Relationship between C.P.V. $\int_{-\infty}^\infty\frac{e^{ist}}{t-i} \,dt$ and $\int_{-\infty}^\infty\frac{e^{ist}}{t-i} \, dt$ This is the problem I am trying to solve: Let $s\in \mathbb{R}$. Compute $$\operatorname{C.P.V.} \int_{-\infty}^\infty \frac{e^{ist}}{t-i} \, dt$$ when it exists. Determine the values of $s$ for which the integral is convergent. 
So far, I've broken the problem down into the cases where $s = 0$ and $s \not = 0$. When $s = 0$ I was able to use the residue theorem to show that $\operatorname{C.P.V.}\int_{-\infty}^\infty \frac{1}{t-i} \, dt = i\pi$. Moreover, since 
$$\operatorname{Re} \int_{0}^\infty \frac{1}{t-i}\,dt = \int_0^\infty \frac{t}{t^2+1} \, dt = \infty$$ I was able to conclude that while the Cauchy Principle Value exists for $s=0$ the integral is not convergent. When $s\not = 0$, I was able to use the residue theorem and Jordan's Lemma to show that 
$$\operatorname{C.P.V.} \int_{-\infty}^\infty \frac{e^{ist}}{t-i} \,dt = \begin{cases} DNE & s < 0 \\ 2i\pi/e^s & s > 0. \end{cases}$$
However, I'm not sure how to show that the integral actually converges when $s> 0$ (I think that it does converge). Also, is there a more simple way to solve the problem that doesn't require breaking up the cases $s=0$ and $s \not = 0$?
 A: 
We first examine the Cauchy Principal Value of the integral of interest.

Let $F(s)=\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ist}}{t-i}\,dt\right)=\lim_{L\to \infty}\int_{-L}^L\frac{e^{ist}}{t-i}\,dt$.
We can evaluate $F(s)$ we apply the residue theorem to the integral $I(s)$, $s\ne 0$, given by 
$$\begin{align}
I(s)&=\lim_{L\to \infty}\oint_{C_L} \frac{e^{isz}}{z-i}\,dz\\\\
&=\underbrace{\lim_{L\to \infty}\int_{-L}^L \frac{e^{ist}}{t-i}\,dt}_{=F(s)}+\underbrace{\lim_{L\to \infty}\int_0^{\text{sgn}(s) \pi} \frac{e^{isLe^{i\phi}}}{Le^{i\phi}-i}\,iLe^{i\phi}\,d\phi}_{=0\,\,\text{applying Jordan's Lemma}}\\\\
&=\begin{cases} 2\pi i e^{-s}&,s>0\\\\0&,s<0\end{cases}
\end{align}$$  
For $s=0$, we have
$$\begin{align}
\lim_{L\to \infty}\int_{-L}^L \frac{1}{t-i}\,dt &=\lim_{L\to \infty}\log\left(\frac{L-i}{-L-i}\right)\\\\
&=\lim_{L\to \infty}\left(\frac12\log\left(\frac{L^2+1}{L^2+1}\right)+i\arctan2(-1,L)-i\arctan2(-1,-L)\right)\\\\
&=i\pi
\end{align}$$
Putting it all together, we see that

$$\bbox[5px,border:2px solid #C0A000]{\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ist}}{t-i}\,dt\right)=\begin{cases}2\pi ie^{-s}&,s>0\\\\i\pi&s=0\\\\0&,s<0\end{cases}}$$


Next, we analyze the convergence properties of the integral of interest.  Obviously, for $s=0$ the integral diverges logarithmically. 
For, $s\ne0$, we can appeal to the Dirichlet-Abel Test.  Note that for all $L$, the integrals $\left|\int_{0}^L \sin(st)\,dt\right|$ and $\left|\int_{0}^L \cos(st)\,dt\right|$ are bounded by $2/s$.  Inasmuch at $\frac{t}{t^2+1}$ monotonically decreases, then $\int_0^\infty \frac{e^{ist}}{t-i}\,dt$ converges for $s\ne 0$.
