Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been struggling with this:

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

share|cite|improve this question

\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*}

share|cite|improve this answer
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 $$ \int_0^{\infty}\frac{1}{x^2+1}\,dx=\frac{\pi}{2}. $$

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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.