Proof of already solved Hard Improper Definite Integral

In summary, the author's friend found an equation that calculates the length of a cosine wave over a given time period. However, the equation is difficult to integrate and may not converge. Several methods were tried, but none of them seemed to work. The equation remains unsolved.
  • #1
Swimmingly!
44
0

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


b460f825d4795dfa285b98575b7c7523.png

3b3bc7b3dad9e055f40c55282b78b5c0.png

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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  • #2
Swimmingly! said:

Homework Statement


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.
Swimmingly! said:
The proof is left for the reader.

Homework Equations


b460f825d4795dfa285b98575b7c7523.png

3b3bc7b3dad9e055f40c55282b78b5c0.png

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.
 
  • #3
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]
 

FAQ: Proof of already solved Hard Improper Definite Integral

1. What is a hard improper definite integral?

A hard improper definite integral refers to a type of mathematical problem that involves finding the area under a curve that is not well-behaved. This means that the curve may have infinite or undefined values at certain points, making it difficult to calculate the integral using traditional methods.

2. How do you know if a hard improper definite integral has been solved?

A hard improper definite integral is considered solved when the calculated value of the integral converges to a finite number. This means that the area under the curve has been accurately determined, despite the challenges posed by the curve's behavior.

3. What techniques are used to solve hard improper definite integrals?

There are several techniques that can be used to solve hard improper definite integrals, such as substitution, integration by parts, and the method of partial fractions. In some cases, advanced techniques like Laplace transforms or contour integration may also be used.

4. Are there any common mistakes to avoid when solving hard improper definite integrals?

One common mistake is forgetting to check for convergence before attempting to solve the integral. It is also important to be aware of any discontinuities or infinite values in the integrand, as these can affect the convergence and solution of the integral.

5. How are hard improper definite integrals used in science?

Hard improper definite integrals are used in various scientific fields, such as physics, engineering, and economics. They are used to calculate important quantities such as work, center of mass, and probability in real-world problems and experiments. They also have applications in signal processing, image analysis, and other areas of data analysis.

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