What is a simple form of this integral? 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?
 A: 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

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

A: 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.
