오늘의 계산 43 - Calculations of Four Integrals

Here I post my solutions to the announced four integrals. I tacitly assumed some formulas to be granted, so to reduce unnecessary calculations that are attainable in regular courses in analysis. One exception to this convention is the detailed proofs for the identity \begin{equation*} \int_{0}^{\infty} \frac{\cos x}{x^2 + \beta^2} \; dx = \frac{\pi}{2\beta} e^{-\beta} \end{equation*} for $\beta \geq 0$, or equivalently \begin{equation*} \int_{0}^{\infty} \frac{\cos \alpha x}{x^2 + \beta^2} \; dx = \frac{\pi}{2\beta} e^{-\alpha \beta} \end{equation*} for $\alpha, \beta > 0$, which you can find in this unfinished document.

 

1^st Integral. Show that for $\alpha \geq 0$, \begin{equation}\label{int1:wts} \int_{-\infty}^{\infty} \frac{\sin (\alpha \tan x)}{x} \; dx = \pi \left( 1 - e^{-\alpha} \right). \end{equation}

Solution.

This integral requires the following identity \begin{equation} \label{int1:cot} \sum_{n=-\infty}^{\infty} \frac{x}{x^2 - n^2\pi^2} = \cot x. \end{equation} Define \begin{equation} \label{int1:in} I(n) := \int_{n\pi -\frac{\pi}{2}}^{n\pi + \frac{\pi}{2}} \frac{\sin (\alpha \tan x)}{x} \; dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin (\alpha \tan x)}{x + n \pi} \; dx. \end{equation} Then \eqref{int1:in} is absolutely convergent, and we have $I(n) = I(-n)$ by symmetry. Thus \begin{equation*} I(n) = \frac{1}{2}\left( I(n) + I(-n) \right) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x \sin (\alpha \tan x)}{x^2 - n^2 \pi^2} \; dx. \end{equation*} Since \begin{equation*} \frac{x}{x^2 - n^2 \pi^2} = O \left( n^{-2} \right) \end{equation*} on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, applying \eqref{int1:cot} shows that \begin{align*} \int_{0}^{\left( N + \frac{1}{2} \right)\pi} \frac{\sin (\alpha \tan x)}{x} \; dx & = \frac{1}{2} \left( I(-N) + \cdots + I(N) \right) \\ & = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum_{n=-N}^{N} \frac{x}{x^2 - n^2 \pi^2} \right) \sin (\alpha \tan x) \; dx \\ & = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \cot x + O \left( N^{-1} \right) \right) \sin (\alpha \tan x) \; dx \\ & = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin (\alpha \tan x)}{\tan x} \; dx + O \left( N^{-1} \right). \end{align*} Taking $N \to \infty$ proves that \eqref{int1:wts} converges, and moreover \begin{align*} \int_{-\infty}^{\infty} \frac{\sin (\alpha \tan x)}{x} \; dx & = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin (\alpha \tan x)}{\tan x} \; dx \\ & = \int_{-\infty}^{\infty} \frac{\sin \alpha t}{t (t^2 + 1)} \; dt \qquad (t = \tan x) \\ & = \int_{-\infty}^{\infty} \frac{\sin \alpha t}{t} \; dt - \int_{-\infty}^{\infty} \frac{t \sin \alpha t}{t^2 + 1} \; dt \\ & = \pi \left(1 - e^{-\alpha} \right), \end{align*} as desired.

2^nd Integral. Show that \begin{equation} \label{int2:wts} \int_{-\infty}^{\infty} \sin x \arctan\left( \frac{1}{x} \right) \; dx = \pi \left( 1 - \frac{1}{e} \right). \end{equation}

Solution. The integral \eqref{int2:wts} is easiest among the four. Simply taking integration by parts, we obtain \begin{align*} \int_{-\infty}^{\infty} \sin x \arctan\left( \frac{1}{x} \right) \; dx & = 2 \int_{0}^{\infty} \sin x \arctan\left( \frac{1}{x} \right) \; dx \\ & = \left. 2 (1 - \cos x) \arctan\left( \frac{1}{x} \right) \right|_{0}^{\infty} + 2 \int_{0}^{\infty} \frac{1 - \cos x}{1 + x^2} \; dx \\ & = \pi \left( 1 - \frac{1}{e} \right). \end{align*}

3^rd Integral. Show that \begin{equation} \label{int3:wts} \int_{0}^{\infty} \frac{\left( 1 - e^{-6x} \right) e^{-x}}{x \left(1 + e^{-2x} + e^{-4x} + e^{-6x} + e^{-8x} \right)} \; dx = \log \left( \frac{3 + \sqrt{5}}{2} \right). \end{equation}

Solution. Let us define \begin{equation} \label{int3:is} I(s) = \int_{0}^{\infty} \frac{\left( 1 - e^{-2x} \right)\left( 1 - e^{-6x} \right)}{1 - e^{-10x}} \frac{e^{-sx}}{x} \; dx. \end{equation} It is clear that the integral \eqref{int3:wts} is equal to $I(1)$. By referring to the identity \begin{equation*} \psi_0 (s) \ = \ -\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+s} \right), \end{equation*} hence differentiating \eqref{int3:is} gives \begin{align*} I'(s) & = - \int_{0}^{\infty} \frac{(1 - e^{-2x})(1 - e^{-6x})e^{-sx}}{1 - e^{-10x}} \; dx \\ & = - \int_{0}^{\infty} (1 - e^{-2x})(1 - e^{-6x})e^{-sx} \left( \sum_{n=0}^{\infty} e^{-10nx} \right) \; dx \\ & = - \sum_{n=0}^{\infty} \left( \frac{1}{10n + s} - \frac{1}{10n + s + 2} - \frac{1}{10n + s + 6} + \frac{1}{10n + s + 8} \right) \\ & = \frac{1}{10} \left[ \psi_0 \left( \frac{s}{10} \right) - \psi_0 \left( \frac{s+2}{10} \right) - \psi_0 \left( \frac{s+6}{10} \right) + \psi_0 \left( \frac{s+8}{10} \right) \right]. \end{align*} Since $I(\infty) = 0$, we have \begin{align*} I(1) & = \int_{\infty}^{1} I'(t) \; dt \\ & = \left. \log \left( \frac{\Gamma \left( \frac{t}{10} \right) \Gamma \left( \frac{t+8}{10} \right)}{\Gamma \left( \frac{t+2}{10} \right) \Gamma \left( \frac{t+6}{10} \right)} \right) \right|_{\infty}^{1} = \log \left( \frac{\Gamma \left( \frac{1}{10} \right) \Gamma \left( \frac{9}{10} \right)}{\Gamma \left( \frac{3}{10} \right) \Gamma \left( \frac{7}{10} \right)} \right) = \log \left( \frac{\sin \left( \frac{3\pi}{10} \right)}{\sin \left( \frac{\pi}{10} \right)} \right) \\ & = \log \left( \frac{ 3 + \sqrt{5}}{2} \right) \end{align*} and therefore \eqref{int3:wts} follows.

4^th Integral. Show that for $\Re \alpha, \Re \beta \geq 0$ and $\alpha \neq 0$, we have \begin{equation} \label{int4:wts} \int_{-\infty}^{\infty} \exp\left( - \alpha x^2 - \frac{\beta}{x^2} \right) = \sqrt{\frac{\pi}{\alpha}} e^{-2\sqrt{\alpha \beta}}. \end{equation}

Solution. First we observe that it suffice to consider the case $\Re \alpha > 0$ and $\Re \beta > 0$ by limiting argument. Being analytic on this domain, we find that it suffices to prove \eqref{int4:wts} for $\alpha, \beta > 0$. Now let \begin{equation*} a = \sqrt{\frac{\alpha}{\beta}} \quad \text{and} \quad b = \sqrt{\alpha \beta}. \end{equation*} Let $I$ denote the integral \eqref{int4:wts}. Then by the substitutions \begin{equation*} s = \sqrt{a} \, x, \quad t = \frac{1}{s}, \quad \text{and} \quad u = t - \frac{1}{t}, \end{equation*} we have \begin{align*} I & = \int_{-\infty}^{\infty} \exp \left( -b\left( a x^{2}+ \frac{1}{ax^2} \right) \right) \; dx \\ & = \frac{1}{\sqrt{a}}\int_{-\infty}^{\infty}\exp\left(-b\left( s^{2}+\frac{1}{s^{2}}\right)\right)\; ds\\ & = \frac{2}{\sqrt{a}}\int_{0}^{\infty}\exp\left(-b\left( s^{2}+\frac{1}{s^{2}}\right)\right)\; ds\\ & = \frac{1}{\sqrt{a}} \bigg[ \int_{0}^{\infty}\exp\left(-b\left( s^{2} + \frac{1}{s^{2}}\right)\right)\; ds + \int_{0}^{\infty}\exp\left(-b\left( t^{2}+\frac{1}{t^{2}}\right)\right)\; \frac{dt}{t^2} \bigg]\\ & = \frac{1}{\sqrt{a}}\int_{0}^{\infty}\exp\left(-b\left( t^{2}+\frac{1}{t^{2}}\right)\right)\,\left( 1+\frac{1}{t^{2}}\right)\; dt\\ & = \frac{1}{\sqrt{a}}\int_{-\infty}^{\infty}\exp\left(-b\left( u^{2}+2\right)\right)\; du\\ & = \sqrt{\frac{\pi}{a b}}\, e^{-2b}, \end{align*} which completes the proof.