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Proof of already solved Hard Improper Definite Integral 
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#1
Dec2912, 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{1cos(\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{1e^{it\omega}}{e^{\omega /C}(e^{\omega /T}1)\omega }= \int_{0}^{+inf}\frac{1e^{A\omega}}{e^{\omega /C}(e^{\omega /T}1)\omega }[/tex] [tex]\frac{d}{dA}I= \int_{0}^{+inf}\frac{\partial }{\partial A}\frac{1e^{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. 


#2
Jan213, 05:42 PM

Mentor
P: 21,216

BTW, the LaTeX code for ∞ is \infty. I replaced your "inf" things throughout your post. 


#3
Jan613, 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{1cos(\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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