Exponential integral transformation

So the second identity is just like the first, except for a phase offset of ##\pi/2##.Also, I just noticed that the first identity I gave is special case (##n=0##) of the second one. So you just need to remember the second one.
  • #1
giokara
9
0
Hello all,
I am searching for an analytic solution to an integral of the following form:

[itex]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[/itex]

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

[itex]\frac{1}{\pi}\int_{0}^{2\pi}e^{jq'\phi}e^{-jk\rho\sin\phi}d\phi[/itex]

Although there seems to be a close link between both expressions, I am unable to transform [itex]I[q',k\rho\,][/itex] 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
 
Physics news on Phys.org
  • #2
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.
 
  • #3
giokara said:
Hello all,
I am searching for an analytic solution to an integral of the following form:

[itex]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[/itex]

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

[itex]\frac{1}{\pi}\int_{0}^{2\pi}e^{jq'\phi}e^{-jk\rho\sin\phi}d\phi[/itex]

Although there seems to be a close link between both expressions, I am unable to transform [itex]I[q',k\rho\,][/itex] 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.
 
Last edited:
  • #4
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
 
  • #5
giokara said:
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:

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

Related to Exponential integral transformation

1) What is the Exponential Integral Transformation?

The Exponential Integral Transformation is a mathematical technique used to transform an exponential function into a simpler form, making it easier to solve or analyze. It is commonly used in fields such as physics, engineering, and finance.

2) How is the Exponential Integral Transformation calculated?

The Exponential Integral Transformation is calculated using the formula Ei(x) = ∫(e^t/t) dt, where the integral is evaluated from t = -∞ to t = x. This is known as the "indefinite integral" of the exponential function.

3) What are the applications of the Exponential Integral Transformation?

The Exponential Integral Transformation has various applications in mathematics, physics, and engineering. It is commonly used in the study of heat transfer, electrical circuits, and diffusion processes. It can also be used to solve differential equations and to calculate probabilities in statistics.

4) What is the relationship between the Exponential Integral Transformation and the Gamma function?

The Exponential Integral Transformation and the Gamma function are closely related, as the Exponential Integral Transformation can be expressed in terms of the Gamma function. Specifically, Ei(x) = -γ - ln|x| - ∫(e^-x/x) dx = -γ - ln|x| - Γ(0,x), where Γ(0,x) is the upper incomplete Gamma function.

5) Are there any limitations to using the Exponential Integral Transformation?

One potential limitation of the Exponential Integral Transformation is that it may not always be possible to find an analytical solution, requiring numerical methods to be used instead. Additionally, the integral may not converge for certain values of x, leading to undefined or complex results.

Similar threads

Replies
3
Views
1K
Replies
3
Views
843
Replies
2
Views
2K
Replies
5
Views
1K
Replies
8
Views
2K
  • Calculus
Replies
3
Views
2K
Replies
14
Views
3K
Replies
4
Views
566
  • Advanced Physics Homework Help
Replies
6
Views
684
Back
Top