Computing $\int_0^{\infty} e^{iax}/(x^2+1)dx$ I am having trouble solving this integral with complex analysis
$$\int\limits^{\infty }_{0}\frac{e^{iax}}{x^{2} +1} dx$$
I have tried two different contours; those being
contour 2
contour 1
with both contours, I got the answer $\int\limits ^{\infty }_{0}\frac{e^{iax}}{x^{2} +1} dx=\frac{\pi }{2e^{a}}$
But according to wolfram alpha's approximation, this is wrong. Wolfram alpha had a real and imaginary part to the answer. So I'm confused. Can someone show the process for solving this with contour integration?
 A: In contour 1, I am guessing you treated the integral along the segment $C1$ as zero, but it is not, and it has a non-zero imaginary part.   
Similarly, the same integral, with imaginary result, is omitted along $C1$ in contour 2.
The integral does not have a real value.
A: The exponential integral $E_1(z):=\int_z^\infty\frac{e^{-t} dt}{t}=e^{-z}U(1,\,1,\,z)$ in terms of a confluent hypergeometric function of the second kind. Since $\frac{1}{x^2+1}=\frac{i}{2}\sum_\pm\frac{\pm 1}{x\pm i}$, your integral is $$\frac{i}{2}\sum_\pm\pm\int_0^\infty\frac{e^{iax}dx}{x\pm i}=\frac{i}{2}\sum_\pm\pm e^{\pm a}\int_{\pm a}^\infty\frac{e^{-z}dz}{z}\\=\frac{i}{2}\sum_\pm\pm e^{\pm a}E_1(\pm a)=\frac{i}{2}\left(U(1,\,1,\,a)-U(1,\,1,\,-a)\right).$$This isn't real, but its real part is $-\frac{\pi}{2}e^{-a}$ for $a>0$. This means I probably have a sign error somewhere editors are welcome to address, since $$\int_0^\infty\frac{\cos axdx}{x^2+1}=\frac12\Re\int_{\Bbb R}\frac{e^{iax}dx}{x^2+1}=\frac{\pi}{2}e^{-|a|}$$is a useful sanity check based on the Cauchy distribution's characteristic function.
A: I can't directly use contour integration to solve this, but I can use Fourier Transforms and their properties.
$$\begin{align*}I(a) &= \int_0^\infty \dfrac{e^{iax}}{x^2+1} dx \\
\\
&= 2\pi \int_0^\infty \dfrac{e^{i2\pi sa}}{\left(2\pi s\right)^2 +1} ds \\
\\
&= \pi \int_{-\infty}^\infty H(s)\dfrac{2}{\left(2\pi s\right)^2 +1} e^{i2\pi sa}ds \\
\\
&= \pi \mathscr{F}^{-1}\left\{H(s)\dfrac{2}{\left(2\pi s\right)^2 +1}\right\}\\
\\
&= \pi \cdot \mathscr{F}^{-1}\left\{H(s)\right\}*\mathscr{F}^{-1}\left\{\dfrac{2}{\left(2\pi s\right)^2 +1}\right\}\\
\\
&= \pi \cdot \dfrac{1}{2}\left[\dfrac{1}{i\pi(-a)} + \delta(-a)\right] * e^{-|a|}\\
\\
&= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2} \dfrac{1}{a} * e^{-|a|}\\
\\
&= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2} \int_{-\infty}^\infty \dfrac{e^{-|\tau|}}{a-\tau} d\tau\\
\\
&= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2} \int_0^\infty \dfrac{e^{-\tau}}{a-\tau} d\tau+ \dfrac{i}{2} \int_{-\infty}^0 \dfrac{e^{\tau}}{a-\tau} d\tau\\
\\
&= \dfrac{\pi}{2} e^{-|a|} - \dfrac{i}{2} \int_{-a}^\infty \dfrac{e^{-(t+a)}}{t} dt - \dfrac{i}{2} \int_{\infty}^a \dfrac{e^{(a-t)}}{t} dt\\
\\
&= \dfrac{\pi}{2} e^{-|a|} - \dfrac{i}{2}e^{-a} \int_{-a}^\infty \dfrac{e^{-t}}{t} dt + \dfrac{i}{2} e^{a}\int_a^{\infty} \dfrac{e^{-t}}{t} dt\\
\\
&= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2}e^{-a} \mathrm{Ei}(a) - \dfrac{i}{2} e^{a}\mathrm{Ei}(-a)\\
\\
&= \dfrac{\pi}{2} e^{-|a|} -i \space \dfrac{e^{a} \mathrm{Ei}(-a) -  e^{-a}\mathrm{Ei}(a)}{2}\\
\end{align*}$$
This answer looks like what @J.G. was driving at.
Note that just deriving the inverse Fourier Transform of the Heaviside unit step, $H(s)$, involves multiple contour integrations and distribution theory.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\mbox{Note that}
\\[2mm] &\
\left.\int_{0}^{\infty}{\expo{\ic ax} \over x^{2} + 1}\,\dd x
\,\right\vert_{\ a\ \in\ \mathbb{R}}
\\[2mm] = &\
\underbrace{\int_{0}^{\infty}{\cos\pars{\verts{a}x} \over x^{2} + 1}\,\dd x}_{\ds{\pi\expo{-\verts{a}} \over 2}}\ +\
\ic\,\mrm{sgn}\pars{a}
\bbox[10px,#ffd]{\Im\int_{0}^{\infty}{\expo{\ic\verts{a}x} \over x^{2} + 1}\,\dd x}\label{1}\tag{1}
\end{align}

Then,
\begin{align}
&\bbox[10px,#ffd]{\Im\int_{0}^{\infty}{\expo{\ic\verts{a}x} \over x^{2} + 1}\,\dd x} =
\overbrace{-\Im\lim_{R \to \infty}\int_{0}^{\pi/2}{\exp\pars{\ic\verts{a}R\expo{\ic\theta}} \over R^{2}\expo{2\ic\theta} + 1}\,R\expo{\ic\theta}\ic\,\dd\theta}
^{\ds{=\ 0}}
\\[2mm] &\
-\Im\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{\infty}^{1 + \epsilon}{\expo{-\verts{a}y} \over
-y^{2} + 1}\,\ic\,\dd y +
\int_{1 - \epsilon}^{0}{\expo{-\verts{a}y} \over
-y^{2} + 1}\,\ic\,\dd y}
\\[8mm] = &\
-\mrm{P.V.}\int_{0}^{\infty}{\expo{-\verts{a}y} \over y^{2} - 1}
\,\dd y =
-\,{1 \over 2}\,\mrm{P.V.}\int_{0}^{\infty}
{\expo{-\verts{a}y} \over y - 1}\,\dd y +
{1 \over 2}\int_{0}^{\infty}
{\expo{-\verts{a}y} \over y + 1}\,\dd y
\\[5mm] = &\
-\,{1 \over 2}\,\expo{-\verts{a}}\mrm{P.V.}\int_{-\verts{a}}^{\infty}
{\expo{-y} \over y}\,\dd y +
{1 \over 2}\,\expo{\verts{a}}\
\underbrace{\int_{\verts{a}}^{\infty}
{\expo{-y} \over y}\,\dd y}_{\ds{\mrm{E}_{1}\pars{\verts{a}}}}
\end{align}

$\ds{\mrm{E}_{1}}$ is the
  Exponential Integral.

Then,
\begin{align}
&\bbox[10px,#ffd]{\Im\int_{0}^{\infty}{\expo{\ic\verts{a}x} \over x^{2} + 1}\,\dd x}
\\[5mm] = &\
-\,{1 \over 2}\,\expo{-\verts{a}}\bracks{%
\mrm{P.V.}\int_{-\verts{a}}^{\verts{a}}{\expo{-y} \over y}\,\dd y +
\mrm{E}_{1}\pars{\verts{a}}} +
{1 \over 2}\,\expo{\verts{a}}\
\mrm{E}_{1}\pars{\verts{a}}
\\[5mm] = &\
-\,{1 \over 2}\,\expo{-\verts{a}}
\int_{0}^{\verts{a}}\pars{{\expo{-y} \over y} + {\expo{y} \over -y}} \,\dd y + \sinh\pars{\verts{a}}\mrm{E}_{1}\pars{\verts{a}}
\\[5mm] = &\
\expo{-\verts{a}}
\int_{0}^{\verts{a}}{\sinh\pars{y} \over y}\,\dd y + \sinh\pars{\verts{a}}\,\mrm{E}_{1}\pars{\verts{a}}
\\[5mm] = &\
\expo{-\verts{a}}\,\mrm{Shi}\pars{\verts{a}} + \sinh\pars{\verts{a}}\,\mrm{E}_{1}\pars{\verts{a}}
\label{2}\tag{2}
\end{align}

$\ds{\mrm{Shi}}$ is the
  Hyperbolic Sine Integral.


\eqref{1} and \eqref{2} lead to:
$$
\bbx{\left.\int_{0}^{\infty}{\expo{\ic ax} \over x^{2} + 1}\,\dd x
\,\right\vert_{\ a\ \in\ \mathbb{R}} =
{\pi\expo{-\verts{a}} \over 2} +
\bracks{\vphantom{\Large A}\expo{-\verts{a}}\,\mrm{Shi}\pars{a} + \sinh\pars{a}\,\mrm{E}_{1}\pars{\verts{a}}}\ic}
$$
