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## Homework Statement

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.

## Homework Equations

## 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.