Re: 0->+inf ln integration

What do you get when you use integration by substitution where the subtitution is x = 1/t?

Re: 0->+inf ln integration

Hi,

Your integral $\displaystyle \int_0^\infty{\ln(x)\over 1+x^2}\,dx$ is an improper integral. So first you have to prove that both $\displaystyle \lim_{a\rightarrow0}\int_a^1{\ln(x)\over 1+x^2}\,dx$ and $\displaystyle \lim_{b\rightarrow\infty}\int_1^b{\ln(x)\over 1+x^2}\,dx$ exist.

Then use the hint to prove that the two limits are negatives of one another.

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)

Re: 0->+inf ln integration

Hi johng

Thanks for your reply.

I'm french native, what is "hint"? :-)

Re: 0->+inf ln integration

Hint means suggestion, so use the substitution as suggested.

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.

Re: 0->+inf ln integration

Quote:

Originally Posted by

**Adi0101** 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,

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

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

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

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.

Re: 0->+inf ln integration

Thanks for the nice advices!

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?

Re: 0->+inf ln integration

Quote:

Originally Posted by

**Adi0101** 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

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

To apply this statement to the integral, you need to show that $\displaystyle \int_0^\infty\frac{\ln x}{1+x^2}\,dx\in\mathbb{R}$, 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 $\displaystyle \int_0^\infty\frac{\ln x}{1+x^2}\,dx =-\int_0^\infty\frac{\ln x}{1+x^2}\,dx$, post #3 suggests showing that

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

using the same substitution.

Re: 0->+inf ln integration

I think that with all these information, I understand much better!

Thanks a lot.

Re: 0->+inf ln integration

Re: 0->+inf ln integration

To see that it converges, you can break it up into pieces:

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

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

So,

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