Contour integral of $ \int_{-\infty}^{+\infty} \frac{k\cdot \sin(kr)}{k^2-k_0^2}dk$

Question

$\require{cancel}$ I am struggling with this integral:

$$\boxed{I = \int_{-\infty}^{+\infty} \dfrac{k\cdot \sin(kr)}{k^2-k_0^2}dk}$$

I’ll try to explain what I’ve done, hoping it’s correct. I started by rewriting the sine function in terms of the exponential function, using the relationship:

$\sin(x)= \dfrac{e^{ix}−e^{−ix}}{2i}$

​To ensure the equivalence between the two equations, I then took the imaginary part of the expression.

$I = \text {Im}\left[\displaystyle\int_{-\infty}^{+\infty} \dfrac{k\cdot e^{ikr}}{k^2-k_0^2}dk\right]$

I chose a semicircle in the upper half. Was this obligatory? Because for example I learned that if I have a function of the type $\displaystyle \int_{-\infty}^{+\infty} f(x) e^{ikx} dx$ with $k>0$ then I must take the contour in the upper half-plane otherwise the integral diverges, but I don't know if this is also the case here. Anyway: the semicircular arc of the chosen contour vanishes by Jordan's lemma. Right? However, the function has two poles on the real axis and the integral diverges as we approach them. To solve this problem, we introduce two small arcs centred at the poles and take the limit for the radius which reduces to zero. In this way we are introducing the concept of principal value?

Drawing of the contour of the integral I chose and thanks to which the result came back:

Regardless, I defined a complex function $f(k)$ and found its poles.

$f(k) = \dfrac{k\cdot e^{ikr}}{k^2-k_0^2}dk$

The function has two simple poles at: $k^2-k_0^2=0 \implies k = \pm k_0$

For each pole, I calculated the residue using the standard formula: $\begin{align*} \mathop{\mathrm{Res}}_{k = k_0} \ f(k) = \lim_{k \to k_0} \dfrac{k\cdot e^{ikr}}{\cancel{(k-k_0)}(k+k_0)}\cdot \cancel{{(k-k_0)}} = \dots=\dfrac{e^{ik_0r}}{2} \end{align*}$

$\begin{align*} \mathop{\mathrm{Res}}_{k = -k_0} \ f(k) = \dots=\dfrac{e^{-ik_0r}}{2} \end{align*}$

My professor mentioned (or at least this is what I understood) that when we indent the contour around poles, we need to account for a "half residue." This means that instead of applying the Cauchy residue theorem in its usual form:

Cauchy residue theorem in its usual form:

$\displaystyle \oint_{\Gamma_1} f(k) dk = 2\pi i \sum \ \text{Res}(f(k))$

we instead have:

$\displaystyle \oint_{\Gamma_1} f(k) dk = \pi i \sum \ \text{Res}(f(k))$

In this way, I found a result that matched the one obtained by my professor. Is it a coincidence or did I do the exercise correctly? We will find out.

$\displaystyle \oint_{\Gamma_1} f(k) dk = {\displaystyle\pi i\cdot \sum \operatorname {Res} (f(k),\pm k_0)}= \pi i\left[\dfrac{e^{ik_0r} + e^{-ik_0r}}{2}\right] = \pi i \cos(k_0 r)$

$\implies I= \text{Im} \left[\displaystyle \oint_{\Gamma_1} f(k) dk\right] = \text{Im} \left[\pi i \cos(k_0 r)\right] \boxed{= \pi \cos(k_0 r)}$

Thank you very much for your attention and time.

Curiosity: could I have also used an outline of this kind? Because I tried, but it doesn't come :(

Answer 1

You appear to be misunderstanding...the residue theorem only applies when the contour encloses the poles. Your chosen contour actually yields;

$$\oint_Cf(z)dz=0$$

by Cauchy's integral theorem.

We can then break this contour down into parts thusly;

$$\oint_C=\int_{k_0+\epsilon}^{R}+\int_{\Gamma_1}+\int_{-R}^{-k_0-\epsilon}-\int_{\gamma_1}+\int_{-k_0+\epsilon}^{k_0-\epsilon}-\int_{\gamma_2}=0$$

Note that I have subtracted the paths for $\gamma_1$ and $\gamma_2$ as we are traveling counter-clockwise rather than clockwise.

Then, as we then let $R \rightarrow \infty$ and $\epsilon \rightarrow0$ we will have;

$$\bigg(\int_{k_0}^{\infty}+\int_{-k_0}^{k_0}+\int_{-\infty}^{-k_0}\bigg)=\int_{-\infty}^{\infty}=\int_{\gamma_1}+\int_{\gamma_2}$$

Here I have taken it on faith that your professor is correct and that $\int_{\Gamma_1}f(z)dz \rightarrow 0$ as $R \rightarrow \infty$. This is typically a very non-trivial step. We now have;

$$\int_{-\infty}^{\infty}f(z)dz=\bigg(\int_{\gamma_1}f(z)dz+\int_{\gamma_2}f(z)dz\bigg)$$

Now, as the two poles for our function $f(z)$ are simple, we can apply the "half-residue" theorem. We will then have;

$$\text{Res}_{k_0}=\frac{e^{ik_0r}}{2}$$ $$\text{Res}_{-k_0}=\frac{e^{-ik_0r}}{2}$$

$$\therefore \int_{-\infty}^{\infty}f(z)dz=i\pi\bigg(\frac{e^{ik_0r}+e^{-ik_0r}}{2}\bigg)=i\pi\cos(k_0r)$$

Now taking the imaginary part yields the expected result;

$$\boxed{\int_{-\infty}^{\infty}\frac{k \cdot \sin(kr)}{k^2-k_0^2}dk=\pi\cos(k_0r)}$$

Additionally, the other contours you have presented would also work perfectly well and you would arrive at the same answer so long as you properly apply the residue theorem to the poles that your contour encloses.

EDIT:

For your third contour, which only encloses one pole, we will have;

$$\oint_C f(z)dz=2i\pi\text{Res}_{-k_0}$$

Then, breaking the total contour down into the individual paths as we did before and taking the limit as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$ will yield;

$$2i\pi\text{Res}_{-k_0}=\int_{-\infty}^{\infty}f(z)dz+i\pi\text{Res}_{-k_0}-i\pi\text{Res}_{k_0}$$

$$\therefore \int_{-\infty}^{\infty}f(z)dz=i\pi(\text{Res}_{k_0}+\text{Res}_{-k_0})$$

$$=i\pi\cos(k_0r)$$

as expected.