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Hard integrals

  • Thread starter dirk_mec1
  • Start date
  • #1
761
13

Homework Statement


[tex]
\int_0^{\infty} \ln \left( \frac{e^x+1}{e^x-1} \right) \mbox{d}x
[/tex]

[tex]
\int_0^{\infty} \frac{1}{x^n+1}\ \mbox{d}x\ \forall n >1
[/tex]

Homework Equations


-


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.
 

Answers and Replies

  • #2
186
0
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.
 
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  • #3
2,967
5
[tex]
\int_0^{\infty} \ln \left( \frac{e^x+1}{e^x-1} \right) \mbox{d}x
[/tex]
Try to make the substitution:
[tex]
t \equiv \frac{e^x + 1}{e^x - 1}
[/tex]

What integral do you get?
 
  • #4
761
13
I get:

[tex] 2 \cdot \int_1^{\infty} \frac{\ln(t)}{1+t}\ \mbox{d}t [/tex]

but I don't see how to solve this one...
 
  • #5
2,967
5
I get:

[tex] 2 \cdot \int_1^{\infty} \frac{\ln(t)}{1+t}\ \mbox{d}t [/tex]

but I don't see how to solve this one...
Your denominator is wrong.
 
  • #6
Char. Limit
Gold Member
1,204
14
I get:

[tex] 2 \cdot \int_1^{\infty} \frac{\ln(t)}{1+t}\ \mbox{d}t [/tex]

but I don't see how to solve this one...
Are you sure about that? I got something different.
 
  • #7
761
13
My bad:

[tex] 2\cdot \int_1^{\infty} \frac{\ln(t)}{(t+1)(t-1)}\ \mbox{d}t [/tex]

I can use partial fraction decomposition but I'm still stuck then. I've also tried a [tex]t =sin(\theta)[/tex] but the what?
 
Last edited:
  • #8
2,967
5
My bad:

[tex] 2\cdot \int_1^{\infty} \frac{\ln(t)}{(t+1)(t-1)}\ \mbox{d}t [/tex]
Ok, so, now your denominator is [itex]t^2 - 1[/itex]. You want it in the form [itex]e^{\alpha x} - 1[/itex]. That is why you have the additional substitution
[tex]
e^{\alpha x} = t^2 \Rightarrow t = e^{\frac{\alpha \, x}{2}}
[/tex]
where [itex]\alpha[/itex] is yet to be determined.

What do you get now?
 
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  • #10
2,967
5
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
761
13
Is the series I calclulated in the correct form? If so how can I exactly calculate:

[tex] \sum_{n=0}^{\infty} \frac{1}{(n-\frac{1}{2} )(n-\frac{3}{2})} [/tex]
 
  • #12
2,967
5
[tex]
\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}}
[/tex]

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

[tex]
\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}
[/tex]
where
[tex]
\zeta(a) \equiv \sum_{n = 1}^{\infty}{\frac{1}{n^{a}}}, \ \mathrm{Re} a > 1
[/tex]
is the Riemann zeta function.
 
  • #13
761
13
How do you get from the single sum with the (2n-1)^2 in the denominator to the two sums?
 
  • #14
Curious3141
Homework Helper
2,843
87
How do you get from the single sum with the (2n-1)^2 in the denominator to the two sums?

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

Sum of reciprocals of odd positives is sum of reciprocals of naturals minus sum of reciprocals of even positives.
 
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  • #15
761
13
Because the union of the set of all odd positive numbers and the set of even positive numbers equals the set of natural numbers. :wink:

Sum of reciprocals of odd positives is sum of reciprocals of naturals minus sum of reciprocals of even positives.
Yes offcourse thanks guys!

For the second integral the hint is use: u =x^n, I don't understand what to do then.
 
  • #16
2,967
5
no, separate it from 0 to 1, and from 1 to [itex]\infty[/itex], then the fraction in the integrand:
[tex]
\frac{1}{x^n + 1}
[/tex]
has different series expansions.
 
  • #17
761
13
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:

[tex]
\int_0^1 \ln(x) e^{-x} (1-x)\ \mbox{d}x
[/tex]

I've tried IBP bu end up with:

[tex]
\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
[/tex]

but what's:
[tex]
e^{-x} \ln(x) ] _0^1
[/tex]
 
  • #18
2,967
5
No, do IBP, but make it like this:
[tex]
u = \ln(x), dv = e^{-x} ( 1 - x) \, dx
[/tex]
What is [itex]v(x)[/itex]?
 
  • #19
761
13
Ah I see it now:

[tex] \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[/tex]
 
  • #21
761
13
Ok and I''m also stuck with this one:

[tex] 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 [/tex]

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

And I'm this one is also difficult:

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

Is tried rewriting as one sine and [tex]u= \frac{1}{2q} \ln(x) [/tex] for this one but didn't seem get anywhere but is it the correct way to do it?
 
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  • #22
Char. Limit
Gold Member
1,204
14
For that first one, the very first thing I'd try is a u=e^(-x) substitution.
 
  • #23
37
0
My bad:

[tex] 2\cdot \int_1^{\infty} \frac{\ln(t)}{(t+1)(t-1)}\ \mbox{d}t [/tex]

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

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.
 
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  • #24
761
13
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
2,967
5
Hint:
[tex]
1 + u^2 + u^4 + u^6 + u^8 = 0
[/tex]
is a symmetric polynomial equation of order 4 w.r.t. [itex]v = u^2[/itex], which may be further reduced. Divide by [itex]v^2[/itex]:
[tex]
\left( v^2 + \frac{1}{v^2} \right) + \left( v + \frac{1}{v} \right) + 1 = 0
[/tex]
and, then, make a substitution [itex]w = v + 1/v[/itex]. It's square is:
[tex]
w^2 = v^2 + 2 + \frac{1}{v^2}
[/tex]
and we finally arrive at the quadratic equation:
[tex]
w^2 + w - 1 = 0
[/tex]
which has two real solutions:
[tex]
w_{1/2} = \frac{-1 \pm \sqrt{5}}{2}
[/tex]
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:
[tex]
2 v^2 + (1 \mp \sqrt{5}) v + 2 = 0
[/tex]
which has a discriminant:
[tex]
D = -10 \mp 2 \sqrt{5} < 0
[/tex]
so the solutions for v come in complex conjugate pairs.
 

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