# Homework Help: Complex analysis integral

1. Aug 3, 2016

### Physgeek64

1. The problem statement, all variables and given/known data
calculate $\int_{-\infty}^{\infty}{\frac{2}{1+x^2} e^{-ikx}dx}$, where k is any positive number

2. Relevant equations

3. The attempt at a solution
So first consider the closed contour $I= \int{\frac{2}{1+z^2} e^{-ikz}dx}$

We can choose the contour to be along the real axis $[-R,R]$ then along the semi circle for which $z=Re^{i \theta}, [0, \pi]$

then $I= \int_{-\infty}^{\infty}{\frac{2}{1+x^2} e^{-ikx}dx} +\int_{0}^{\pi} {\frac{2}{1+(Re^{i \theta})^2} e^{-ikRe^{i\theta}}dx}$

I calculated the residue to be $\pi e^{k}$, with the poles being at $z=i$ and $z=-i$

I think the integral around the semi-circle is supposed to go to zero in the limit of R going to infinity, but I am struggling to see why. I can't use Jordan's lemma (?) since it is not in the usual formal, and if i expand

$e^{-ikRe^{i\theta}} = e^{-ikR(cos(\theta)+isin(\theta))}$
$= e^{-ikRcos(\theta)}e^{kRsin(\theta)}$

but since $\theta$ is between $[0, \pi], sin(\theta)$ is greater than, or equal to zero, meaning that the integrand does not drop off as R goes to infinity..

Any help would be greatly appreciated- thanks in advance :)

Last edited: Aug 3, 2016
2. Aug 3, 2016

3. Aug 3, 2016

### Physgeek64

Noted and corrected. Thank you :)

4. Aug 3, 2016

### Ray Vickson

For $k > 0$ completing in the upper x-plane won't work, because for $x = t + i y$ we have $e^{-ikx}= e^{-ikt}\, e^{ky}$. which behaves badly for large $y > 0$. However, we can complete in the lower x-plane instead.

Alternatively, we can see that for real $x$ the integrand has a real part that is an even function of $x$ and an imaginary part that is odd (so will integrate to 0); that is, we could obtain the same result if we replace $e^{-ikx}$ by $\cos(kx)$. But then, we might as well extend this to $e^{ikx}$ without changing its value.

So, let's look at the integral with integrand $f(x) = e^{ikx}/(1+x^2)$. Now we can complete in the upper x-plane, because$e^{ik(t+iy)} = e^{ikt} e^{-ky}$. However, completing along a semi-circle is not very convenient; using a rectangle is much easier. So, let's take the original integral from $x = -R$ to $x = +R$, taking $R \to \infty$ eventually. Complete in a rectangle as follows: append the integral from $R+i0$ to $R + iR$ along the vertical, then from $R + iR$ to $-R + iR$ along an upper horizontal, then from $-R + iR$ down to $-R + i0$ along the other vertical. It is pretty easy to get bounds on each of the vertical and upper-horizontal integrals, and to show that they $\to 0$ as $R \to \infty$.

Last edited: Aug 3, 2016
5. Aug 3, 2016

### Physgeek64

Closing in the lower half plane produces the same problem. Though the exponential power will be negative, over the limit $[ 0, -\pi], sin(\theta)$ is negative, meaning the resultant power is positive, and the same problem occurs. (This may also be seen by making the substitution $x \to (-x)$ in my initial working).

Edit:
But x (more accurately z) itself is complex, not real, over either the semi-circle or rectangle so taking the real part of the function will not simply be a matter of changing the exponential into its $cosx+isinx$ form. For example the integrand is actually $\frac{1}{1+R^2e^{2i\theta}} e^{-ikRe^{i\theta}}Rie^{i\theta}$ taking the real part of this is not as straightforward, surely ?

Sorry if I'm missing something obvious- thanks for the help

Last edited: Aug 3, 2016
6. Aug 3, 2016

### Ray Vickson

No: we start with the integral over the real line, and show that we can replace $e^{-ikx}$ by $e^{ikx}$ in that integral. Then we extend $x$ into the upper complex plane. Alternatively, complete in the lower x-plane, and use arguments similar to the one below.

Taking $e^{ikx} = e^{ik(t + iy)} = e^{ikt} e^{-ky}$ for $t = R$ and $0 \leq y \leq R$, the right-hand vertical integral becomes
$$I_1 = \int_0^R \frac{e^{ikR} e^{-ky}}{(R + iy)^2 + 1} i \, dy,$$.
Use the facts that $|\int f| \leq \int |f|$, $|a/b| = |a|/|b|$ and $|e^{ikt} e^{-ky}| = |e^{ikt}| e^{-ky} = e^{-ky}$. The magnitude of the denominator in $|f|$ is
$$|(R+iy)^2 + 1| = \sqrt{R^4+2R^2 y^2+y^4+2 R^2-2y^2+1} \geq \sqrt{R^4 + 2 R^2},$$
since $2R^2 y^2 \geq 0$ and $y^4 - 2 y^2 \geq -1$ for $y \geq 0$. We thus have
$$|I_1| \leq \int_0^R \frac{e^{-ky}}{ \sqrt{R^4+2R^2}} dy \to 0$$
as $R \to \infty$.

Similar types of arguments apply to the other vertical integral and to the upper horizontal integral.

Trying something like the above on a semi-circular arc of radius $R$ would be much harder, so that is why I suggested not doing it that way.

Last edited: Aug 3, 2016