# 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.
 Quote by Swimmingly! 1. The problem statement, all variables and given/known data Some friend of mine found this on a book: $$\int_{0}^{+\infty}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}]$$
 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: $$\int_{0}^{\infty}\frac{1-cos(\omega t)}{e^{\omega /C}(e^{\omega /T}-1)\omega }dω=ln[\frac{(\frac{T}{C})!}{|(\frac{T}{C}-iTt)!|}]$$