Evaluating Definite Integrals: A Comparison and Explanation

[Mandelbroth]

I finally got the answer from Math Forum by Mathman

v=\pi-u\;\Rightarrow \;u=\pi-v,\;du=-dv

\int_{0}^{\pi}e^{jx\cos (u)}\cos(mu)du=-\int_{\pi}^{0}e^{jx\cos (\pi-v)}\cos(m\pi-mv)dv=\int_{0}^{\pi}e^{-jx\cos (v)}(-1)^m\cos(mv)dv

\theta=v\;\Rightarrow \;\int_{0}^{\pi}e^{-jx\cos (v)}(-1)^m\cos(mv)dv=\int_{0}^{\pi}e^{-jx\cos (\theta)}(-1)^m\cos(m\theta)d\theta

This is what I am looking for, a step by step proof that the two change of variable give the same exact answer. I know there got to be a way to proof identity.

Your first post in this thread proved the identity, actually.

 

[yungman]

Your first post in this thread proved the identity, actually.

Yeh, hind sight show I was close! But I am not that good in math, still have problem twisting the substitution around every which way to see it.

It's one thing working through the class and textbooks, it's another thing to get into this Bessel function derivation that really twisting the substitution every which way! I am just glad I finally got my derivations for the Bessel functions needed for me to go back to my antenna theory. This and electromagnetics really take the calculus, ODE, PDE and numerical analysis through the ringer. Been stuck for two week in here struggling through this and I am only on chapter 5 of the antenna theory! One thing for sure, I will be a lot better off in math after this...and I don't know whether I am laughing or crying at this point!

 

[andrien]

In another thread in which you asked about how to derive a different representation of bessel function,I told there that you should show it by yourself.The point is that only even m will contribute,all integrals involving odd m will vanish.For that you can use the expansion of cos[xcos(theta)].The thread is some below in this forum.

 

[yungman]

The thread is some below in this forum.

What do you mean?

 

[andrien]

What do you mean?

I mean the thread is some below in this section of the forum.Also if you will use expansion of Sin(xCosθ),you will find that only odd m contribute and the extra minus sign outside Sin(xCosθ) will do the job.Just show it.

 

[yungman]

I mean the thread is some below in this section of the forum.Also if you will use expansion of Sin(xCosθ),you will find that only odd m contribute and the extra minus sign outside Sin(xCosθ) will do the job.Just show it.

What do you mean by some below in this section?

 

[andrien]

I mean this one
https://www.physicsforums.com/showthread.php?t=702141
My horrible english.

 

[yungman]

I mean this one
https://www.physicsforums.com/showthread.php?t=702141
My horrible english.

Yes, I just visit and responded to the post. It's been a long road on this problem and I am glad I finally got my answer.

Thanks