Thread: integrate Sin (z Sin t) , etc

1. integrate Sin (z Sin t) , etc

hi all, i need some help to solve this problem..
i knew we have to make the equation with jacobi-anger expansion,
so that we get the series form of that equation, and then integrate it

but i cant find the exact jacobi-anger expansion in any web for sin(z sin t) sin(z cos t) cos(z sin t) sin(z sin t)
i tried wiki but i think it doesnt correct

so, please help me find the RIGHT expansion, or any of you guys have a different method to solve it?

thank you very much

2. Integrate sin(z sin(t)) with respect to what variable?

3. Originally Posted by HallsofIvy
Integrate sin(z sin(t)) with respect to what variable?
ow, sorry.. my mistake.

integrate it with respect variable t

4. Originally Posted by romanistarz
ow, sorry.. my mistake.

integrate it with respect variable t
There is no answer in terms of standard mathematical functions.

5. whoa, really?? so i cant solve this integration??

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| Sin (z Sin t) dt
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i knew a way that we have to change it to a kind of Series first,
and then we can integrate it..

6. Ok, let's try that. Gotta' sometimes just jump in there you know:

$\displaystyle \int \sin(z\sin(t))dt=\int \sum_{k=0}^{\infty}\frac{(-1)^n z^{2n+1}\sin^{2n+1}(t)}{(2n+1)!}dt$

Now, I haven't checked by hand, but Mathematica can evaluate:

$\displaystyle \int \sin^{n}(x)dx$ in closed form:

$\displaystyle \int \text{Sin}[t]{}^{\wedge}n dt= -\text{Cos}[t] \text{Hypergeometric2F1}\left[\frac{1}{2},\frac{1-n}{2},\frac{3}{2},\text{Cos}[t]^2\right] \text{Sin}[t]^{1+n} \left(\text{Sin}[t]^2\right)^{\frac{1}{2} (-1-n)}$

7. sorry, but i cant get the idea.. please explain your idea...

or anyone can give different solution??

8. If you look up the Jacobi - Anger expansion, you get this page Jacobi - Anger expansion. At the bottom is the expansion of $\displaystyle sin(z sin(t))$:

$\displaystyle sin(z sin(t)) = 2 \sum_{n=1}^{\infty} J_{2n-1}(z) sin[(2n-1)t]$

Do you know how to go from the left hand side to the right hand side? If you do, then do that. Then you need to integrate the series with respect to t, which shouldn't be too difficult.

9. Originally Posted by Vlasev
If you look up the Jacobi - Anger expansion, you get this page Jacobi - Anger expansion. At the bottom is the expansion of $\displaystyle sin(z sin(t))$:

$\displaystyle sin(z sin(t)) = 2 \sum_{n=1}^{\infty} J_{2n-1}(z) sin[(2n-1)t]$

Do you know how to go from the left hand side to the right hand side? If you do, then do that. Then you need to integrate the series with respect to t, which shouldn't be too difficult.
that's the problem , i cant make the left side into the right side...
so i can't be sure the series form in wiki are right..

10. Ahh, I see. Well, what have you tried so far? What are some obvious identities that you'll have to use?

11. well, i haven't tried anything but to make the left side into the right side in here Jacobi - Anger expansion
and so far i couldn't get anything..

i would appreciate whoever can make the equation in the left side into the right side..

12. As a start let's show the correctness of some of the formulas on that page. Let's start with the following identity:

$\displaystyle e^{\frac{z}{2}(t-1/t)} = \sum\limits_{n=-\infty}^{\infty}t^{n}J_n(z)$

It is correct. Now we want to make the left hand side look like the one provided in the article, so we want to make:

$\displaystyle \frac{1}{2}(t-1/t) = i\cos(\theta)$

Note that$\displaystyle \frac{1}{2}(t-1/t) = \frac{t - t^{-1}}{2}$ which looks a lot like the complex definition for cosine, except that we need t to be $\displaystyle e^{i\phi}$ and instead of a minus in the middle, we need a plus. After a few tries, just plug in $\displaystyle i e^{i\phi}$, because we need a factor of $\displaystyle i$, in order to obtain the desired exponent.

$\displaystyle e^{\frac{z}{2}(t-1/t)} = exp(z \frac{t - t^{-1}}{2}) = exp(z\frac{i e^{i \theta} - (i e^{i \theta})^{-1}}{2}) = exp(i z\frac{ e^{i \theta} + e^{-i \theta}}{2}) = e^{i z \cos(\theta)}$

Here we've used that $\displaystyle i^{-1} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$ and hence the plus sign between the exponential functions. Now, let's plug in $\displaystyle t = i e^{i\theta}$ in the doubly infinite sum:

$\displaystyle \sum\limits_{n=-\infty}^{\infty}t^{n}J_n(z) = \sum\limits_{n=-\infty}^{\infty}(ie^{i\theta}})^{n}J_n(z) = \sum\limits_{n=-\infty}^{\infty}i^ne^{i n\theta}}J_n(z)$

Thus we have the first identity $\displaystyle e^{i z \cos(\theta)} = \sum\limits_{n=-\infty}^{\infty}i^n J_n(z) e^{i n\theta}}$

For reasons I will explain later, we need to find an analogous identity for $\displaystyle e^{i z \sin(\theta)}$(It has to do with the complex definitions of sine and cosine). Let's use the initial identity. This time, however, we want to make $\displaystyle \frac{1}{2}(t-1/t) = i\sin(\theta)$, so just use $\displaystyle t = e^{i\theta}$ to get:

$\displaystyle e^{i z \sin(\theta)} = \sum\limits_{n=-\infty}^{\infty} J_n(z) e^{i n\theta}}$

13. For the next step, we need to exploit the symmetry of the Bessel function $\displaystyle ( J_{-n}(z) = (-1)^n J_n(z))$ for any integer n. Take the doubly infinite series and split it into 2 parts:

$\displaystyle \sum\limits_{n=-\infty}^{\infty}i^n J_n(z) e^{i n\theta}} = \sum\limits_{n=-\infty}^{-1}i^n J_n(z) e^{i n\theta}} + J_0(z) + \sum\limits_{n=1}^{\infty}i^n J_n(z) e^{i n\theta}}$

Lets work on the first sum. We need to flip the limits to make them look like the ones on the right sum. We do this by flipping the sign on every n in the summand:

$\displaystyle \sum\limits_{n=-\infty}^{-1}i^n J_n(z) e^{i n\theta}} = \sum\limits_{n=1}^{\infty}i^{-n} J_{-n}(z) e^{-i n\theta}}$

Similarly to before, $\displaystyle i^{-n} = \frac{1}{i^n} = \frac{i^n}{i^{2n}} = \frac{i^n}{(-1)^{n}} = (-1)^ni^n}$. Then, using the Bessel function identity, we get:

$\displaystyle \sum\limits_{n=1}^{\infty}i^{-n} J_{-n}(z) e^{-i n\theta}} = \sum\limits_{n=1}^{\infty}(-1)^n i^n (-1)^n J_{n}(z) e^{-i n\theta}} = \sum\limits_{n=1}^{\infty}i^n J_{n}(z) e^{-i n\theta}}$

Now we are ready to combine the two sums:

$\displaystyle \sum\limits_{n=1}^{\infty}i^n J_{n}(z) e^{-i n\theta} + J_0(z) + \sum\limits_{n=1}^{\infty}i^n J_n(z) e^{i n\theta} = J_0(z) + \sum\limits_{n=1}^{\infty}i^n J_n(z)( e^{i n\theta}+e^{-i n\theta})$

$\displaystyle = J_0(z) + \sum\limits_{n=1}^{\infty}i^n J_n(z) 2cos(n\theta) = J_0(z) + 2\sum\limits_{n=1}^{\infty}i^n J_n(z) cos(n\theta) = e^{iz\cos(\theta)}$

In a similar fashion, we can derive the other identity and I suggest you do it as practice. It is:

$\displaystyle e^{iz\sin(\theta)} = J_0(z) + 2\sum\limits_{n=1}^{\infty} J_{2n}(z)\cos(2n\theta) + 2i\sum\limits_{n=1}^{\infty} J_{2n-1}(z) \sin[(2n-1)\theta]$ (I'm not 100% sure my derivation and expression are correct)

14. Now, finally, let's derive the expression for $\displaystyle \sin(z\sin(\theta))$. We need to use the complex number definition for sin(t). Then we plug in $\displaystyle t = z\sin(\theta)$:

$\displaystyle \sin(t) = \frac{1}{2i}(e^{i t} - e^{-i t}) = \frac{1}{2i}(e^{i z\sin(\theta)} - e^{-i z\sin(\theta)}) = \frac{1}{2i}(e^{i z\sin(\theta)} - e^{-i z\sin(-\theta)})$

Now, you see that we need to use the second series to expand this.

$\displaystyle e^{i z\sin(\theta)} - e^{i z\sin(-\theta)}) =$

$\displaystyle = J_0(z) + 2\sum\limits_{n=1}^{\infty} J_{2n}(z)\cos(2n\theta) + 2i\sum\limits_{n=1}^{\infty} J_{2n-1}(z) \sin[(2n-1)\theta] -J_0(z) - 2\sum\limits_{n=1}^{\infty} J_{2n}(z)\cos(-2n\theta) - 2i\sum\limits_{n=1}^{\infty} J_{2n-1}(z) \sin[-(2n-1)\theta]) =$

$\displaystyle = 2\sum\limits_{n=1}^{\infty} J_{2n}(z)\cos(2n\theta) + 2i\sum\limits_{n=1}^{\infty} J_{2n-1}(z) \sin[(2n-1)\theta] - 2\sum\limits_{n=1}^{\infty} J_{2n}(z)\cos(2n\theta) + 2i\sum\limits_{n=1}^{\infty} J_{2n-1}(z) \sin[(2n-1)\theta]) =$

Note that $\displaystyle \sin(-\theta) = -\sin(\theta)$ and $\displaystyle \cos(-\theta) = \cos(\theta)$.

$\displaystyle = 4i\sum\limits_{n=1}^{\infty} J_{2n-1}(z) \sin[(2n-1)\theta]$

Let's not forget the factor 1/(2i). After the division, we get

$\displaystyle \sin(z\sin(\theta)) = 2\sum\limits_{n=1}^{\infty} J_{2n-1}(z) \sin[(2n-1)\theta]$

15. whoa man, thank you so very much...
it really helps me,

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