# Checking my values for a complex integral

1. Aug 10, 2008

### majesticman

I am going to provide my answer to a complex integral and i was just seeking a few pointers as to weather i was on the right track or was there something i completely forgot...happens quite a bit...lol

$$\oint exp(z+(1/z))$$ around the path $$\left |z|\right=1$$

now i converted that to a Laurent series....to get

$$\sum ^{inf} _{0} (1/n!) (z+(1/z))^n$$

then using the residue theorem i can have that the integral is equal to 2*pi*i given that b1=1 for taking the series around z=0

am i right???

2. Aug 10, 2008

### Count Iblis

No, you must expand each of the powers and then extract the coefficient of 1/z and add them up. I think the sum of the series can be expressed in terms of the Bessel function of order 1 at -2 i (but I have to check to verify this)

Now, you can find that result directly simply by putting
z = exp(i theta), and integrating from theta = 0 to 2 pi. You then get an integral that is the same up to some factor to the Bessel function of order 1 of imaginary argument.