Using contour integration techniques, evaluate the integral $$f(t)=\int^{\infty}_{-\infty}d\omega \frac{e^{-i\omega t}}{\omega-\omega_0+i\gamma}$$ where the integral is along the real axis and $t$, $\omega_0$, and $\gamma$ are real constants with $\gamma > 0$. Be sure to consider the integral for both

a) $t>0$ and

b) $t<0$

HINT: Apply the residue theorem around any poles.

So I believe that the function $f(\omega) = \frac{e^{-i\omega t}}{\omega-\omega_0+i\gamma}$ has a simple pole at $\omega = \omega_0 -i\gamma$. Therefore, to find the residue at that pole I use method $$R(z_0)=\lim_{z\to z_0}(z-z_0)f(z)$$ and i get $$R(\omega_0-i\gamma)=\lim_{\omega\to \omega_0-i\gamma}\frac{({\omega-\omega_0+i\gamma})e^{-i\omega t}}{\omega-\omega_0+i\gamma}=e^{-i(\omega_0-i\gamma)t}=e^{-(\gamma+i\omega_0)t}$$ Then by the Residue Theorem $$\oint_C f(z)=2\pi i \sum residues $$ I get $$f(t) = 2\pi i e^{-(\gamma+i\omega_0)t}$$

I don't know if this solution is even close or that I even applied the theorem correctly. I would think that if we're only integrating along the real axis we should get a real valued solution. Also, how can we only integrate along the real axis if the pole is not on the axis? Should i take the real part of this solution ($2\pi e^{-\gamma t}\sin\omega_0 t$)? What about the 2 cases for t? Can i just say that if $t<0$ then $f(t)=2\pi i e^{(\gamma+i\omega_0)t}$? Or is there something that I'm missing entirely?

Thanks for taking the time to help.


1 Answer 1


Okay the residue is computed correctly.

You should use the upper counterclockwise semicircular contour, with radius $\rho$, and center $z=0$, oriented that the base is traversed from left to right. Let the flat contour be $\displaystyle s_1$ and the arc be $s_2$.

Now, I'll prove it later if you ask, but, if $t<0$, then we have $\displaystyle \lim_{\rho \to \infty}\int_{s_2}f(z)\,dz=0$. And so then we must have $\displaystyle \lim_{\rho \to \infty}\int_{s_1}f(z)\,dz=\int_{-\infty}^{\infty}f(z)\,dz=2\pi i e^{-(\gamma+i\omega_0)t}$.

On the other hand, if $t>0$, then $\displaystyle \lim_{\rho \to \infty}\int_{s_2}f(z)\,dz$ does not converge. We must use the lower counterlockwise semicircular contour. No poles are enclosed in this contour, (because $\gamma>0$), and so knowing that the integral over the bottom semicircle is $0$, so is $\displaystyle \int_{-\infty}^{\infty}f(z)\,dz$.

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  • $\begingroup$ I now see that there are no poles in the lower region and so any contour in that region would be zero. But I don't understand the relation between t and which contour we decide to use. Why do we choose the upper for t<0 and the lower for t>0? Thanks for your patience. $\endgroup$ Commented Jul 21, 2019 at 2:33
  • $\begingroup$ Haha okay. replacing $\omega$ with $z$ for my own comfort. So let's note $|e^{-itz}|=|e^{-i(tx+tiy)}=|e^{-tix+ty}|=e^{ty}$. Obviously $e^{ty}$ is bounded in the upper plane if $t<0$, and bounded in the lower plane if $t>0$. Now, your aren't just integrating the e factor, but you're dividing by $\omega-\omega_0+i\gamma$, so that causes the limit as $\rho \to \infty$ to become $0$. i can send you a picture of a proof and a theorem of the book i'm reading right now. we only covered this a few days ago $\endgroup$ Commented Jul 21, 2019 at 2:41
  • $\begingroup$ Give me a couple days to think this over. Physics is my main subject so a pic from a math book probably wouldn't help much. Haha! Thanks for the help, I'll discuss this problem with other students and see if they can fill in the more subtle parts of your replies for me. $\endgroup$ Commented Jul 21, 2019 at 2:50
  • $\begingroup$ the math book explanation isn't too hard, i promise. definitely easier than physics i promise $\endgroup$ Commented Jul 21, 2019 at 3:01
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    $\begingroup$ I added it in anyways $\endgroup$ Commented Jul 21, 2019 at 3:05

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