1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.