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Proof of already solved Hard Improper Definite Integral

by Swimmingly!
Tags: definite, hard, improper, integral
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Swimmingly!
#1
Dec29-12, 08:35 PM
P: 45
1. The problem statement, all variables and given/known data
Some friend of mine found this on a book:
[tex]\int_{0}^{+inf}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}][/tex]
The proof is left for the reader.
2. Relevant equations


3. The attempt at a solution
First very safe step:
cos(ωt)=Re(e^(iωt))

Second.1: A possibility is using a substitution of: x=e^ω But now instead of 1/ω we have 1/ln(x) which is difficult to handle in integration.

Second.2: I tried using derivation under the integral sign which I've basically never used before, assuming the legality of my move. If it=A.
[tex]I=\int_{0}^{+inf}\frac{1-e^{it\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }=
\int_{0}^{+inf}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }[/tex]
[tex]\frac{d}{dA}I=
\int_{0}^{+inf}\frac{\partial }{\partial A}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }=\int_{0}^{+inf}\frac{-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)}[/tex]
Which may not be convergent. At least as ω goes to zero the function goes to infinite. And as it goes to infinite 1/T must be greater than the exponent of the upper part of the fraction. And I still have to integrate with respect to A after.

Alternative methods:
Use of representation by series. Maybe with the help of integration by parts.
Assuming the result is similar to the derivative result and just try differentiating.
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Mark44
#2
Jan2-13, 05:42 PM
Mentor
P: 21,216
Quote Quote by Swimmingly! View Post
1. The problem statement, all variables and given/known data
Some friend of mine found this on a book:
[tex]\int_{0}^{+\infty}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}][/tex]
What's the variable of integration? You omitted it. Is it ##\omega##? If so, the integral should have d##\omega## in it.

BTW, the LaTeX code for ∞ is \infty. I replaced your "inf" things throughout your post.
Quote Quote by Swimmingly! View Post
The proof is left for the reader.
2. Relevant equations


3. The attempt at a solution
First very safe step:
cos(ωt)=Re(e^(iωt))

Second.1: A possibility is using a substitution of: x=e^ω But now instead of 1/ω we have 1/ln(x) which is difficult to handle in integration.

Second.2: I tried using derivation under the integral sign which I've basically never used before, assuming the legality of my move. If it=A.
[tex]I=\int_{0}^{+\infty}\frac{1-e^{it\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }=
\int_{0}^{+\infty}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }[/tex]
[tex]\frac{d}{dA}I=
\int_{0}^{+\infty}\frac{\partial }{\partial A}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }=\int_{0}^{+\infty}\frac{-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)}[/tex]
Which may not be convergent. At least as ω goes to zero the function goes to infinite. And as it goes to infinite 1/T must be greater than the exponent of the upper part of the fraction. And I still have to integrate with respect to A after.

Alternative methods:
Use of representation by series. Maybe with the help of integration by parts.
Assuming the result is similar to the derivative result and just try differentiating.
Swimmingly!
#3
Jan6-13, 02:38 PM
P: 45
Thank you, I didn't know that about latex and I forgot to write dω. ALL INTEGRALS ARE WITH RESPECT TO dω.

The problem is still open. If anyone can help here it is better written:
[tex]\int_{0}^{\infty}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }dω=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}][/tex]


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