# Chaitanya's Random Pages

## October 30, 2014

### Mountain ranges of the world

Filed under: geography — ckrao @ 1:06 pm

As a geography lesson for myself, here is a list of major mountain ranges of the world, where those that are either long or have a significantly high peak are shown (hence several near Tibet are given). Maps are shown below the table.

## October 25, 2014

### An integral related to Bessel polynomials

Filed under: mathematics — ckrao @ 12:46 am

In this post I want to share this cute result that I learned recently:

$\displaystyle \int_{-\infty}^{\infty} \frac{\cos x}{x^2+1}\ dx = \int_{-\infty}^{\infty} \frac{\cos x}{(x^2+1)^2}\ dx = \frac{\pi}{e} \quad\quad (1)$

Let us see how the first integral is derived and then generalised. The integrand has poles at $\pm i$ in the complex plane and we may apply contour integration to proceed. Note that as $\sin x /(x^2+1)$ is an odd function, $\int_{-\infty}^{\infty} \sin x /(x^2+1)\ dx = 0$ and so

$\int_{-\infty}^{\infty}\frac{\cos x}{x^2+1}\ dx = \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1}.\quad\quad(2)$

It therefore suffices to consider the integrand $f(z) = e^{iz}/(z^2+1)$. We consider the semicircular contour of radius $R$ (to be traversed anticlockwise) in the upper half plane centred at 0. It encloses the pole at $i$.

Along this closed contour we use the residue theorem to compute

\begin{aligned} \oint f(z)\ dz &= 2\pi i \text{Res}[f(z)]_{z=i}\\ &= 2\pi i \lim_{z \rightarrow i} (z-i)\frac{e^{iz}}{z^2 +1}\\ &= 2\pi i \lim_{z \rightarrow i} \frac{e^{iz}}{z +i}\\ &= 2\pi i \frac{e^{-1}}{2i}\\ &= \frac{\pi}{e}.\quad\quad (3) \end{aligned}

On the semicircular arc $z = R(\cos \theta + i\sin \theta) = Re^{i\theta}$ for $0 \leq \theta \leq \pi$ and so

\begin{aligned} |f(z)| &= \left| \frac{e^{iz}}{z^2+1} \right|\\ &= \frac{|e^{iR(\cos \theta + i\sin \theta)}|}{|R^2 e^{i2\theta} + 1|}\\ &= \frac{|e^{-R \sin \theta}|}{|R^2 e^{i2\theta} + 1|}\\ &\leq \frac{e^{-R\sin \theta}}{R^2-1}\\ &\leq \frac{1}{R^2-1} \quad \text{for } 0 \leq \theta \leq \pi,\quad\quad (4) \end{aligned}

where in the second last step we use the fact that distance to the origin of a circle of radius $R^2$ centred at -1 is at least $R^2-1$. Then

\begin{aligned} \lim_{R \rightarrow \infty} \int_C f(z)\ dz &\leq \lim_{R \rightarrow \infty} \int_C |f(z)| \ |dz| \\ &\leq \lim_{R \rightarrow \infty} \pi R \sup_{z \in C} |f(z)|\\ &\leq \lim_{R \rightarrow \infty} \pi R \frac{1}{R^2-1} \quad \text{(by (4))}\\ &= 0.\quad\quad(5) \end{aligned}

It follows that $\lim_{R \rightarrow \infty} \int_C f(z)\ dz = 0$ and we are left with

\begin{aligned} \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1}\ dx &= \lim_{R \rightarrow \infty} \int_{-R}^R f(z)\ dz \\ &= \lim_{R \rightarrow \infty} \oint f(z)\ dz - \int_C f(z) \ dz\\ &= \lim_{R \rightarrow \infty} \oint f(z)\ dz\\ &= \frac{\pi}{e} \quad\text{(by (3)).}\quad \quad(6) \end{aligned}

To generalize this result to integrals of the form $\int_{-\infty}^{\infty} \frac{\cos x}{(x^2+1)^{n+1}}\ dx = \int_{-\infty}^{\infty} \frac{e^{ix}}{(x^2+1)^{n+1}}\ dx$ for non-negative integers $n$, we choose the same contour as above and use the residue limit formula for poles of order $(n+1)$:

$\displaystyle \text{Res} [f(z)]_{z=z_0} = \frac{1}{n!} \lim_{z \rightarrow z_0} \left[\frac{d^n}{dz^n} (z-z_0)^{n+1} f(z) \right]. \quad\quad (7)$

To apply (7) we make use of the General Leibniz rule for the $n$‘th derivative of a product:

$\displaystyle (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)} g^{(n-k)}.\quad \quad (8)$

Hence

\begin{aligned} \oint \frac{e^{iz}}{(z^2+1)^{n+1}}\ dz &= 2\pi i \text{Res} \left[ \frac{e^{iz}}{(z^2+1)^{n+1}} \right]_{z=i}\\ &= \frac{2\pi i}{n!} \left[\frac{d^n}{dz^n}\frac{e^{iz}}{(z+i)^{n+1}} \right]_{z=i}\\ &= \frac{2\pi i}{n!} \sum_{k=0}^n \left[\binom{n}{k}\left(e^{iz}\right)^{(n-k)} \left(\frac{1}{(z+i)^{n+1}}\right)^{(k)}\right]_{z=i} \quad\text{(applying (8))}\\ &= \frac{2\pi i}{n!} \sum_{k=0}^n \binom{n}{k} (i)^{n-k} e^{-1} (-n-1)(-n-2)\ldots (-n-k) \frac{1}{(2i)^{n+k+1}}\\ &= \frac{2\pi i}{n!e} \sum_{k=0}^n \binom{n}{k}\frac{1}{i^{2k+1}}\frac{(n+k)!}{n!}(-1)^k\\ &= \frac{\pi}{n!e} \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{(n+k)!}{n!}\frac{1}{2^{n+k}}\\ &= \frac{\pi}{e} \frac{1}{2^n n!}\sum_{k=0}^n \frac{(n+k)!}{k!(n-k)!}\frac{1}{2^k}\\ &= \frac{\pi}{e} \frac{1}{2^n n!} y_n(1), \quad\quad(8)\end{aligned}

where $y_n(x)$ is the $n$‘th order Bessel polynomial defined by

$\displaystyle y_n(x) = \sum_{k=0}^n \frac{(n+k)!}{k!(n-k)!} \left(\frac{x}{2}\right)^k.\quad\quad(9)$

For example, for $n = 1$, the integral is given by

\begin{aligned} 2\pi i \left[ \frac{d}{dz} \frac{e^{iz}}{(z+i)^2}\right]_{z=i} &= \left[ \frac{-2e^{iz}}{(z+i)^3} + \frac{ie^{iz}}{(z+i)^2}\right]_{z=i}\\ &= 2\pi i\left(\frac{-2e^{-1}}{8i^3} + \frac{ie^{-1}}{-4}\right)\\ &= \frac{\pi}{e}. \quad\quad (10)\end{aligned}

The general sum in (8) can be also be written as

$\displaystyle \sum_{k=0}^n \frac{(k+n)!}{2^k k!(n-k)!} = e \sqrt{\frac{2}{\pi}}K_{n+1/2}(1),\quad \quad (11)$

where $K_{\alpha}(x)$ is the modified Bessel function of the second kind.

Proceeding similarly to (4), $\left| e^{iz}/(1+z^2)^{n+1} \right| \leq 1/(R^2-1)^{n+1}$ and so similar to (5)  the integral on the arc converges to 0 as $R \rightarrow \infty$. Hence following the same argument as in (6) the desired line integral along the real axis is equal to the contour integral in (8).

Evaluating (8) for $n = 0, 1, 2,3,4,5,\ldots$ the first terms of  $(e/\pi)\int_{-\infty}^{\infty} \frac{\cos x}{(x^2+1)^2}\ dx$ are given by

$\displaystyle \frac{1}{1}, \frac{2}{2}, \frac{7}{8}, \frac{37}{48}, \frac{266}{384}, \frac{2431}{3840}, \cdots\quad\quad(12)$

In this sequence $a_n/b_n$, the denominators $b_n$ are related to the previous ones by multiplication by $2n$, while curiously the numerators are related by the second order recurrence

$a_{n} = (2n-1)a_{n-1} + a_{n-2}.\quad\quad(13)$

This follows from the following recurrence relation for the Bessel polynomials:

$y_n(x) = (2n-1)xy_{n-1}(x) + y_{n-2}(x).\quad\quad (14)$

This can be proved using (9). We have

\begin{aligned} (2n-1)xy_{n-1}(x) + y_{n-2}(x) &= (2n-1)x\sum_{k=0}^{n-1} \frac{(n+k)!}{k!(n-k)!} \left(\frac{x}{2}\right)^k + \sum_{k=0}^{n-2} \frac{(n+k)!}{k!(n-k)!} \left(\frac{x}{2}\right)^k\\ &=(2n-1)\sum_{k=1}^n \frac{(n+k-2)!}{(k-1)!(n-k)!}2\frac{x^k}{2^k} + \sum_{k=0}^{n-2} \frac{(n+k-2)!}{k!(n-2-k)!}\left(\frac{x}{2}\right)^k\\ &= \frac{(n-2)!}{0!(n-2)!} + \sum_{k=1}^{n-2} (n+k-2)! \left[ \frac{2(2n-1)}{(k-1)!(n-k)!} + \frac{1}{k!(n-k-2)!}\right] \left(\frac{x}{2}\right)^k \\ & \quad \quad + (2n-1) \left[\frac{n + n -1-2)!}{(n-2)!1!}2 \left(\frac{x}{2}\right)^{n-1} + \frac{(2n-2)!}{(n-1)!0!}2\left(\frac{x}{2}\right)^n\right]\\ &= 1 + \sum_{k=1}^{n-2} (n+k-2)! \left[\frac{2(2n-1)k + (n-k-1)(n-k)}{k!(n-k)!}\right]\left(\frac{x}{2}\right)^k\\ & \quad \quad + (2n-1) \left[ \frac{(2n-3)!}{(n-2)!}2\left(\frac{x}{2}\right)^{n-1} + \frac{(2n-2)!}{(n-1)!}2\left(\frac{x}{2}\right)^n\right].\quad\quad(15) \end{aligned}

Now

\begin{aligned} 2(2n-1)k + (n-k-1)(n-k) &= 2k(2n-1) + (n + k - 1 - 2k)(n-k)\\ &= 2k(2n-1-n+k) + (n+k-1)(n-k)\\ &= (n+k-1)(2k + n-k)\\ &= (n+k-1)(n+k).\quad\quad(16) \end{aligned}

Substituting this into (15),

\begin{aligned} (2n-1)xy_{n-1}(x) + y_{n-2}(x) &= 1 + \sum_{k=1}^{n-2}\frac{(n+k)!}{k!(n-k)!}\left(\frac{x}{2}\right)^k + \frac{(2n-1)!(n-1)2}{(n-1)!(2n-2)}\left(\frac{x}{2}\right)^{n-1} + \frac{(2n)!n2}{n! 2n}\left(\frac{x}{2}\right)^{n}\\ &= \sum_{k=0}^n \frac{(n+k)!}{k!(n-k)!}\left(\frac{x}{2}\right)^k\\ &= y_n(x), \end{aligned}

thus verifying (14).

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