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I was going through the derivation of the relationship between current density and electric field within the Drude model. Here is the step that I am having trouble to understand:

$$\int_{-\infty}^{\infty}\frac{\mathrm d\omega}{2\pi}\frac{\sigma_{0}}{1-\mathrm i\omega\tau}e^{-\mathrm i\omega \left(t-t^{\prime}\right)} = \frac{\sigma_{0}}{\tau}e^{-\frac{\left(t-t^{\prime}\right)}{\tau}}\Theta\left(t-t^{\prime}\right) $$

Here, $\omega$ is the variable, $\mathrm i$ is the imaginary number, $t$ and $t^{\prime}$ are the dummy variables, $\Theta$ is the Heaviside step function, and the rest of them are constants.

Any thought how this complex integral was evaluated?

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    $\begingroup$ Probably contour integration, one picks up the pole at omega ~1/tau. $\endgroup$ – Your Majesty Oct 28 '16 at 9:32
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Oct 28 '16 at 9:40
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    $\begingroup$ @Qmechanic, I was considering it before posting the question. However, this derivation is related to a solid state physics topic and a physicist who is familiar with frequency-dependent Drude model parameters might be more familiar with this specific integral than a mathematician. $\endgroup$ – M Asgar Oct 28 '16 at 14:36
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The integral in question can be written as a Fourier transform, that is,

$$\mathcal{F}_{\omega \to s} \left\{ \frac{\sigma_0}{2\pi} \frac{1}{1-i\tau \omega}\right\}(s) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \frac{\sigma_0}{1-i\tau \omega}e^{-is\omega}$$

where $s \equiv t-t'$. This has a pole at $\omega = -i/\tau$, and from the Laurent series,

$$f(\omega)= \frac{\sigma_0}{2\pi} \left(\frac{i}{\tau}e^{-s/\tau}(\omega + i/\tau)^{-1} + \frac{s}{\tau}e^{-s/\tau} + \mathcal{O}((\omega+i/\tau)^2) \right)$$

one finds the residue is,

$$\mathrm{Res}(f,-i/\tau) = \frac{i\sigma_0}{2\pi\tau}e^{-s/\tau}.$$

Now consider the same integral, over the original real line contour $C_1$, plus an additional contour $C_2$ which lies in the complex plane, enclosing the pole to form a Jordan curve. We then have,

$$\int_{C_1 + C_2} \frac{d\omega}{2\pi} \frac{\sigma_0}{1-i\tau \omega}e^{-is\omega} = 2\pi i \, \mathrm{Res}(f,-i/\tau) = -\frac{\sigma_0}{\tau}e^{-(t-t')/\tau}.$$

We choose $C_2$ as the circular contour in the lower half plane, with a radius $R$, parametrised as,

$$\omega(\theta) = Re^{-i\theta}, \quad \theta \in [0,\pi].$$

Now, we seek a bound using the estimation lemma for the contour integral, given by,

$$\left| \int_{C_2} d\omega \, f(\omega)\right| \leq \mathcal{l}(C_2) M$$

where $M := \mathrm{max}_{\omega \in C_2} |f(\omega)|$ and $\mathcal{l}(C_2) = \pi R$, the contour length. Solving,

$$\frac{d}{d\theta} |f(Re^{-i\theta})| = 0$$

one finds that there is a maximum at $\theta = \pi/2$ and that,

$$|f(Re^{-i\pi/2})| = \frac{\sigma_0}{2\pi} \frac{e^{-Rs}}{|R\tau - 1|}$$

and thus the integral is bounded by,

$$ \left| \int_{C_2} d\omega \, f(\omega) \right| \leq \frac{\sigma_0 R}{2} \frac{e^{-Rs}}{|R\tau-1|}$$

which tends to zero as $R \to \infty$ providing that $s>0$, that is, $t>t'$. Thus the original integral we wanted to evaluate is given by,

$$\int_{-\infty}^\infty \frac{d\omega}{2\pi} \frac{\sigma_0}{1-i\tau \omega}e^{-is\omega} = -\frac{\sigma_0}{2\pi}e^{-(t-t')/\tau}\Theta(t-t').$$

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  • $\begingroup$ Thank you very much and I appreciate your effort. I am not familiar with some of the steps (my complex algebra is still quite weak). I will get back to you as questions arise. $\endgroup$ – M Asgar Oct 28 '16 at 14:41

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