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Homework Help: Integration of an exponential function

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to integrate the following function:
    [tex]
    y = \int e^{-\frac{n}{\omega}cos \omega t} dt
    [/tex]



    2. Relevant equations

    How should I solve this? Will this lead me to Bessel function?

    2. My attempt

    [tex]
    a=\frac{n}{\omega}
    [/tex]
    [tex]
    u=-a cos (\omega t)
    [/tex]
    [tex]
    du = a\omega sin (\omega t) dt
    [/tex]
    [tex]
    y = \int e^{-\frac{n}{\omega}cos \omega t} dt = \int \frac{1}{a\omega sin(\omega t)} e^u du
    [/tex]
    [tex]
    y = \frac{1}{a\omega sin(\omega t)} \int e^u du
    [/tex]
    [tex]
    y = \frac{1}{a \omega sin (\omega t)} e^u
    [/tex]
    [tex]
    y = \frac{1}{- a \omega sin (\omega t)} e^{-a cos(\omega t)}
    [/tex]

    Is my solution correct?

    Thanks
     
    Last edited: Sep 28, 2010
  2. jcsd
  3. Sep 28, 2010 #2


    When you define [itex]u=-a\cos(w t)[/tex], u becomes a function of t and t is a function of u so you can't move that quantity outside of the integral sign.

    It's like having:

    [tex]\int t(u)e^u du[/tex]

    and then moving the function t(u) outside of the integral which is not permissible. Your integral is a non-elementary integral. That is, there is no elementary function that has as a derivative, the integrand. How about this though: Suppose we have:

    [tex]e^{iz\cos(t)}=\sum_{-\infty}^{\infty}i^n J_n(z)e^{int}[/tex]

    Then I get:

    [tex]\int{e^{-a\cos(wt)}}dt=\frac{1}{w}\left\{J_0(-a/i)+\sum_{\substack{n=-\infty\\n\ne 0}}^{\infty} i^n J_n(-a/i)\frac{1}{in}e^{inwt}\right\}[/tex]

    However, I'm not sure that is correct and will leave it to you to figure out if you're interested. :)
     
    Last edited: Sep 28, 2010
  4. Sep 28, 2010 #3
    I really do not know how did you get that..
    Actually, I got this integral when I try to solve a 1st order inhomogeneous non-linear differential equation with variable coefficients..

    So.. I am stuck with that integration. Suppose that your solution is correct, I really do not know how to employ the initial condition into that solution.
     
  5. Sep 28, 2010 #4
    Please post the IVP and we can go through it if you wish.
     
  6. Sep 28, 2010 #5
    Excuse me, but what is IVP?
     
  7. Sep 28, 2010 #6
    It's "Initial Value Problem". That's the differential equation and initial conditions like this is an IVP:

    [tex]y''+y=0,\quad y(0)=1,\, y'(0)=2[/tex]
     
  8. Sep 28, 2010 #7
    this is the original equation:

    [tex]
    e^{-a cos wt} \ T(t) = m \int e^{-a cos wt} dt + p \int sin wt \ e^{-a cos wt}
    [/tex]

    I want to have solution for [tex]T(t)[/tex]

    [tex]
    e^{-a cos wt} \ T(t) = m \int e^{-a cos wt} dt + p e^{-a cos wt}
    [/tex]

    The initial condition is:
    [tex]
    T(t=0) = T_0
    [/tex]
     
  9. Sep 28, 2010 #8
    That's not a differential (or integral equation). Why not just solve for T(t) but first, don't get the dummy-variable "t" in the integral confused with the "t" in T(t). They are not the same ok. It's better if you change it to say u and for example write:

    [tex]T(t)=e^{a\cos(wt)}\left\{m\int_0^t e^{-a \cos(wu)}\left(m+p\sin(wu)\right)du\right\}[/tex]

    Now, that is just an integral function and there's nothing wrong with just leaving it like that and in this particular case, T(0)=0. And whenever you need to work with actual numerical values, just compute the integral numerically. That which I did above with that infinite sum was just a formality and you probably wouldn't use it for actual applications.
     
  10. Sep 28, 2010 #9
    well it starts with a differential equation, I am using the method from a calculus book written by Adams. My idea was to calculate that "unsolvable" integral part with matlab.

    Thanks for the tips,
    Out of curiosity, how did you get a solution as a series function?
     
  11. Sep 28, 2010 #10
    We're given (see Mathworld on Bessel functions):

    [tex]e^{iz\cos(t)}=\sum_{n=-\infty}^{\infty} i^n J_n(z)e^{int}[/tex]

    and we have:

    [tex]\int e^{-a \cos(wt)}dt=\frac{1}{w}\int e^{-a\cos(u)}du[/tex]

    So let:

    [tex]-a=iz[/tex]

    and then:

    [tex]
    \begin{aligned}
    \frac{1}{w}\int e^{-a\cos(u)}du&=\frac{1}{w}\int e^{iz\cos(u)}du\\
    &=\frac{1}{w}\int \sum_{n=-\infty}^{\infty} i^n J_n(-a/i)e^{inu} du \\
    &=\lim_{T\to\infty}\frac{1}{w}\int \sum_{n=-T}^{T} i^n J_n(-a/i)e^{inu} du
    \end{aligned}
    [/tex]

    where I've taken the sum in it's Fourier sense (that limit) and the results then follows.

    However there may be convergence issues, branch-cut issues, as well as justifying switching the order of summation and integration.

    Edit: Just wanted to check the antiderivative above against numerical results. Below I just checked the integral of e^{icos(t)} over (1,5):

    Code (Text):

    In[113]:=
    tstart = 1;
    tend = 5;
    val1 = NIntegrate[Exp[I*Cos[t]],
       {t, tstart, tend}]
    myf[t_] := N[BesselJ[0, 1]*t +
        Sum[(I^(n - 1)/n)*BesselJ[n, 1]*
          Exp[I*n*t], {n, 1, 10}] +
        Sum[(I^(-n - 1)/(-n))*BesselJ[-n, 1]*
          Exp[(-I)*n*t], {n, 1, 10}]]
    val2 = N[myf[tend] - myf[tstart]]

    Out[115]=
    3.2298547320753195 - 1.5910876205729956*I

    Out[117]=
    3.2298547320752578 - 1.5910876205726603*I
     
    That's not bad for a quick check but of course inadequate in a Real Analysis class.
     
    Last edited: Sep 28, 2010
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