How can I solve $\int_{-\infty}^{+\infty}\frac{e^{iky}}{(1 + ik)^2}\,dk$? I want to solve the integral above
\begin{equation}
I = \int_{-\infty}^{+\infty}\frac{e^{iky}}{(1 + ik)^2}dk
\end{equation}
I only know that I have a pole of second order in $k = i$ but I don't know how to proceed. 
 A: Note that 
$$I=-\int_{-\infty}^{\infty} dk \frac{e^{i k y}}{(k-i)^2} $$
To evaluate, let's assume that $y \gt 0$ and consider the complex integral
$$\oint_C dz \frac{e^{i y z}}{(z-i)^2} $$
where $C$ is a semicircle of radius $R$ in the upper half plane.  This contour integral is equal to
$$\int_{-R}^R dk \frac{e^{i k y}}{(k-i)^2} + i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta} y}}{\left ( R e^{i \theta}-i \right )^2} $$
As $R \to \infty$, the second integral vanishes because its magnitude is bounded by
$$\frac{2 R}{(R-1)^2} \int_0^{\pi/2} d\theta \ e^{-R y \sin{\theta}} \le \frac{2 R}{(R-1)^2} \int_0^{\pi/2} d\theta \ e^{-2 R y \theta/\pi} \le \frac{\pi}{(R-1)^2 y}$$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=i$.  Thus,
$$\int_{-\infty}^{\infty} dk \frac{e^{i k y}}{(k-i)^2} = i 2 \pi (i y) e^{-y}$$
Thus, for $y \gt 0$, $I = 2 \pi y \, e^{-y}$.  
When $y \lt 0$, we cannot use the above contour because the second integral will increase without bound as $R \to \infty$.  Rather, we close the contour below the real axis.  In this case, there is no pole so by the residue theorem, the integral is zero.  Accordingly,

$$I = 2 \pi y\, e^{-y} H(y) $$

where $H$ is the Heaviside step function, one for $y \gt 0$, zero for $y \lt 0$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
I & \equiv \int_{-\infty}^{+\infty}
{\expo{\ic ky} \over \pars{1 + \ic k}^{2}}\,\dd k
\,\,\,\stackrel{k\ic\ \mapsto\ -k}{=}\,\,\,
-\ic\int_{-\infty\ic}^{\infty\ic}{\expo{-ky} \over \pars{k - 1}^{2}}\,\dd k =
-\ic\expo{-y}\int_{-1 - \infty\ic}^{-1 + \infty\ic}{\expo{-ky} \over k^{2}}
\,\dd k
\end{align}
The last integral is similar to an Inverse Laplace transform and it has a double pole at $\ds{k = 0}$:


*

*$\ds{\large y < 0}$: In this case, we 'close' the contour with a
semi-circumference $\pars{~\mbox{where}\ \Re\pars{k} < 0~}$ whose radius $\to \infty$. Such contour doesn't enclose the above mentioned pole such that the integral vanishes out:
$$
I = 0\quad\mbox{whenever}\qquad y < 0
$$



*$\ds{\large y > 0}$: The above mentioned semi-circunference closes, in this case, the contour with $\ds{\Re\pars{k} > 0}$:
$$
I = -\ic\expo{-y}\pars{-2\pi\ic\,\partiald{\expo{-ky}}{k}}
_{\ k\ =\ 0} = 2\pi y\expo{-y}\qquad y > 0
$$

$$
I \equiv
\int_{-\infty}^{+\infty}{\expo{\ic ky} \over \pars{1 + \ic k}^{2}}\,\dd k =
\bbx{\ds{2\pi y\expo{-y}\Theta\pars{y}}}\qquad
\pars{~\Theta\ \mbox{is the}\ Heaviside\ Step\ Function.~}
$$



Note that the integration along the semi-circumferences vanishes out as its radius $\ds{R \to \infty}$:
  \begin{align}
0 & < \verts{\int_{-\pi/2}^{\pi/2}{\expo{-\verts{y}R\expo{\ic\theta}} \over R^{2}\expo{2\ic\theta}}\,R\expo{\ic\theta}\ic\,\dd\theta} <
{2 \over R}\int_{0}^{\pi/2}\exp\pars{-\verts{y}R\sin\pars{\theta}}\,\dd\theta
\\[5mm] & <
{2 \over R}\int_{0}^{\pi/2}\exp\pars{-2\verts{y}R\theta/\pi}\,\dd\theta =
{\pi \over \verts{y}}\,{1 - \exp\pars{-\verts{y}R} \over R^{2}}
\,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\to}\,\,\, 0
\end{align}

