# Homework Help: 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

### Dickfore

21. Jan 3, 2012

### dirk_mec1

Ok and I''m also stuck with this one:

$$I = \int_{0}^{\infty} \frac{{{e^{ - x}}\left( {1 - {e^{ - 6x}}} \right)}}{{x\left( {1 + {e^{ - 2x}} + {e^{ - 4x}} + {e^{ - 6x}} + {e^{ - 8x}}} \right)}}dx$$

I suspect an smart subtitution will do(since IBP fails here, right?).

And I'm this one is also difficult:

$$I = \int\limits_0^1 {\frac{{\sin \left( {p\ln x} \right) \cdot \cos \left( {q\ln x} \right)}}{{\ln x}}}dx$$

Is tried rewriting as one sine and $$u= \frac{1}{2q} \ln(x)$$ for this one but didn't seem get anywhere but is it the correct way to do it?

Last edited: Jan 3, 2012
22. Jan 3, 2012

### Char. Limit

For that first one, the very first thing I'd try is a u=e^(-x) substitution.

23. Jan 3, 2012

### HACR

I think the sign is wrong. oops never mind. ok, to compensate, how to partial decompose ((u^3+1)(u^3-1))/(ln(u)(1+u^2+u^4+u^6+u^8)) might be a key.

Last edited: Jan 3, 2012
24. Jan 5, 2012

### dirk_mec1

yes I was of course immediately thinking about u=e^{-x} the problem is then that the partial fraction decomposition gets tough or not?

25. Jan 5, 2012

### Dickfore

Hint:
$$1 + u^2 + u^4 + u^6 + u^8 = 0$$
is a symmetric polynomial equation of order 4 w.r.t. $v = u^2$, which may be further reduced. Divide by $v^2$:
$$\left( v^2 + \frac{1}{v^2} \right) + \left( v + \frac{1}{v} \right) + 1 = 0$$
and, then, make a substitution $w = v + 1/v$. It's square is:
$$w^2 = v^2 + 2 + \frac{1}{v^2}$$
and we finally arrive at the quadratic equation:
$$w^2 + w - 1 = 0$$
which has two real solutions:
$$w_{1/2} = \frac{-1 \pm \sqrt{5}}{2}$$
Then, for each solution for w, we have a quadratic equation w.r.t. v, which, in turn, gives 2 solutions. You have to further take the square root to get the 8 solutions for u.

Just to get you started, the equations for v are:
$$2 v^2 + (1 \mp \sqrt{5}) v + 2 = 0$$
which has a discriminant:
$$D = -10 \mp 2 \sqrt{5} < 0$$
so the solutions for v come in complex conjugate pairs.