# Integration of an exponential function

## 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

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

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.

It's like having:

$$\int t(u)e^u du$$

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:

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

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. :)

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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.

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

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

Excuse me, but what is IVP?

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

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

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:
$$T(t=0) = T_0$$

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.

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?

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} \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}

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.

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