User Tools

Site Tools



The Contour Integration Trick

When we try to solve a complicated integral like

$$ \int_{-\infty}^\infty \frac{1}{t^2 - \tau^2} \mathrm{e}^{-i \omega t}$$

it can be helpful to consider the following “complification”, which actually makes things much simpler: we treat the real parameter $t$ as a complex variable and then rewrite our integral along the real axis as a contour integral in the complex plane.

This is possible, because an integral along a curve can be split into several parts and the complete integral is just the sum of the integrals along the parts:

$$ \int_a^c f(z) dz = \int_a^b f(z) dz + \int_b^c f(z) dz .$$

We make use of this fact in the following way:

Therefore, if the integral along the upper arc is zero for some reason, we can indeed write an integral along the real axis as an integral along a closed contour in the complex plane.

Why is this useful?

If we succeed to rewrite our integral along the real axis as a contour integral in the complex plane, we can make use the incredibly useful residue theorem.

First, we note that the integral of a complex function along a closed contour is zero, if there are no poles inside the closed contour. However, in our example above there are two poles at $t = \pm i \tau$ (= at this points the function blows up and becomes infinity). Now, the residue theorem tells us that if we want to compute the integral along a closed contour which encloses poles, the result of the integral is the sum of all residues times $2 \pi i$:

$$ \oint_C f(z) dz = 2 \pi i \sum \mathrm{Res}(f,a_k),$$

where $a_k$ denotes the poles inside the closed contour $C$.

The residuum of a function $f$ is the coefficient in front of the $n^{-1}$ term in the Laurent series expansion of $f$ the around the pole.

For our example integral $ \int_{-\infty}^\infty \frac{1}{t^2 - \tau^2} \mathrm{e}^{-i \omega t}$, we therefore have (if the integral along the upper arc is zero)

\begin{align} \int_{-\infty}^\infty \frac{1}{t^2 - \tau^2} \mathrm{e}^{-i \omega t} = \oint_{C}\frac{1}{t^2 - \tau^2} \mathrm{e}^{-i \omega t} &= 2 \pi i \mathrm{Res}\left(\frac{1}{t^2 - \tau^2} \mathrm{e}^{-i \omega t},t=i\tau \right) \notag \\ &= 2 \pi i \frac{\mathrm{e}^{-i \omega i \tau} }{i\tau + i \tau} \end{align}

Our closed contour only encloses one pole (at $t= i \tau$) and thus we have just one residuum. It is, of course, possible to close the contour in the lower complex plane and thus get a contour that encloses the other pole (at $t= -i \tau$). The result is the same. The thing is that when we rewrite our integral along the real line as an integral along a closed contour, we always enclose only of the two poles.

Why does the integral along the upper arc vanish?

Without the $\mathrm{e}^{-i \omega t} $ factor, the rest of the integral $ \frac{1}{t^2 - \tau^2} $ vanishes for $|t| \to \infty$. To investigate the $\mathrm{e}^{-i \omega t} $ factor, we need to split the now complex variable $t$ into its real and imaginary parts:

$$ \mathrm{e}^{-i \omega t} = \mathrm{e}^{-i \omega (\mathrm{Re}(t)+ i \mathrm{Im}(t)) } = \mathrm{e}^{-i \omega \mathrm{Re}(t)+ \mathrm{Im}(t) }. $$ Thus, while the real part simply oscillates and thus gets damped by the $ \frac{1}{t^2 - \tau^2} $ factor, the imaginary part could be troublesome for $|t| \to \infty$. However, we close the contour along the upper arc, which means $ t \to +i \infty$ and therefore this factor goes to zero, too. Therefore the integral along the upper arc vanishes, because $ \mathrm{e}^{-\infty} = 0 $.

The Residue Theorem

The idea is that expand the function that we integrate as a Laurent series around the pole. The Laurent series is similar to the Taylor series, but includes additionally negative powers and is the appropriate mathematical tool to describe a complex function in the neighborhood of a pole.

We deform our contour in the following way:


The Laurent series expansion of our function $f(z)$ around the pole $z_0$ can be written as

$$ f(z) = \sum_{n=-\infty}^\infty a_n (z-z_0)^n $$

We are only interested in a tiny circle around the pole and thus write $z-z_0= \epsilon \mathrm{e}^{i\theta}$ and call this for brevity $\xi \equiv \epsilon \mathrm{e}^{i\theta}$.

Therefore, we have

$$ \oint f(z) dz= \oint \left( \sum_{n=-\infty}^\infty a_n \xi^n \right)d\xi = \sum_{n=-\infty}^\infty \oint a_n \xi^n d\xi .$$

The thing is now that this integral vanishes for all terms except for $n=-1$.

Good Reads

quantum_field_theory/methods/contour_integration_trick.txt · Last modified: 2017/12/06 09:33 (external edit)