hi everybody,

I have to integrate the function : ln(x)/(1+x²) between 0 and +inf.
I can't find the solution.
When I try an integration by part, I feel like running around in circles...
On the exercice statement, there is a suggestion: set x=(1/t) ...this don't help me...
The solution I have to find is 0.
Could you help me to resolve this?

• Nov 16th 2013, 08:33 AM
emakarov
Re: 0->+inf ln integration
What do you get when you use integration by substitution where the subtitution is x = 1/t?
• Nov 16th 2013, 08:52 AM
johng
Re: 0->+inf ln integration
Hi,
Your integral $\int_0^\infty{\ln(x)\over 1+x^2}\,dx$ is an improper integral. So first you have to prove that both $\lim_{a\rightarrow0}\int_a^1{\ln(x)\over 1+x^2}\,dx$ and $\lim_{b\rightarrow\infty}\int_1^b{\ln(x)\over 1+x^2}\,dx$ exist.
Then use the hint to prove that the two limits are negatives of one another.
• Nov 16th 2013, 08:52 AM
Re: 0->+inf ln integration
When I substitute:

x=1/t
=> dx=-1/t²dt

[0int+inf] [[ln(1/t)/[(1/t)²+1]]*-1/t²]*dt
= - [0int+inf] [ln(1/t)/(1+t)²]*dt

Then, I try a by party:
f= ln(1/t) -> df=-1/t
dg=1/(1+t²) -> g= atan(t)

=> ln(1/t)*atan(t)- - [0int+inf] (1/t)*atan(t)Then, i try a new by party on [0int+inf] (1/t)*atan(t) and I find a similar expression as the statement one...

(sorry for the "[0int+inf]" I don't know how to write it on a computer)
• Nov 16th 2013, 08:55 AM
Re: 0->+inf ln integration
Hi johng

I'm french native, what is "hint"? :-)
• Nov 16th 2013, 09:03 AM
johng
Re: 0->+inf ln integration
Hint means suggestion, so use the substitution as suggested.
• Nov 16th 2013, 09:11 AM
Re: 0->+inf ln integration
Ok tanks.

I can't get a solution with this substitution (see above).
Sure, I do something wrong but I do not know what.
• Nov 16th 2013, 09:25 AM
emakarov
Re: 0->+inf ln integration
Quote:

When I substitute:

x=1/t
=> dx=-1/t²dt

[0int+inf] [[ln(1/t)/[(1/t)²+1]]*-1/t²]*dt
= - [0int+inf] [ln(1/t)/(1+t)²]*dt

You did not change the limits of integration. We have x = 0 when t = ∞ and vice versa. So,

$\int_0^\infty\frac{\ln x}{1+x^2}\,dx= \int_\infty^0\frac{\ln(1/t)}{1+1/t^2} \left(-\frac{1}{t^2}\right)\,dt=$ (swapping the limits and removing - before $\frac{1}{t^2}$)

$\int_0^\infty\frac{\ln(1/t)}{1+t^2}\,dt=$ (since $\ln(1/t)=-\ln t$)

$-\int_0^\infty\frac{\ln t}{1+t^2}\,dt$

which is exactly the same integral as you started with, but with the opposite sign.

Click "Reply With Quote" to see the LaTeX code.

Also, here is a nice online French-English dictionary that I used to study French.
• Nov 16th 2013, 10:11 AM
Re: 0->+inf ln integration

We find an exact opposite integral after x=1/t substitution, ok but does it prove immediatly as compared to the first integral that the result is 0?
Do I have to resolve the integral (that I can't manage) to show that?
• Nov 16th 2013, 10:48 AM
emakarov
Re: 0->+inf ln integration
Quote:

We find an exact opposite integral after x=1/t substitution, ok but does it prove immediatly as compared to the first integral that the result is 0?

Yes, it does, but you may need to verify some auxiliary facts. First, the statement that we use is

$\forall x\in\mathbb{R}.\,x=-x\to x=0$.

To apply this statement to the integral, you need to show that $\int_0^\infty\frac{\ln x}{1+x^2}\,dx\in\mathbb{R}$, or that the integral converges, as John said in post #3. Also, you need to double-check that integrating by substitution works for improper integrals (I just don't remember the relevant theorems) and that it is done correctly (for example, x = 1/t = 0 corresponds to t = +∞ and not t = -∞). In general, the level of strictness depends on the course you are taking.

Note also that instead of proving $\int_0^\infty\frac{\ln x}{1+x^2}\,dx =-\int_0^\infty\frac{\ln x}{1+x^2}\,dx$, post #3 suggests showing that

$\int_0^1\frac{\ln x}{1+x^2}\,dx = -\int_1^\infty\frac{\ln x}{1+x^2}\,dx$

using the same substitution.
• Nov 16th 2013, 11:11 AM
Re: 0->+inf ln integration
I think that with all these information, I understand much better!

Thanks a lot.
• Nov 16th 2013, 03:36 PM
SlipEternal
Re: 0->+inf ln integration
Edit: Post deleted.
• Nov 16th 2013, 05:59 PM
SlipEternal
Re: 0->+inf ln integration
To see that it converges, you can break it up into pieces:

\begin{align*}\int_1^\infty \dfrac{\ln(x)}{1+x^2}dx & = \sum_{n\ge 0} \int_{e^n}^{e^{n+1}} \dfrac{\ln(x)}{1+x^2}dx \\ & \le \sum_{n\ge 0}\max\left\{ \dfrac{\ln(x)}{1+x^2} \mid e^n \le x \le e^{n+1} \right\}(e^{n+1}-e^{n})\end{align*}

Since $\dfrac{\ln(x)}{1+x^2}$ is decreasing on $(e,\infty)$, for all $n>0$, we know $\int_{e^n}^{e^{n+1}} \dfrac{\ln(x)}{1+x^2}dx \le \dfrac{\ln(e^n)}{1+e^{2n}}(e^{n+1}-e^n) = \dfrac{ne^n}{1+e^{2n}}(e-1) \le \dfrac{n}{e^n}(e-1)$. Also, since $0 \le \ln(x) \le 1$ for all $1 \le x \le e$ and $0 < \dfrac{1}{1+x^2} < 1$ for all $1 \le x \le e$, we know that $\int_{e^0}^{e^1} \dfrac{\ln(x)}{1+x^2}dx \le e-1$.

So,

\begin{align*}\int_1^\infty \dfrac{\ln(x)}{1+x^2}dx & = \sum_{n\ge 0} \int_{e^n}^{e^{n+1}} \dfrac{\ln(x)}{1+x^2}dx \\ & \le \left(1+\sum_{n\ge 1} \dfrac{n}{e^n} \right)(e-1) \\ & = \left(1+\dfrac{e}{(e-1)^2}\right)(e-1) \\ & = \dfrac{e^2-e+1}{e-1}\end{align*}.