# Complex integral via complex contour

1. Jul 12, 2009

### ab959

Hi, I am using the Fourier transform to price a European put option. I have obtained the following integral:

$$q(x,\tau)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{-[\omega^2-i\omega(\gamma-3)+2(\gamma-1)]\tau}}{(1-i\omega)(2-iw)}d\omega$$

which I need to solve. I have solved this through the use of the convolution theorem, finding the inversion for

$$G(x)=e^{-[\omega^2-i\omega(\gamma-3)+2(\gamma-1)]\tau}}$$ and

$$\frac{1}{(1-i\omega)(2-iw)}$$

To obtain the result

$$q(x,\tau)=e^{-\gamma\tau+x} N(-d_1)-e^{2x} N(-d_2)$$ with
$$d_{1,2}=\frac{(\gamma\mp1)\tau+x}{\sqrt{2\tau}}$$

Where N(*) corresponds to the cumulative normal distribution.

I want to be able to obtain this result not using the convolution theorem but rather constructing a complex contour. I am having difficulty. I attempted to construct a rectangular contour connecting the real axis between -R and R, the line Im(w)=-3/2 between -R and R and the sides of the rectangle closing off the two lines. I then took R->infinity. However when I calculate 2*pi*i Residue(w=-i) I obtain a complex number which doesnt make sense... I am not sure what I am doing wrong.

Any suggestions would be very welcome. It is also interesting to note that the Residues (not 2*pi*i*residues) at the two poles (w=-i, w=-2i) correspond to the coefficients in the solution of N(-d1) and N(-d_2).