# Proof of already solved Hard Improper Definite Integral

by Swimmingly!
Tags: definite, hard, improper, integral
 P: 45 1. The problem statement, all variables and given/known data Some friend of mine found this on a book: $$\int_{0}^{+inf}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}]$$ 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. $$I=\int_{0}^{+inf}\frac{1-e^{it\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }= \int_{0}^{+inf}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }$$ $$\frac{d}{dA}I= \int_{0}^{+inf}\frac{\partial }{\partial A}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }=\int_{0}^{+inf}\frac{-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)}$$ 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.
Mentor
P: 19,758
 Quote by Swimmingly! 1. The problem statement, all variables and given/known data Some friend of mine found this on a book: $$\int_{0}^{+\infty}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}]$$
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 by Swimmingly! 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. $$I=\int_{0}^{+\infty}\frac{1-e^{it\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }= \int_{0}^{+\infty}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }$$ $$\frac{d}{dA}I= \int_{0}^{+\infty}\frac{\partial }{\partial A}\frac{1-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)\omega }=\int_{0}^{+\infty}\frac{-e^{A\omega}}{e^{\omega /C}(e^{\omega /T}-1)}$$ 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.
 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: $$\int_{0}^{\infty}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }dω=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}]$$

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