Integration of an exponential function

  • Thread starter bancux
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  • #1
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Homework Statement



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



Homework 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:

Answers and Replies

  • #2
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53


[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]


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:
  • #3
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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. :)

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.
 
  • #4
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53
Please post the IVP and we can go through it if you wish.
 
  • #5
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Please post the IVP and we can go through it if you wish.

Excuse me, but what is IVP?
 
  • #6
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53
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]
 
  • #7
12
0
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]
 
  • #8
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53
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.
 
  • #9
12
0
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?
 
  • #10
1,800
53
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:
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:

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