What is a simple form of this integral?

Question

This integral reminds me of something familiar but I cannot remember the rule to make it simple.

$$\int_{-\infty}^{+\infty} \frac{\exp(i a \cdot v)}v \mathrm d v$$

where $a$ is a scalar for simplicity now.

What is the method to expand this integral?

Answer 1

By Residue Theorem and Cauchy Integral Formula

$$\int_{-\infty}^{+\infty} \frac{\exp(i a \cdot v)}v \mathrm d v = -i \pi.$$

The function has a singularity on the contour at $z = 0$. Make a small detour around $z = 0$ to avoid the problem. Pick a small $\rho > 0$ and consider a contour consisting of the arc $A_{R}$ followed by a line segment along a real axis between $-R$ and $-\rho$, followed by an upper semi-circle centered at $0$ with radius $\rho$ and, finally, a line segment along the positive part of the real exit from $\rho$ to $R$. Call the contour $B_{R,\rho}$ and denote the line segments by $L_{-}$ and $L_{+}$, respectively and the small semi-circle by $A_{\rho}$. The function is analytic with $B_{R,\rho}$. Its integral along this contour is 0: \begin{equation} \int_{A_{R}} \frac{e^{iz}}{z} dz = 0 + \int_{-R}^{-\rho} \frac{e^{ix}}{x} dx + \int_{A_{\rho}} \frac{e^{iz}}{z} dz + \int_{\rho}^{R} \frac{e^{ix}}{x} dx. \end{equation} Because \begin{equation} \int_{-R}^{-\rho} \frac{e^{ix}}{x} dx = - \int_{\rho}^{R} \frac{e^{-ix}}{x} dx, \end{equation} by combining the second the fourth integral \begin{equation} \int_{\rho}^{R} \frac{e^{ix} - e^{-ix}}{x} dx + \int_{A_{R}} \frac{e^{iz}}{z} dz + \int_{A_{\rho}} \frac{e^{iz}}{z} dz = 0. \end{equation} Because the integrand in the leftmost integral is \begin{equation} 2i \cdot \frac{e^{ix} - e^{-ix}}{2ix} = 2i \cdot \frac{sin(x)}{x} \end{equation} so we get \begin{equation} 2i \int_{\rho}^{R} \frac{sin(x)}{x} dx = - \int_{A_{R}} \frac{e^{iz}}{z} dz - \int_{A_{\rho}} \frac{e^{iz}}{z} dz = 0 \end{equation} and dividing by $2i$ \begin{equation} \int_{\rho}^{R} \frac{sin(x)}{x} dx = \frac{i}{2} \Big( \int_{A_{R}} \frac{e^{iz}}{z} dz + \int_{A_{\rho}} \frac{e^{iz}}{z} dz \Big). \end{equation} So the integral over $A_{R}$ vanishes as $R \rightarrow \infty$ and got the result.

Source: Hitczenko P. Some applications of the residue theorem MATH322, 2005.

Answer 2

I am working on this. Perhaps Fourier Transform best tool here as suggested by Stephen? Anyway below some novel ideas!

Related to Abel's polynomial?

Consider Abel's polynomial where

$$f(t)=t e^{at}$$

and now divide by $t^{-2}$ so

$$\frac{f(t)}{t^2}=\frac 1 t e^{at}$$

and now your equation is

$$\frac 1 v e^{iav}:=\frac 1 v e^{z v}$$

that is perhaps related to the questions

1. Using the Lambert W to express a solution of a differential equation.

2. its DY $y''=(y')^{3} e^{y}$, some easy way to solve this non-linear differential equation?.

where I encountered the Lambert W that is in the generating function of $\frac 1 v e^{z v}$.

Possible to reformulate the function somehow to make it easier to solve?

"This integral reminds me of something familiar -- $\int_{-\infty}^{+\infty} \frac{\exp(i a \cdot v)}v \mathrm d v$".

You are probably thinking of Laurent's series and Cauchy's Integral theorem

where the path $\gamma$ needs to be connected. Now let $\frac{\exp(i a > v)}v=\frac{\exp(i a v)}{\exp(\log(v))}=\exp(iav-log(v))$. I cannot yet see how to reformulate this question to use such theorems because the function should be analytic, investigating What is $iav-\log(v)$? Any series expansion or inequality for it?.