오늘의 계산 46 - A Simple Trigonometric Integral

Problem. Show that \begin{equation}\label{prob:wts} \int_{0}^{\infty} \frac{\cos ax}{x} \, \frac{\sinh\beta x}{\cosh\gamma x} \, dx = \frac{1}{2} \, \log \left( \frac{\cosh\frac{a\pi}{2\gamma}+\sin\frac{\beta\pi}{2\gamma}}{\cosh\frac{a\pi}{2\gamma}-\sin\frac{\beta\pi}{2\gamma}} \right) \end{equation} for $|\Re \beta| < \Re \gamma$.

(Original Problem by Doe John, in his posting [1].)

To prove \eqref{prob:wts}, we need some preliminaries.

Lemma 1. Let $\alpha_1, \cdots, \alpha_k$ and $\beta_1, \cdots, \beta_k$ be complex numbers such that \begin{equation}\label{lem1:cond} \alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k. \end{equation} Then \begin{equation}\label{lem1:wts} \lim_{n\to\infty} \frac{\Gamma(n+\alpha_1) \cdots \Gamma(n+\alpha_k)}{\Gamma(n+\beta_1) \cdots \Gamma(n+\beta_k)} = 1. \end{equation}

Proof. By Stirling's formula, \begin{align*} \Gamma(n + z + 1) & \sim \sqrt{2\pi (n+z)} \, \left( \frac{n+z}{e} \right)^{n+z} \\ & = \sqrt{2\pi (n+z)} \, \left( 1 + \frac{z}{n} \right)^{n+z} n^{n+z} e^{-(n+z)} \\ & = \sqrt{2\pi n} \, \left( 1 + \frac{z}{n} \right)^{z+(1/2)} \left( 1 + \frac{z}{n} \right)^{n} e^{-z} n^{n+z} e^{-n} \\ & \sim \sqrt{2\pi n} \, n^{n+z} e^{-n}. \end{align*} Therefore we have $$ \frac{\Gamma(n+\alpha_1) \cdots \Gamma(n+\alpha_k)}{\Gamma(n+\beta_1) \cdots \Gamma(n+\beta_k)} \sim \frac{n^{\alpha_1 - 1} \cdots n^{\alpha_k - 1}}{n^{\beta_1 - 1} \cdots n^{\beta_k - 1}} = 1. $$

Lemma 2. For any $z \in \mathbb{C}\setminus\mathbb{Z}$, we have \begin{equation}\label{lem2:wts} \frac{\Gamma(n+z)}{\Gamma(1-n+z)} \sim (-1)^{n-1} \frac{\sin \pi z}{\pi}\Gamma(n)^{2}. \end{equation}

Proof. From Euler's reflection formula, \begin{align*} \frac{\Gamma(n+z)}{\Gamma(1-n+z)} &= \frac{\Gamma(n+z)\Gamma(n-z)}{\Gamma(1-n+z)\Gamma(n-z)} \\ &= \frac{\sin \pi (1-n+z)}{\pi} \Gamma(n+z)\Gamma(n-z) \\ &= (-1)^{n-1} \frac{\sin \pi z}{\pi} \Gamma(n+z)\Gamma(n-z) \\ &\sim (-1)^{n-1} \frac{\sin \pi z}{\pi} \Gamma(n)^2, \end{align*} where the approximation relation follows from \eqref{lem1:wts}.

Lemma 3. Let $\alpha_1, \cdots, \alpha_k$ and $\beta_1, \cdots, \beta_k$ be complex numbers away from integers satisfying \eqref{lem1:cond}. Then \begin{equation}\label{lem3:wts} \prod_{n=-\infty}^{\infty} \frac{(n+\alpha_1) \cdots (n+\alpha_k)}{(n+\beta_1) \cdots (n+\beta_k)} = \frac{\sin (\pi \alpha_1) \cdots \sin (\pi \alpha_k)}{\sin (\pi \beta_1) \cdots \sin (\pi \beta_k)}. \end{equation}

Proof. From \eqref{lem2:wts}, \begin{align*} \prod_{n=-\infty}^{\infty} \frac{(n+\alpha_1) \cdots (n+\alpha_k)}{(n+\beta_1) \cdots (n+\beta_k)} &= \lim_{N\to\infty} \prod_{n=-N+1}^{N-1} \frac{(n+\alpha_1) \cdots (n+\alpha_k)}{(n+\beta_1) \cdots (n+\beta_k)} \\ & =\lim_{N\to\infty} \frac{\dfrac{\Gamma(N+\alpha_1)}{\Gamma(1-N+\alpha_1)} \cdots \dfrac{\Gamma(N+\alpha_k)}{\Gamma(1-N+\alpha_k)}}{\dfrac{\Gamma(N+\beta_1)}{\Gamma(1-N+\beta_1)} \cdots \dfrac{\Gamma(N+\beta_k)}{\Gamma(1-N+\beta_k)}} \\ & = \frac{\sin (\pi \alpha_1) \cdots \sin (\pi \alpha_k)}{\sin (\pi \beta_1) \cdots \sin (\pi \beta_k)}. \end{align*}

Lemma 4. Let $\alpha_1, \cdots, \alpha_k$ and $\beta_1, \cdots, \beta_k$ be complex numbers away from integers. Then \begin{equation}\label{lem4:wts} \prod_{n=-\infty}^{\infty} \left( \frac{(n+\alpha_1) \cdots (n+\alpha_k)}{(n+\beta_1) \cdots (n+\beta_k)} \right)^{(-1)^n} = \frac{\tan(\pi \alpha_1 /2) \cdots \tan(\pi \alpha_k /2)}{\tan(\pi \beta_1 / 2) \cdots \tan(\pi \beta_k /2)}. \end{equation}

Proof. From \eqref{lem3:wts}, \begin{align*} \prod_{n=-\infty}^{\infty} \left( \frac{n+\alpha}{n+\beta}\right)^{(-1)^n} &= \prod_{n=-\infty}^{\infty} \left( \frac{2n+\alpha}{2n+\beta} \cdot \frac{2n+1+\beta}{2n+1+\alpha} \right) \\ &= \prod_{n=-\infty}^{\infty} \frac{\left( n+\frac{\alpha}{2} \right)\left( n+\frac{1+\beta}{2} \right)}{\left( n+\frac{\beta}{2} \right)\left( n+\frac{1+\alpha}{2} \right)} \\ &= \frac{\sin ( \pi \alpha / 2 ) \sin (\pi (1+\beta) / 2 )}{\sin ( \pi \beta / 2 ) \sin ( \pi (1+\alpha) / 2 )} \\ &= \frac{\tan ( \pi \alpha / 2 )}{\tan ( \pi \beta / 2 )}. \end{align*}

Now we return to the original problem.

Proof of the Problem. For $x > 0$, we have $$ \left|\cosh (\gamma x)\right| \geq \frac{e^{\Re \gamma x} - e^{-\Re \gamma x}}{2} = \sinh (\Re \gamma x)$$ and likewise $$ |\sinh (\beta x)| \leq \frac{e^{\Re \beta x} + e^{-\Re \beta x}}{2} = \cosh (\Re \beta x). $$ Thus $$ \left| \frac{\cos ax}{x} \frac{\sinh \beta x}{\cosh \gamma x} \right| \leq \frac{1}{x} \frac{\cosh (\Re \beta x)}{\sinh (\Re \gamma x)} = O\left( \exp \left( -(\Re \gamma - |\Re \beta|) x \right) \right), $$ and hence \eqref{prob:wts} is analytic in the prescribed region. Therefore it suffices to consider the case $\beta, \gamma > 0$. For $a, p, q > 0$, Corollary 4 of [2] shows that \begin{align*} \int_{0}^{\infty} \frac{\cos ax}{x} \, (e^{-px} - e^{-qx}) \, dx &= \int_{0}^{\infty} \frac{\cos x}{x} \, (e^{-px/a} - e^{-qx/a}) \, dx \quad (x \mapsto x/a)\\ &= \mathrm{ctr} \; \left(\cos x \, e^{-px/a}\right) - \mathrm{ctr} \; \left(\cos x \, e^{-qx/a}\right) \\ &= \frac{1}{2} \log \left( \frac{a^2 + q^2}{a^2 + p^2} \right). \end{align*} Thus we have \begin{align*} \int_{0}^{\infty} \frac{\cos ax}{x} \, \frac{\sinh \beta x}{\cosh \gamma x} \, dx &= \int_{0}^{\infty} \frac{\cos ax}{x} \, \frac{e^{-(\gamma-\beta)x } - e^{-(\gamma+\beta)x}}{1 + e^{-2\gamma x}} \, dx \\ &= \int_{0}^{\infty} \frac{\cos ax}{x} \, \left[ \sum_{n=0}^{\infty} (-1)^{n} \left( e^{-((2n+1)\gamma-\beta)x } - e^{-((2n+1)\gamma+\beta)x} \right) \right] \, dx \\ &= \frac{1}{2} \sum_{n=0}^{\infty} (-1)^{n} \log \left( \frac{a^2 + ((2n+1)\gamma+\beta)^2}{a^2 + ((2n+1)\gamma-\beta)^2} \right) \\ &= \frac{1}{4} \sum_{n=-\infty}^{\infty} (-1)^{n} \log \left( \frac{a^2 + ((2n+1)\gamma+\beta)^2}{a^2 + ((2n+1)\gamma-\beta)^2} \right) \\ &= \frac{1}{4} \log \prod_{n=-\infty}^{\infty} \left( \frac{a^2 + ((2n+1)\gamma+\beta)^2}{a^2 + ((2n+1)\gamma-\beta)^2} \right)^{(-1)^n}. \end{align*} Now letting $u = \frac{a}{2\gamma}$ and $v = \frac{\beta}{2\gamma}$, we obtain \begin{align*} \int_{0}^{\infty} \frac{\cos ax}{x} \, \frac{\sinh \beta x}{\cosh \gamma x} \, dx & = \frac{1}{4} \log \prod_{n=-\infty}^{\infty} \left[ \frac{\left( u^2 + \left(n+\frac{1}{2}+v\right)^2 \right)}{\left( u^2 + \left(n+\frac{1}{2}-v\right)^2 \right)} \right]^{(-1)^n} \\ & = \frac{1}{4} \log \prod_{n=-\infty}^{\infty} \left[ \frac{\left(n+\frac{1}{2}+v + iu\right)\left(n+\frac{1}{2}+v - iu\right)}{\left(n+\frac{1}{2}-v + iu\right)\left(n+\frac{1}{2}-v - iu\right)} \right]^{(-1)^n} \\ & = \frac{1}{4} \log \left[ \frac{\tan \left(\dfrac{\pi}{4}+\dfrac{\pi(v + iu)}{2}\right) \tan \left(\dfrac{\pi}{4}+\dfrac{\pi(v - iu)}{2}\right)}{\tan \left(\dfrac{\pi}{4}-\dfrac{\pi(v + iu)}{2}\right) \tan \left(\dfrac{\pi}{4}-\dfrac{\pi(v - iu)}{2}\right)} \right]. \end{align*} Now we have the following trigonometric identities $$ \cot \left( \frac{\pi}{4} - \theta \right) = \tan \left( \frac{\pi}{4} + \theta \right) \quad \text{and} \quad \tan \left (\frac{\alpha+\beta}{2} \right) \tan\left( \frac{\alpha-\beta}{2} \right) = \frac{\cos \beta - \cos \alpha}{\cos \beta + \cos \alpha},$$ from which we deduce \begin{align*} \int_{0}^{\infty} \frac{\cos ax}{x} \, \frac{\sinh \beta x}{\cosh \gamma x} \, dx & = \frac{1}{2} \log \left[ \tan \left(\frac{\pi}{4}+\frac{\pi(v + iu)}{2}\right) \tan \left(\frac{\pi}{4}+\frac{\pi(v - iu)}{2}\right) \right] \\ & = \frac{1}{2} \log \left( \frac{\cos \left( i \pi u \right) - \cos \left( \frac{\pi}{2} + \pi v \right)}{\cos \left( i \pi u \right) + \cos \left( \frac{\pi}{2} + \pi v \right)} \right) \\ & = \frac{1}{2} \log \left( \frac{\cosh \left( \pi u \right) + \sin \left( \pi v \right)}{\cosh \left( \pi u \right) - \sin \left( \pi v \right)} \right). \end{align*} This proves \eqref{prob:wts} as desired.

References

1. Doe John, 'AoPS Forum - another integral' in "Art of Problem Solving"
2. sos440, '오늘의 계산 41 - Realizing Nonsensical (in progress)' in "수학과 잡담을 위한 소박한 장소"