Integrating exp(k1*cos (t + k2)) from 0 to 2*pi

  • #1

Main Question or Discussion Point

Hi

I'm trying to compute the following integral (in LaTeX notation; * denotes multiplication)

\int_0^{2\pi} exp (k_1 * cos (t + k_2)) d t

with k_1 and k_2 being known constants. Furthermore k_2 is between 0 and 2 pi.

From Wikipedia [1] I get the following formula

\int_0^{2\pi} exp (k * cos (t)) d t = 2\pi I_0 (k)

where I_0 is a Bessel function of the first kind. I must admit I can't figure if I can change my original problem into one that is solvable using the equation from Wikipedia. If not, does anybody have any ideas on how to solve this?

Thanks
Søren

[1] http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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Maybe I'm missing something, but if you substitute t' = t + k2, aren't you done already?

By the way, since you seem to know LaTeX, this forum supports it. You can use tex (or itex, for inline) tags like so (click on the formula or quote my post to view the source):
[tex] \int_0^{2\pi} \exp (k_1 * cos (t + k_2)) d t [/tex]
Or inline: [itex] \int_0^{2\pi} \exp (k_1 * cos (t + k_2)) d t [/itex]
 
  • #3
What confused me was that if I make the substitution then the limits also change. This is, however, irrelevant as the cosine is periodic; the only thing that matters is that the length of the interval over which the integration is performed is 2 pi. I'm sorry for not seeing this myself. I just got lost in the details and forgot to look at the actual contents of the equation.

Thanks a bunch,
Søren

P.S. Cool with the LaTeX trick. I actually looked for this functionality, but I couldn't find it.
 

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