Trying to solve improper integral I've been trying to solve this 
$$ \int_{-\infty}^\infty {\sin(x)\over x+1-i }dx $$
using residue theorem. I've tried using a square contour pi, pi+pii, -pi+pii, pi
and half a circle but with the former had trouble with the upper contour (pii+pi to pii-pi) and with the circle I couldn't prove the upper contour when integrated to be 0
 A: 
Assuming $\text{Im}(a)>0$, we have:
  $$ \int_{-\infty}^{+\infty}\frac{\sin z}{z-a}\,dz = \color{red}{\pi e^{ia}} \tag{1}$$
  by the residue theorem. Write $\sin z$ as $\text{Im}(e^{iz})$, consider a semicircular contour in the upper half-plane, prove the ML inequality, profit.

A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\int_{-\infty}^{\infty}{\sin\pars{x} \over x + 1 - \ic}\,\dd x} =
{1 \over 2\ic}\sum_{\sigma = \pm 1}\sigma
\int_{-\infty}^{\infty}{\expo{\sigma x\ic} \over x - \pars{-1 + \ic}}\,\dd x =
{1 \over 2i}\times\pars{1}\times\bracks{2\pi\ic\expo{\pars{-1 + i}\ic}} =
\color{#f00}{\expo{-1 - \ic}\pi}
\end{align}

The term with $\ds{\sigma = 1}$ yields the whole contribution to the sum because the pole $\ds{\pars{-1 + \ic}}$ rests in the upper complex half-plane.

A: A different approach could be this one: First multiply top and bottom by $x+1+i$ to arrive at:$$ \int_{-\infty}^\infty {(x+1+i)\sin(x)\over (x+1)^2+1 }dx $$ Now let $x+1=t$ to get $$ \int_{-\infty}^\infty {(t+i)\sin(t-1)\over t^2+1 }dx $$ Applying Sum formula for sine, we get  $$ \int_{-\infty}^\infty {(t+i)(\sin t\cos(-1)+\cos t\sin(-1))\over t^2+1 }dx $$ Now distribute the $(t+i)$ split the integral from the $+$ sign and realize that due to even-odd properties, some parts cancel, and that was the first reason of my post. Looks like there is a $t\sin t$ and a $i\cos t$ that is EVEN in the numerator but these are all "standard" complex integrals evaluated elsewhere on this site (and that is the second reason of my post) and I am sure you must have seen those too. Of course there are those $\cos 1$ and $\sin(-1)$ that are part of the answer. Can you work it out from here?
