# Complex contour integration, Cauchy's theorem

I am new at this and I am trying to get a hang of complex contour integration.

I would like to use Cauchy's residue theorem to evaluate the following integral (with real values of w):

where k,y > 0 and are real.

I know that in order to solve the line integral in question, by using this method, I need to find the sum of the residues at the poles and multiply it by 2*pi*i. I also know that one can go about computing the residues at the poles in one of two ways; by doing grunt-work derivatives or by doing series expansion.

I would like to take the most simple path, so I want to expand this function into a series. I have been trying to follow the example here: https://en.wikipedia.org/wiki/Residue_theorem, but I am not sure how to expand this in a useful way, that will lead to me find the residues.

Could somebody point me in the right direction?

Thanks in advance.

## 1 Answer

For $$t<0$$, the exponential function $$e^{iwt}$$ exponential decays in the lower-half of the $$w$$-plane.

Moreover, given $$k>0$$ and $$y>0$$, the denominator $$w-k+iy$$ has a simple zero at $$w=k-iy$$, which lies in the fourth quadrant of the $$w$$-plane.

We choose a number $$R>\sqrt{k^2+y^2}$$ and define the function $$I(t;k,y)$$ as

$$I(t;k,y)=\oint_{C_R}\frac{e^{iwt}}{w-k+iy}\,dw$$

where $$C_R$$ is the contour comprised of $$(i)$$ the real line segment from $$-R$$ to $$R$$ and (ii) the semicircular arc $$|w|=R$$, $$\pi\le \arg(w)\le 2\pi$$.

Applying the residue theorem, we can evaluate the contour integral in $$(1)$$ for $$t<0$$ as

$$I(t;k,y)=-2\pi i \text{Res}\left(\frac{e^{iwt}}{w-k+iy}, w=k-iy\right)=-2\pi i e^{i(k-iy)t}$$

where the minus sign accounts for the clockwise orientation of the contour $$C_R$$.

Similarly for $$t>0$$, we close the contour in the upper-half $$w$$-plane in which there is no pole singularity. Hence, we find $$I(t;k,y)=0$$.

Finally, it is straightforward to show that for $$t<0$$, the limit as $$R\to \infty$$ of the contribution from integrating over the semi-circle component of $$C_R$$ to the contour integral is zero. That is to say,

$$\lim_{R\to \infty}\int_{\pi}^{2\pi}\frac{e^{itRe^{i\phi}}}{Re^{i\phi}-k+iy}\,Re^{i\phi}\,d\phi=0$$

Putting it all together, we find

$$I(t;k,y)=-2\pi i e^{(y+ik)t}H(-t)$$

where $$H(\cdot)$$ is the Heaviside function.

And we are done!

• Please let me know how I can improve my answer. I really want to give you the best answer I can. – – Mark Viola Jan 9 at 23:11