1. ## help with integration please!!!

Hi all,

I have a problem integrating the following function. I would appreciate if someone can help me. The equation is attached.

I would really appreciate if someone can give me a thorough answer so that I can fully understand the process.

Thanks.

$\displaystyle \int_0^{2\pi} \cos (p\theta) e^{k\cos \theta} d\theta$
Do rather,
$\displaystyle \int_0^{2\pi} e^{ip\theta}e^{k\cos \theta} d\theta$.
Now assuming that $\displaystyle p,k$ are positive integers,
$\displaystyle \oint_{|z|=1} z^p e^{\frac{k}{2}(z+z^{-1})} \frac{1}{iz} \ dz$.

Now apply the residue theorem ...

3. Hi,

Thankss for your reply. However, I cannot see how you got to the second part of your equations. Indeed p and theta are integer values.

I was looking into other methods such as integration by parts etc. Do you think that integration by parts will work?

Thanks

Regards

4. Originally Posted by tecne
Hi,

Thankss for your reply. However, I cannot see how you got to the second part of your equations. Indeed p and theta are integer values.

I was looking into other methods such as integration by parts etc. Do you think that integration by parts will work?

Thanks

Regards
Integration by parts won't work. (At least any substitution I can think of.) TPH's method is the only way I know how to integrate this.

How can $\displaystyle \theta$ be an integer if you are integrating over it??

Oh, and the real part of $\displaystyle e^{i \theta}$ is just $\displaystyle cos(\theta)$, so do TPH's integral, then take the real part of the answer.

-Dan

5. Originally Posted by topsquark
Integration by parts won't work. (At least any substitution I can think of.) TPH's method is the only way I know how to integrate this.

How can $\displaystyle \theta$ be an integer if you are integrating over it??

Oh, and the real part of $\displaystyle e^{i \theta}$ is just $\displaystyle cos(\theta)$, so do TPH's integral, then take the real part of the answer.

-Dan

I apologise. I meant k and p are integers not theta. In particular p=1/2.

TPH stands for what? Since you know how to solve this with The TPH method can you help me to achieve this?

Thanks

6. Originally Posted by tecne
I apologise. I meant k and p are integers not theta. In particular p=1/2.

TPH stands for what? Since you know how to solve this with The TPH method can you help me to achieve this?

Thanks
TPH stands for "ThePerfectHacker." I meant for you to do the problem using his method. Sorry for the confusion.

-Dan

7. The only bad thing about my approach is that you need to compute the residue which is a little time consuming. But it works out. You need to use Laurent series over here. Find the Taylor expansions and take the Cauchy product. I leave those details for you.

What do you not understand about the contour integral?