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Could you please help solving the following integral,

\begin{equation} \int_{-\infty}^{\infty}{d\omega \frac{\gamma (\gamma+2i\omega)}{(\gamma-2i\omega)^3}} e^{-i\omega t} \end{equation} I could solve it analytically in mathematica but the solution isn't very intelligible.

Thank you

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  • $\begingroup$ This does not look like a contour integral... $\endgroup$ – Fabian Dec 21 '16 at 17:51
  • $\begingroup$ depending on the sign of $t$, close the integral in the upper or lower half plane. then calculate the corresponding residues and you are done $\endgroup$ – tired Dec 21 '16 at 18:47
  • $\begingroup$ It's been a while since I took a complex analysis class. My guess solution to the integral was $2\pi i \gamma* 2\gamma e^{-\gamma/2*t}$ for t>0 but wasn't sure if the residual was correct. Could you please elaborate your solution @tired $\endgroup$ – crossingsymmetry Dec 21 '16 at 19:02
  • $\begingroup$ Is that a convolution in the integrand? $\endgroup$ – GFauxPas Dec 21 '16 at 19:18
  • $\begingroup$ @GFauxPas Why did you ask that? It is not a convolution. $\endgroup$ – crossingsymmetry Dec 22 '16 at 2:12
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First, we rewrite the integral slightly.

\begin{align} I:&=\int_{-\infty}^\infty d\omega\frac{\gamma(\gamma + 2i\omega)}{(\gamma-2i\omega)^3}e^{-i\omega t}\\ &=\int_{-\infty}^\infty d\omega\frac{\gamma(\gamma + 2i\omega)}{-8i^3\left(\omega - \frac{\gamma}{2i}\right)^3}e^{-i\omega t}\\ &=\int_{-\infty}^\infty d\omega\frac{\gamma(\gamma + 2i\omega)}{8i\left(\omega+i\frac{\gamma}{2}\right)^3}e^{-i\omega t}\\ &=-\frac{i}{8}\int_{-\infty}^\infty d\omega\frac{\gamma(\gamma + 2i\omega)}{\left(\omega+i\frac{\gamma}{2}\right)^3}e^{-i\omega t}\\ &=-\frac{i}{4}\int_{-\infty}^\infty d\omega\frac{\alpha(\alpha+i\omega)}{(\omega + i\alpha)^3}e^{-i\omega t}\\ &=\frac{1}{4}\int_{-\infty}^\infty d\omega\frac{\alpha(\omega-i\alpha)}{(\omega + i\alpha)^3}e^{-i\omega t} \tag1 \end{align}

where I have set $\alpha = \frac{\gamma}{2}$ in the last step for simplicity's sake.

We want to turn this into a contour integral so that we can use the residue theorem. The easiest contour would be one of which the integral along the real axis is a part of, where the integrals along the other parts of the contour evaluate to $0$.

To do this, first consider the case that $t<0$.

Let $\Gamma$ be a contour that runs along the real axis from $-R$ to $R$ and is then closed along the arc of a semi-circle of radius $R$, centred at the origin, in the upper half plane.

Then

\begin{align} \oint_\Gamma dz\frac{\alpha(z-i\alpha)}{(z + i\alpha)^3}e^{-iz t}=\int_{-R}^R d\omega\frac{\alpha(\omega-i\alpha)}{(\omega + i\alpha)^3}e^{-i\omega t} + \int_\text{Arc} dz\frac{\alpha(z-i\alpha)}{(z + i\alpha)^3}e^{-iz t} \tag2 \end{align}

We can calculate the integral on the LHS of $(2)$ using the Residue Theorem easily - as the integrand has no poles in the upper half plane, it simply evaluates to $0$.

As we take the limit $R\rightarrow\infty$, the positive imaginary part of $z$ along the semi-circle arc ensures that the second integral of the RHS of $(2)$ goes to $0$.

Thus, for $t<0$, we have

\begin{align} \int_{-\infty}^\infty d\omega\frac{\alpha(\omega-i\alpha)}{(\omega + i\alpha)^3}e^{-i\omega t} = 0 \tag3 \end{align}

Now let $t>0$.

Let $\Psi$ now be a contour along the real axis from $-R$ to $R$ so that $|R|>\alpha$ and then closed along the arc of a semi-circle of radius $R$, centred at the origin, in the lower half plane.

Since this contour is now traversed in clockwise direction, we have

\begin{align} -\oint_\Psi dz\frac{\alpha(z-i\alpha)}{(z + i\alpha)^3}e^{-iz t}=\int_{-R}^R d\omega\frac{\alpha(\omega-i\alpha)}{(\omega + i\alpha)^3}e^{-i\omega t} + \int_\text{Arc} dz\frac{\alpha(z-i\alpha)}{(z + i\alpha)^3}e^{-iz t} \tag4 \end{align}

By the Residue Theorem, we have

\begin{align} \oint_\Psi dz\frac{\alpha(z-i\alpha)}{(z + i\alpha)^3}e^{-iz t}&=2\pi i \text{Res}(f,-i\alpha)\\ &=\alpha\pi i\lim\limits_{z\rightarrow -i\alpha}\frac{d^2}{dz^2}\left[(z-i\alpha)e^{-izt}\right]\\ &=\alpha\pi i\lim\limits_{z\rightarrow -i\alpha}ite^{-izt}(\alpha t + itz -2)\\ &=-2\alpha\pi te^{-\alpha t}(\alpha t - 1) \end{align}

Plugging this into $(4)$, we obtain

\begin{align} 2\alpha\pi te^{-\alpha t}(\alpha t - 1) = \int_{-R}^R d\omega\frac{\alpha(\omega-i\alpha)}{(\omega + i\alpha)^3}e^{-i\omega t} + \int_\text{Arc} dz\frac{\alpha(z-i\alpha)}{(z + i\alpha)^3}e^{-iz t} \tag5 \end{align}

As we take the limit as $R\rightarrow\infty$, the integral along the semi-circle arc goes to zero (see the argument above).

Thus, for $t>0$, we have

\begin{align} \int_{-\infty}^\infty d\omega\frac{\alpha(\omega-i\alpha)}{(\omega + i\alpha)^3}e^{-i\omega t} = 2\alpha\pi te^{-\alpha t}(\alpha t - 1) \tag6 \end{align}

Putting these two results together and plugging $\gamma$ back in to get the answer in terms of the original variables, we have the final result

\begin{align} \int_{-\infty}^\infty d\omega\frac{\gamma(\gamma + 2i\omega)}{(\gamma-2i\omega)^3}e^{-i\omega t} = \frac{\pi\gamma}{4}\theta(t)te^{-\frac{\gamma}{2}t}\left(\frac{\gamma}{2}t - 1\right) \tag7 \end{align}

where $\theta(t)$ is the Heaviside step function.

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  • $\begingroup$ Thank you for the solution. I actually just solved the problem and got to a similar solution but forgot about the Heaviside step function. Nonetheless it is still good to have the formal solution that I can use as a reference in future! $\endgroup$ – crossingsymmetry Dec 22 '16 at 16:25
  • $\begingroup$ You're welcome! Glad I could help. $\endgroup$ – Tom Dec 22 '16 at 16:43

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