How to calculate this integral?

[neptunecs]

Hi,How to calculate this integral:

which

I have ever calculated it and the result(R(s)) is ∞,but I don't think it's right,I'm not sure with it.
Thank you for your help!
neptunecs.

 

[Curious3141]

Hi,How to calculate this integral:

which

I have ever calculated it and the result(R(s)) is ∞,but I don't think it's right,I'm not sure with it.
Thank you for your help!
neptunecs.

It'll take forever for the attachments to be approved, why not just use LaTex?

 

[neptunecs]

It'll take forever for the attachments to be approved, why not just use LaTex?

where is LaTex in this forum?I didn't find it.Tell me,please.Thanks.

 

[Curious3141]

https://www.physicsforums.com/showthread.php?t=8997

 

[neptunecs]

R(s)=\int_{-\infty}^{\infty} f(t)f(t+s) dt
which f(t)=Acos(wt+f0)
Thank for Curious3141's help!
neptunecs.

 

[J77]

\int\cos mx \cos nx dx=\frac{\sin(m-n)x}{2(m-n)}+\frac{\sin(m+n)x}{2(m+n)}+c

Of course the c will go because you have limits.

(This integral can be found in the front or back of most Calculus textbooks.)

 

[Curious3141]

R(s)=\int_{-\infty}^{\infty} f(t)f(t+s) dt
which f(t)=Acos(wt+f0)
Thank you for Curious3141's help!
neptunecs.

You're welcome. If you're not that interested in the method, a quick way to get an answer is to try this link : http://integrals.wolfram.com/index.jsp

Just input the integrand as "b Cos[w x+k]b Cos[w(x+l)+k]" since the script uses x as the integrand by default. The constant I changed from A to b because Acos has a special meaning of arccosine. f0 corresponds to k and s corresponds to l.

EDIT : I had previously made a small error in the expression which changes the indef. integral but does not affect the convergence of the integral for the given bounds. Now the integrand expression is correct.

 

[Curious3141]

And you're right, for the bounds given, the integral does not converge.

 

[Pseudo Statistic]

\int\cos mx \cos nx dx=\frac{\sin(m-n)x}{2(m-n)}+\frac{\sin(m+n)x}{2(m+n)}+c

Of course the c will go because you have limits.

(This integral can be found in the front or back of most Calculus textbooks.)

Or he could've converted the integrand to complex form.. :D

 

[neptunecs]

And you're right, for the bounds given, the integral does not converge.

Thank you.
But the integral does not converge seems strange.Maybe my teacher made a mistake.

 

[J77]

I made a mistake in giving you that standard solution due to the phase shifts in your integrands...

Using the mathematica link, the solution should be:

\frac{A^2}{2}\left[t\cos(\omega s)+\frac{\cos(2\omega t)\sin(2f_0+\omega s)}{2\omega}+\frac{\cos(2f_0+\omega s)\sin(2\omega t)}{2\omega}\right]^\infty_{-\infty}

Though you should check it too... the non-convergence thing still stands...

 

[neptunecs]

I made a mistake in giving you that standard solution due to the phase shifts in your integrands...

Using the mathematica link, the solution should be:

\frac{A^2}{2}\left[t\cos(\omega s)+\frac{\cos(2\omega t)\sin(2f_0+\omega s)}{2\omega}+\frac{\cos(2f_0+\omega s)\sin(2\omega t)}{2\omega}\right]^\infty_{-\infty}

Though you should check it too... the non-convergence thing still stands...

Thank very much.This solution is particular.
my teacher said It's non-convergence thing, too.
neptunecs.