Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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. –  sos440 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

2 Answers 2

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 $C$)

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

so

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

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

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

share|improve this answer
    
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 $$ \begin{align} \int_{-\infty}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx&=2\cdot\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-a^2y-\frac{p^2}{4h^2y}\right)}{\sqrt{y}}\,dy\\ &=\sqrt{\pi}\cdot\sqrt{\frac{\pi}{a^2}}\;e^{-2\sqrt{a^2\cdot\frac{p^2}{4h^2}}}\\ &=\frac{\pi}{a}\;e^{-\frac{pa}{h}}. \end{align} $$


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

share|improve this answer

Your Answer

 
discard

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.