Integration of an exponential function

[bancux]

Homework Statement

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

Homework Equations

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

2. My attempt

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

Is my solution correct?

Thanks

 

[jackmell]

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

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.<br /> <br /> It's like having:<br /> <br /> $$\int t(u)e^u du$$<br /> <br /> 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:<br /> <br /> $$e^{iz\cos(t)}=\sum_{-\infty}^{\infty}i^n J_n(z)e^{int}$$<br /> <br /> Then I get:<br /> <br /> $$\int{e^{-a\cos(wt)}}dt=\frac{1}{w}\left\{J_0(-a/i)+\sum_{\substack{n=-\infty\\n\ne 0}}^{\infty} i^n J_n(-a/i)\frac{1}{in}e^{inwt}\right\}$$<br /> <br /> However, I'm not sure that is correct and will leave it to you to figure out if you're interested. :)[/itex]

 

[bancux]

Suppose we have:

$$e^{iz\cos(t)}=\sum_{-\infty}^{\infty}i^n J_n(z)e^{int}$$

Then I get:

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

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.

 

[jackmell]

Please post the IVP and we can go through it if you wish.

 

[bancux]

Please post the IVP and we can go through it if you wish.

Excuse me, but what is IVP?

 

[jackmell]

It's "Initial Value Problem". That's the differential equation and initial conditions like this is an IVP:

$$y''+y=0,\quad y(0)=1,\, y'(0)=2$$

 

[bancux]

this is the original equation:

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

I want to have solution for $$T(t)$$

$$e^{-a cos wt} \ T(t) = m \int e^{-a cos wt} dt + p e^{-a cos wt}$$

The initial condition is:
$$T(t=0) = T_0$$

 

[jackmell]

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:

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

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.

 

[bancux]

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?

 

[jackmell]

We're given (see Mathworld on Bessel functions):

$$e^{iz\cos(t)}=\sum_{n=-\infty}^{\infty} i^n J_n(z)e^{int}$$

and we have:

$$\int e^{-a \cos(wt)}dt=\frac{1}{w}\int e^{-a\cos(u)}du$$

So let:

$$-a=iz$$

and then:

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

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):

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.