Exponential integral transformation

[giokara]

Hello all,
I am searching for an analytic solution to an integral of the following form:

$I[q',k\rho\,]=\frac{1}{\pi}\int_{0}^{2\pi}e^{jq'(\phi-\phi_0)}e^{-jk\rho\sin(\phi-\phi_0)}d\phi$

In this equation, $q'$ is real and $k\rho$ is real and positive.
Also, the following integral is closely related to the definition of Anger and Weber functions:

$\frac{1}{\pi}\int_{0}^{2\pi}e^{jq'\phi}e^{-jk\rho\sin\phi}d\phi$

Although there seems to be a close link between both expressions, I am unable to transform $I[q',k\rho\,]$ in order to use the known expression for the second integral. The reason is that the period of the exponential in the first expression is arbitrary, which does not allow a simple translation of the integrand. Has someone any ideas how to tackle this problem?

Lots of thanks in advance,
Giorgos

 

[JJacquelin]

Hi !

The relevant special functions are the "Incomplete Bessel Functions"
With this key word, one can find on the WEB publications and methods for evaluation.

 

[Mute]

Hello all,
I am searching for an analytic solution to an integral of the following form:

$I[q',k\rho\,]=\frac{1}{\pi}\int_{0}^{2\pi}e^{jq'(\phi-\phi_0)}e^{-jk\rho\sin(\phi-\phi_0)}d\phi$

In this equation, $q'$ is real and $k\rho$ is real and positive.
Also, the following integral is closely related to the definition of Anger and Weber functions:

$\frac{1}{\pi}\int_{0}^{2\pi}e^{jq'\phi}e^{-jk\rho\sin\phi}d\phi$

Although there seems to be a close link between both expressions, I am unable to transform $I[q',k\rho\,]$ in order to use the known expression for the second integral. The reason is that the period of the exponential in the first expression is arbitrary, which does not allow a simple translation of the integrand. Has someone any ideas how to tackle this problem?

Lots of thanks in advance,
Giorgos

Do you mean you can't transform the first into the second because the phase $\phi_0$ is arbitrary? Unless $q'(\phi-\phi_0)$ were denoting an arbitrary function, which it looks like it isn't, it looks like the arguments have the same period.

Anywho, there's a nice identity for exponentials of sines or cosines that relates them to Bessel functions:

$$e^{z\cos \theta} = I_0(z) + 2\sum_{n=1}^\infty I_n(z) \cos(n\theta),$$
where $I_n(z)$ is the modified Bessel function of the first kind.

There is a related identity for the regular Bessel function J:

$$e^{iz\sin\theta} = \sum_{n=-\infty}^\infty J_n(z) e^{in\theta}.$$

EDIT: It just occurred to me that your "j" is probably the imaginary unit and you are using an engineering notation, which means you should use the second identity (in which the imaginary unit is written as "i", in standard math and physics notation).

With these identities you may be able to write the integral in a more convenient form, perhaps relating it to the functions JJacquelin mentioned.

 

[giokara]

Hi

Thanks for the hints, it has helped me a lot. I hadn't heard about the incomplete Bessel functions until today. However, I think the identity for the regular Bessel functions is easier to use in this case, as it results in an integral of a single exponential function (I was indeed using the engineer notation for the complex variable).

Thanks again!
Giorgos

 

[Mute]

Hi

Thanks for the hints, it has helped me a lot. I hadn't heard about the incomplete Bessel functions until today. However, I think the identity for the regular Bessel functions is easier to use in this case, as it results in an integral of a single exponential function (I was indeed using the engineer notation for the complex variable).

Thanks again!
Giorgos

Hey, I want to point out that I made a typo in the second identity in my earlier post (which is corrected now). I wrote cosine where I should have written sine:

$$e^{iz\sin\theta} = \sum_{n=-\infty}^\infty J_n(z) e^{in\theta}.$$