# Derviation of bessel function of first kind via contour integration

1. Jun 6, 2009

### jack5322

Hi everyone, I have a question concerning the derivation of the J_0(t). In my book, it states that the inverse laplace transform of (s^2+1)^-1/2 is this function. It gives me a contour to integrate around and derive it. The problem is this: I always get an extra I in the answer. This is the contour: Take an infinite semicircle starting from 1 all the way to negative infinity. Inside the semicircle there is a dogbone contour. Now, since there are no singularities inside the contour, it is zero. Thus, the Integral of the dogbone + the vertical line = 0, because the rest vanishes identically. Next, we divide by 2pi*i to get the old function back because by definition the function equals the vertical integral divided by 2pi*i. After this, we only need to find the limits between the branch points and we can write it as the poisson integral representation of the bessel function. I get an extra i because when I parametrize the dogbone, we have the integrand multiplied by i. What I got for the limits on each side of the contour is exp(-pi*i/2)integrand for the limit on the left and exp(-3pi*i/2) . Thus negative Integral divided by pi is the desired answer. I looked to confirm my answer on wikipedia under bessel functions, scrolled down, and I was a constant off. Where did I go wrong?

2. Jun 7, 2009

### Count Iblis

You should connect the dogbone to the Bromwich contour. At some arbitrary point on the Bromwich, you move to the dogbone, complete the dogbone and then move back to the Bromwich and complete the Bromwich. The integral along the two connection lines will, of course, cancel. But the point is that this makes sure you choose the polar angles on the dogbone correctly.

3. Jun 7, 2009

### jack5322

yes but what difference does that make? can you show me how it fixes the angles and how the arguments are found? I think mine are right but the answer is wrong, how is this possible
?

4. Jun 7, 2009

### Count Iblis

If you do it wrong, it would only yield an erroneous minus sign (because the ambiguity in a square root acan only be a minus sign). In case of the inverse Laplace transform, we are given the function
1/sqrt(s^2+1) for real positive s. There we take the usual positive square root. This then fixes the definition of the complex function 1/sqrt(z^2+1) that we need to use on the right of the imaginary axis.

There are two ways to proceed. Either you do the dogbone contour at once, then you need to place the brach cut in the segment from minus i to i. Or you split it in two and using analytical continuation to define the function on the lft side of the imaginary axis. This has the advantage of not having to bother about the branch cuts.

In the former case you go from minus i to i infinitessimally to the right of the imaginary axis and when you return you are infinitessimally to the left of the inmginary axis. In the later case, you don't need to do that and you can integrate exactly over the imaginary axis.

Anyway, if we do it in the first way, when you go from minus i to i you are integrating the function:

exp(i z t)/sqrt(z^2+1)

the square root is defined such that it agrees with the positive square root on the real axis. This means that the polar angle relative to the point i must be chosen to range from -pi/2 to 3/2 pi and relative to -i to range from -pi/2 to 3/2 pi as well. So, at z = 0, the net phase is
exp(-i pi/2)*exp(i pi/2) = 1. The branch cut of the branch point at i moves from i to below on the imaginary axis. The branch cut of the branch point at -i moves down from -i. Below -i on the imaginary axis, the two branch cuts can be omitted as doing so doesn't make a difference to the produsct of the square roots of z-i and z+i.

So, we see that the integral is:

$$\int_{-i}^{i}\frac{\exp(z t)}{\sqrt{1+z^2}}dz$$

on the way up on the right hand side of the imaginary axis. Putting z = x i gives:

$$i \int_{-1}^{1}\frac{\exp(ix t)}{\sqrt{1-x^2}}dx$$

And we know that we need to take the positive square root here as argued above.

On the way down, all that changes is that the sign of the square root changes. So, we can write that integral as:

$$-i\int_{1}^{-1}\frac{\exp(ix t)}{\sqrt{1-x^2}}dx$$

Changing x to minus x gives:

$$i\int_{-1}^{1}\frac{\exp(-ix t)}{\sqrt{1-x^2}}dx$$

Adding up the first part gives for the total integral:

$$2i\int_{-1}^{1}\frac{\cos(x t)}{\sqrt{1-x^2}}dx$$

Dividing by 2 pi i gives you the inverse Laplace transform:

$$\frac{1}{\pi}\int_{-1}^{1}\frac{\cos(x t)}{\sqrt{1-x^2}}dx$$

You can rewrite this in a more familiar form by substituting x = cos(u):

$$\frac{1}{\pi}\int_{0}^{\pi}\cos\left[t\cos(u)\right]du$$