Hard integrals

1. Dec 19, 2011

dirk_mec1

1. The problem statement, all variables and given/known data
$$\int_0^{\infty} \ln \left( \frac{e^x+1}{e^x-1} \right) \mbox{d}x$$

$$\int_0^{\infty} \frac{1}{x^n+1}\ \mbox{d}x\ \forall n >1$$

2. Relevant equations
-

3. The attempt at a solution
I've tried IBP and separating the ln into two terms and failed. I've also tried a subtitution for exp(x)+1 and exp(x)-1 with no succes. For the second one I've tried using a trigoniometric substitution but it didn't work out.

2. Dec 19, 2011

Ivan92

If you notice, these integrals can't be evaluated with standard integration techniques. These are called Improper Integrals. I would read up on this as I can't teach you the entire thing itself. You would need to understand the types of improper integrals as well as know the comparison theorem.

Last edited: Dec 19, 2011
3. Dec 19, 2011

Dickfore

Try to make the substitution:
$$t \equiv \frac{e^x + 1}{e^x - 1}$$

What integral do you get?

4. Dec 19, 2011

dirk_mec1

I get:

$$2 \cdot \int_1^{\infty} \frac{\ln(t)}{1+t}\ \mbox{d}t$$

but I don't see how to solve this one...

5. Dec 19, 2011

Dickfore

6. Dec 19, 2011

Char. Limit

Are you sure about that? I got something different.

7. Dec 20, 2011

dirk_mec1

$$2\cdot \int_1^{\infty} \frac{\ln(t)}{(t+1)(t-1)}\ \mbox{d}t$$

I can use partial fraction decomposition but I'm still stuck then. I've also tried a $$t =sin(\theta)$$ but the what?

Last edited: Dec 20, 2011
8. Dec 20, 2011

Dickfore

Ok, so, now your denominator is $t^2 - 1$. You want it in the form $e^{\alpha x} - 1$. That is why you have the additional substitution
$$e^{\alpha x} = t^2 \Rightarrow t = e^{\frac{\alpha \, x}{2}}$$
where $\alpha$ is yet to be determined.

What do you get now?

Last edited: Dec 20, 2011
9. Dec 20, 2011

dirk_mec1

Last edited by a moderator: May 5, 2017
10. Dec 20, 2011

Dickfore

yes, those are the necessary tricks to evaluate the integral, although I don't know why you only gave an upper bound estimate. I can't zoom in to your picture to see the details.

The second integral can be brought to a similar form.

11. Dec 21, 2011

dirk_mec1

Is the series I calclulated in the correct form? If so how can I exactly calculate:

$$\sum_{n=0}^{\infty} \frac{1}{(n-\frac{1}{2} )(n-\frac{3}{2})}$$

12. Dec 21, 2011

Dickfore

$$\frac{e^{-u/2}}{1 - e^{-u}} = \sum_{n = 1}^{\infty}{e^{-\frac{u}{2}} \, (e^{-u})^{n - 1}} = \sum_{n = 1}^{\infty}{e^{-(n - \frac{1}{2}) u}}$$

$$\int_{0}^{\infty}{u \, e^{-(n - \frac{1}{2}) \, u} \, du} = \frac{1}{(n - 1/2)^2} \, \int_{0}^{\infty}{t \, e^{-t} \, dt}$$

$$\begin{array}{l} \sum_{n = 1}^{\infty}{\frac{1}{(n - 1/2)^2} } = \sum_{n = 1}^{\infty}{\frac{4}{(2 n - 1)^2}} \\ \\ = 4 \, \left[ \sum_{n = 1}^{\infty}{\frac{1}{n^2}} - \sum_{n = 1}^{\infty}{\frac{1}{(2 n)^2}} \right] \\ = 4 \, \left( 1 - \frac{1}{4} \right) \, \sum_{n = 1}^{\infty}{\frac{1}{n^2}} = 3 \, \zeta(2) \end{array}$$
where
$$\zeta(a) \equiv \sum_{n = 1}^{\infty}{\frac{1}{n^{a}}}, \ \mathrm{Re} a > 1$$
is the Riemann zeta function.

13. Dec 23, 2011

dirk_mec1

How do you get from the single sum with the (2n-1)^2 in the denominator to the two sums?

14. Dec 23, 2011

Curious3141

Because the union of the set of all odd positive numbers and the set of even positive numbers equals the set of natural numbers.

Sum of reciprocals of odd positives is sum of reciprocals of naturals minus sum of reciprocals of even positives.

Last edited: Dec 23, 2011
15. Dec 23, 2011

dirk_mec1

Yes offcourse thanks guys!

For the second integral the hint is use: u =x^n, I don't understand what to do then.

16. Dec 23, 2011

Dickfore

no, separate it from 0 to 1, and from 1 to $\infty$, then the fraction in the integrand:
$$\frac{1}{x^n + 1}$$
has different series expansions.

17. Dec 25, 2011

dirk_mec1

But if you split this integral into two geometric series how do you solve for the integral on the infinite interval?

And how do I calculate this integral:

$$\int_0^1 \ln(x) e^{-x} (1-x)\ \mbox{d}x$$

I've tried IBP bu end up with:

$$\int_0^1 \ln(x) e^{-x} (1-x)\ \mbox{d}x = -e^{-x} (1-x) \ln(x) ] _0^1 + \int_0^1 e^{-x} \left( \frac{(1-x)}{x} +\ln(x) (-1) \right) \mbox{d}x$$

but what's:
$$e^{-x} \ln(x) ] _0^1$$

18. Dec 25, 2011

Dickfore

No, do IBP, but make it like this:
$$u = \ln(x), dv = e^{-x} ( 1 - x) \, dx$$
What is $v(x)$?

19. Dec 26, 2011

dirk_mec1

Ah I see it now:

$$\int_0^1 \ln(x)(1-x)e^{-x}\ \mbox{d}x = \ln(x)x e^{-x} ]_0^1 - \int_0^1 e^{-x}\ \mbox{d}x = e^{-x}]_0^1 = \frac{1}{e}-1$$

20. Dec 26, 2011