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):

enter image description here

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.


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


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!

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

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