Proof integral:$\lim_{x \rightarrow +\infty} \int_0^{x} dx'e^{-i(k-k_0)x'} = \pi\delta(k-k_0) - P(\frac{i}{k-k_0})$,where $P$ stands for...

Question

Where can I find the full derivation or the proof of the following integral? $$\lim_{x \rightarrow +\infty} \int_0^{x} dx'e^{-i(k-k_0)x'} = \pi\delta(k-k_0) - P(\frac{i}{k-k_0})$$, where $P$ stands for "Cauchy principal value".

I saw this formula in a research paper. I would like to make sense of it. Especially I don't understand the Cauchy principal part (the second term). Any help or the reference will be appreciated, given that I have some general understanding of the applied complex analysis, such as the residue theorem.

Answer 1

Extending Fourier Transforms to include distributions, the Fourier Transform of the Dirac Delta is

$$\begin{align} \mathscr{F}\{\delta(k-k_0)\}&=\int_{-\infty}^{\infty} \delta(k-k_0)e^{ikx}\,dk\\\\ &=e^{ik_0x} \end{align}$$

Invoking the inverse Fourier Transform reveals that the Dirac Delta can be represented as

$$\delta(k-k_0)=\frac1{2\pi}\int_{-\infty}^\infty e^{-i(k-k_0)x}\,dx \tag1$$

Next, note that Fourier Transform of the function $\frac{1}{k-k_0}$ can be interpreted as the Cauchy-Principal Value

$$\begin{align} \mathscr{F}\left(\frac{1}{k-k_0}\right)&=\text{PV}\int_{-\infty}^{\infty} \frac{1}{k-k_0}e^{ikx}\,dk \tag 2\\\\ \end{align}$$

We can evaluate the integral in $(2)$ using contour integration. To do this we analyze the integral $I(x)$ given by

$$I(x)=\oint_C \frac{1}{z-k_0}e^{izx}\,dz \tag 3$$

where for $x>0$ ($x<0$), $C$ is the contour comprised of (i) the real line segment from $z=-R$ to $k_0-\epsilon$, (ii) a semi-circular contour centered at $k_0$ with radius $\epsilon$ in the upper (lower) half plane, (iii) the line segment from $k_0+\epsilon$ to $R$, and (iv) a semi-circular contour centered at the origin with radius $R$ in the upper (lower) half plane.

Since the integrand in $(3)$ is analytic in and on $C$, its value is zero. Therefore we can write for $x>0$

$$\begin{align} I(x)&=0\\\\ &=\left(\int_{-R}^{k_0-\epsilon}\frac{1}{k-k_0}e^{ikx}\,dk+\int_{k_0+\epsilon}^{R}\frac{1}{k-k_0}e^{ikx}\,dk\right) \tag 4\\\\ &+\int_\pi^0 \frac{1}{\epsilon e^{i\phi}}e^{i(k_0+\epsilon e^{i\phi})\,x}\,\left(i\epsilon e^{i\phi}\right)\,d\phi \tag 5\\\\ &+\int_0^\pi\frac{1}{R e^{i\phi}-k_0}e^{i(k_0+R e^{i\phi})\,x}\,\left(iR e^{i\phi}\right)\,d\phi \tag 6\\\\ \end{align}$$

As $R\to \infty$ and $\epsilon \to 0$, the right-hand side of $(4)$ becomes equal to the integral of interest given in $(2)$, the integral in $(5)$ approaches $-i\pi e^{ik_0x}$ and the integral in $(6)$ approaches $0$. Therefore, we find for $x>0$

$$\text{PV}\int_{-\infty}^{\infty} \frac{1}{k-k_0}e^{ikx}\,dk=i\pi e^{ik_0x}$$

A parallel development shows that for $x<0$

$$\text{PV}\int_{-\infty}^{\infty} \frac{1}{k-k_0}e^{ikx}\,dk=-i\pi e^{ik_0x}$$

Invoking the Fourier Inverse Transform reveals that the function $\frac1{k-k_0}$ can be represented as

$$\text{PV}\left(\frac{1}{k-k_0}\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}i\pi\text{sgn}(x)\,e^{-(k-k_0)x}\,dx \tag 7$$

Multiplying $(1)$ by $\pi$, multiplying $(7)$ by $-i$, and adding the results yields

$$\begin{align} \pi \delta(k-k_0)-i\text{PV}\left(\frac{1}{k-k_0}\right)&=\int_{-\infty}^\infty\frac{1+\text{sgn}(x)}{2}e^{-i(k-k_0)x}\,dx\\\\ &=\int_0^\infty e^{-i(k-k_0)x}\,dx \end{align}$$

as was to be shown!

Answer 2

Let $k-k_0 \equiv k$ (so we simplify our terms...). In order for the integral to converge, we modify the exponent $k \to k -i \epsilon$, with $\epsilon \to 0$. The result. is $-\frac{i}{k-i\epsilon}$. Now, from the theory of distributions, using $ \lim_{\epsilon \to 0} \frac{1}{k-i\epsilon}=P\frac{1}{k}+i\pi\delta(k)$ (

V.S. Vladimirov (2002). Methods of the theory of generalized functions. Taylor & Francis. ISBN 0-415-27356-0

), and reverting to our initial substitution of variable (ie, $k \equiv k-k_0$) one gets the final result: $\pi\delta(k-k_0)-P\frac{i}{k-k0}$.