# Integration of the function exp(cos(x))

1. Mar 3, 2008

Hi. Im trying to integrate the function exp(a*cos(x)) from zero to infinit, where a is a constant. First Ive rewritten it to Bessel functions of the first kind like this:
exp(z*cos(x))=BesselI(0,z)+sum(BesselI(v,z)*cos(v*x),v=1..infinit)
where BesselI is the modified Bessel function of the first kind. The last term is the sum over v from 1 to infinit and v is an integer. Is this correct?
Secondly Im trying to get Maple to do the integral using the modified Bessel functions, but I cant get it to work. Can anybody help me? Thank you for your time.
Best regards
Anne

2. Mar 4, 2008

### coomast

Gladbo, if the integral is the following:

$$\int_{0}^{\infty}e^{a cos(x)}dx$$

then I believe it is diverging, meaning it is infinite. This can be shown by the following argument.

step 1
Draw a picture with the function $$e^{a cos(x)}$$, you will see that is is periodic with period $$2\cdot \pi$$ and has a lowest value for positive a at $$x=\pi$$ equal to $$e^{-a}$$ or for negative a at $$x=0$$ equal to $$e^{-a}$$. This value is always positive.

step 2
Consider the constant function $$f(x)=\frac{1}{2}e^{-a}$$, valid for all x and lower than the lowest value of the function to be integrated.

step 3
The integral can geometrically be interpreted as the area between the function and the X-axes. A positive value for certain because the function does not go below the X-axes itself. This is also true for the constant function.

step 4
The integral of the constant function from 0 to infinity is now:

$$\int_{0}^{\infty}\frac{1}{2}e^{-a}dx= \frac{1}{2}e^{-a}\int_{0}^{\infty}dx= \frac{1}{2}e^{-a}\left[ x \right]_{0}^{\infty}= \infty$$

It is obviously diverging to infinity.

step 5
Because the function you would like to integrate is always larger then the constant one, it's area will also be larger and because the integral for the smaller one is infinity it will certainly be so for the larger one.

Hope this helps, it is not written with firm math, only in words, but I am convinced that the argument is clear.

3. Mar 4, 2008

### Mute

That method works because of that theorem that goes like "if $f(x) < g(x)$ for all x, then $\int_a^b dx~f(x) < \int_a^b dx~g(x)$".

You can also show that the integral is divergent as follows: the integrand is indeed $2\pi$ periodic, so if you write it as a sum of integrals over intervals of $2\pi$:

$$\int_0^{\infty}dx~e^{a\cos x} = \sum_{k=0}^{\infty} \int_{2k\pi}^{2(k+1)\pi}dx~e^{a\cos x}$$

then because you are integrating over an entire period in each case, in each of these integrals you could change variables to get the bounds of the integral to be 0 and $2\pi$. Because the integrand is 2pi periodic, all of the integrands are the same:

$$\int_0^{\infty}dx~e^{a\cos x} = \sum_{k=0}^{\infty} \int_{0}^{2\pi}dx~e^{a\cos x}$$

The summand now no longer depends on k (and can be evaluated numerically to be about 7.955 for a = 1 - possibly even analytically by contour integration, but it really won't matter in a second), and so what you have is

$$\left(\int_{0}^{2\pi}dx~e^{a\cos x}\right)\sum_{k=0}^{\infty}1$$

which diverges.

Of course, perhaps there is a result in terms of distributions, since the integral over cos(x) from -\infty to \infty is a sum of delta functions at +/- 2pi.

Last edited: Mar 4, 2008
4. Mar 5, 2008

Thank you very much both of you!

5. Apr 20, 2008

### udara

I have same problem but my limits are from 0 to pi/2 is any body know how to do this

Last edited: Apr 20, 2008
6. Apr 21, 2008

### Philcorp

Smells like Bessel functions. Go to the library and look for a fat russian book, called "table of integrals series and products". It will never let you down.

7. Apr 21, 2008

### Pere Callahan

:rofl:

8. Apr 22, 2008

### udara

Thank you

Thanks both of you

Udara