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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.

semicircular contour

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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