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I would like to solve the integral $$A\int_{-\infty}^\infty\frac{e^{-ipx/h}}{x^2+a^2}dx$$ where h and a are positive constants. Mathematica gives the solution as $\frac\pi{a}e^{-|p|a/h}$, but I have been trying to reduce my reliance on mathematica. I have no idea what methods I would use to solve it.

Is there a good (preferably online) resource where I could look up methods for integrals like this fairly easily?

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Are you familiar with contour integration and the method of residues from complex analysis? – Gerry Myerson Oct 23 '12 at 0:47
Of course there exist some methods which only involve real analysis technique, but undoubtedly the contour integration is the easiest. – Sangchul Lee Oct 23 '12 at 1:25
I for some reason chose to take real analysis for fun this quarter rather than the more pragmatic (as a physics major) complex analysis class. – ari Oct 23 '12 at 2:10
up vote 4 down vote accepted

Consider the positively-oriented contour $C$ that spans the real axis from $-R$ to $R$ and then around the semicircle $Re^{i\theta}$ for $0\le \theta\le \pi$. Let

$$f(x) := \frac{e^{-ipx/h}}{x^2+a^2} = \frac{e^{-ipx/h}}{(x+ia)(x-ia)}$$

Now, we have (if $z_n$ are the poles of $f$ in $C$)

$$\oint_C f(z)\, dz = \int_{-R}^R f(z)\, dz + \oint_{\text Arc} f(z)\, dz = 2\pi i \sum \operatorname*{Res}_{z = z_n} f(z)$$

Letting $R \to \infty$, we see, for $p/h < 0$

$$\int_{\text Arc}\frac{e^{-ipx/h}}{x^2+a^2}\,dz = \int_0^\pi \frac{e^{-ipRe^{i z}/h}}{(Re^{i z})^2+a^2}\,dz = 0$$

$$\oint_C f(z)\, dz = \int_{-\infty}^\infty f(z)\, dz$$

so, because $ia$ lies in $C$ (and is the only pole in $C$)

$$ z_0 = \operatorname*{Res}_{z = ia} f(z) = \lim_{z\to ia}(z-ia)f(z) = \frac{e^{-ip(i a)/h}}{2ia} = \frac{e^{-pa/h}}{2ia} $$


$$\int_{-\infty}^\infty f(z)\, dz = 2\pi i z_0 = 2\pi i\frac{e^{-pa/h}}{2ia} = \frac{\pi e^{-pa/h}}{a}$$

for $p/h < 0$

Considering the new contour $\Gamma$ which is the same as $C$ except that it traverses $Re^{i\theta}$ for $\pi\le \theta\le 2\pi$, we see that

$$\int_{\text Arc}\frac{e^{-ipx/h}}{x^2+a^2}\,dz = \int_0^\pi \frac{e^{ipRe^{i z}/h}}{(Re^{i z})^2+a^2}\,dz = 0$$

when $p/h > 0$ and $R \to \infty$. Using the method above, we now have

$$\int_{-\infty}^\infty f(z)\, dz = 2\pi i z_0 = 2\pi i\frac{e^{-pa/h}}{2ia} = \frac{\pi e^{pa/h}}{a}$$

for $p/h > 0$

Putting our results together, we obtain the complete answer

$$\int_{-\infty}^\infty f(z)\, dx = \frac{\pi e^{\left|\frac{p}{h}\right|a}}{a}$$

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I have to read up on contours a bit more before I fully understand your answer. I get that you are doing a closed curve integral who's only contributory member is the real integral I'm concerned with. I don't really know the methods you used to find the solution to the closed integral. – ari Oct 23 '12 at 2:33
in your answer you have $e^{-pa/h}$ is that assuming p is positive? Because mathematica has $e^{-|p|a/h}$, a very different answer when p is negative. – ari Oct 23 '12 at 2:35
@ari I have added the cases when $p/h$ is positive or negative. I'm not sure why Mathematica did not put absolute value signs around the $h$, because I assume that is a constant as well? Why would $p$ be absolute but not $h$? – Argon Oct 23 '12 at 20:12
This can also be done using Fourier transforms, as $\frac{1}{x^2 + a^2}$ is the inverse Fourier transform of a function of the form $Ce^{-A|\omega|}$ (I don't know what $C$ and $A$ are off the top of my head). – Stefan Smith Oct 23 '12 at 21:17

$\mathbf{Method\;1: }$ Integral Fourier Transform

Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by $$ \begin{align} F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{t=0}^v\\ &=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\ &=\frac{2a}{\omega^2+a^2}. \end{align} $$ 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 yields $t=-\frac{p}{h}$. Thus, $$ \int_{-\infty}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx=\frac{\pi e^{-a\left|\frac{p}{h}\right|}}{a}. $$ $\mathbf{Method\;2: }$

Note that: $$ \int_{y=0}^\infty e^{-(x^2+a^2)y}\,dy=\frac{1}{x^2+a^2}, $$ therefore $$ \int_{x=0}^\infty\int_{y=0}^\infty e^{-(x^2+a^2)y}\;e^{-\frac{ipx}{h}}\,dy\,dx=\int_{x=0}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx $$ Rewrite $$ \begin{align} \int_{x=0}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx&=\int_{y=0}^\infty\int_{x=0}^\infty e^{-(yx^2+\frac{ip}{h}x+a^2y)}\,dx\,dy\\ &=\int_{y=0}^\infty e^{-a^2y} \int_{x=0}^\infty e^{-\left(yx^2+\frac{ip}{h}x\right)}\,dx\,dy. \end{align} $$ In general $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ \end{align} $$ Let $u=x+\frac{b}{2a}\;\rightarrow\;du=dx$, then $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\ \end{align} $$ The last form integral is Gaussian integral that equals to $\frac{1}{2}\sqrt{\frac{\pi}{a}}$. Hence $$ \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right). $$ Thus $$ \int_{x=0}^\infty e^{-(yx^2+\frac{ip}{h}x)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(\frac{\left(\frac{ip}{h}\right)^2}{4y}\right)=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(-\frac{p^2}{4h^2y}\right). $$ Next $$ \int_{x=0}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-a^2y-\frac{p^2}{4h^2y}\right)}{\sqrt{y}}\,dy. $$ In general $$ \begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\ &=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ \end{align} $$ The trick to solve the last integral is by setting $$ I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv. $$ Let $t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then $$ I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt. $$ Let $t=v\;\rightarrow\;dt=dv$, then $$ I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Adding the two $I_t$s yields $$ 2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then $$ I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}. $$ Thus $$ \begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ &=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}\\ \end{align} $$ and $$ \begin{align} \int_{-\infty}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx&=2\cdot\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-a^2y-\frac{p^2}{4h^2y}\right)}{\sqrt{y}}\,dy\\ &=\sqrt{\pi}\cdot\sqrt{\frac{\pi}{a^2}}\;e^{-2\sqrt{a^2\cdot\frac{p^2}{4h^2}}}\\ &=\frac{\pi}{a}\;e^{-\frac{pa}{h}}. \end{align} $$

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

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