# Complex Integral with exponential

I've been struggling with this:

$$\int_{0}^{\infty }\frac{e^{-px}}{x^{2}+1}\mathrm{d}x, \; \; p\ge 0.$$

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\begin{align*} \int_0^\infty \frac{e^{-px}}{x^2 + 1}dx &\overset{(1)}{=} \int_0^\infty \int_0^\infty e^{-px} e^{-sx} \sin(s)ds dx \\ &\overset{(2)}{=} \int_0^\infty \int_0^\infty e^{-(p+s)x} \sin (s)dx ds\\ &\overset{(3)}{=} \int_0^\infty \frac{\sin(s)}{(p+s)} ds \\ &\overset{(4)}{=} \text{Ci}(p) \sin (p)+\frac{1}{2} (\pi -2 \text{Si}(p)) \cos (p)\\ \end{align*}

$\displaystyle(1): \int_0^{\infty} e^{-sx} \sin(s)dx = \frac{1}{1+x^2}$

$(2):$ change of order of integration.

$\displaystyle(3): \int_{0}^\infty e^{-(p+s)x} dx = \frac{1}{(p+s)}$

$(4):$ \begin{align*} \int_0^\infty \frac{\sin(s)}{(p+s)} ds &= \int_p^\infty \frac{\sin(y - p)}{y }dy \\ &= \int_p^\infty \frac{\cos(p)\sin(y) - \cos(y) \sin(p)}{y }dy \\ &= - \sin(p) \int_p^\infty \frac{\cos(y)}{y}dy + \cos(p)\int_p^\infty \frac{\sin(y)}{y}dy\\ &= \sin(p) \text{Ci}(p) + \cos(p) \left( \int_0^{\infty } \frac{\sin(y)}{y}dy - \int_0^{p } \frac{\sin(y)}{y}dy\right )\\ &= \text{Si}(p)\cos(p) + \frac \pi 2 \cos(p) - \sin(p) \text{Ci}(p) \end{align*}

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The RHS of (3) should be $\frac 1 {p+s}.$ Also (2) should be based. – user64494 Sep 20 '13 at 15:35
@user64494 sorry ... i'll fix in that in next edit. – Santosh Linkha Sep 20 '13 at 15:37
Mathematica outputs $$ConditionalExpression[ CosIntegral[p] Sin[p] + 1/2 Cos[p] (Pi - 2 SinIntegral[p]), p > 0] .$$ – user64494 Sep 20 '13 at 16:03
@user64494 try this Integrate[E^(-p x)/(x^2 + 1), {x, 0, Infinity}, Assumptions :> {p > 0}] – Santosh Linkha Sep 20 '13 at 16:04
So it seems that it isnt possible to find an explicit solution in terms of elementary functions. I tried solving it by Residue-calculus, but i stumbled on some problems finding an appropriate counter to integrate by. I suppose there might be some tricks to it – TheOscillator Sep 20 '13 at 19:24

This definite integral can be 'solven' by using integral Fourier transform or Laplace transform. Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by \begin{align} F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{0}^{t=v}\\ &=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\ &=\frac{2a}{\omega^2+a^2}. \end{align} Next, the inverse Fourier transform of $F(\omega)$ is \begin{align} f(t)=\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\\ e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{\omega^2+a^2}e^{i\omega t}\,d\omega\\ \frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2+a^2}\,d\omega. \end{align} Comparing the last integral to the problem, we have $a=1$ and $it=-p\;\rightarrow\;t=ip$. Therefore $$\int_{-\infty}^{\infty}\frac{e^{-px}}{x^2+1}\,dx=\pi e^{-ip}.$$ I cannot assure that $$\int_{-\infty}^{\infty}\frac{e^{-px}}{x^2+1}\,dx=2\int_0^{\infty}\frac{e^{-px}}{x^2+1}\,dx,$$ except for $p=0$ that yields $$\int_0^{\infty}\frac{1}{x^2+1}\,dx=\frac{\pi}{2}.$$

$$\text{# }\mathbb{Q.E.D.}\text{ #}$$

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Very nice method. +1 – Integrals Jun 2 '14 at 19:59
@Integrals Thanks Jeff but your upvote doesn't increases my reps. I have reached the daily limit votes. :D – Tunk-Fey Jun 2 '14 at 20:03
very beautiful!! – user153330 Mar 1 at 20:38

The answer, done with Maple, is not simple: $$-1/2\, \left( -\sin \left( 2\,p \right) {\it Ci} \left( p \right) + \cos \left( 2\,p \right) \left( {\it Si} \left( p \right) -1/2\,\pi \right) +{\it Si} \left( p \right) -1/2\,\pi \right) \cos \left( p \right) +1/2\, \left( -\cos \left( 2\,p \right) {\it Ci} \left( p \right) -\sin \left( 2\,p \right) \left( {\it Si} \left( p \right) - 1/2\,\pi \right) +{\it Ci} \left( p \right) \right) \sin \left( p \right)$$ See MapleHelp concerning the notation.

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