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$ \newcommand{\sech}{\operatorname{sech}} $ Is there any analytic/asymptotic way to estimate the value of the integral:

$$ \int_{0}^{\infty} \sech\left(\varepsilon x\right)\cdot \cos\left(kx\right)\,dx , \qquad k,\varepsilon > 0, $$ where $\varepsilon $ is a small parameter.

Thank you.

Martin Sleziak
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lisa
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1 Answers1

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$$\newcommand{\sech}{\operatorname{sech}} \begin{align*}\int_{0}^{+\infty}\frac{\cos(nx)}{\cosh x}\,dx &= 2\sum_{m\geq 0}(-1)^m \int_{0}^{+\infty}\cos(n x)e^{-(2m+1)x}\,dx\\&=2\sum_{m\geq 0}\frac{(-1)^m (2m+1)}{n^2+(2m+1)^2}\\&=\frac{\pi}{2}\,\sech\frac{\pi n}{2}\end{align*}$$ since: $$ \text{Res}\left(\sec\frac{\pi z}{2},z=2m+1\right)=\frac{2}{\pi}(-1)^{m+1} $$ hence it follows that: $$ \int_{0}^{+\infty}\sech(\varepsilon x)\cos(kx)\,dx = \frac{\pi}{2\varepsilon}\,\sech\left(\frac{\pi k}{2\varepsilon}\right)$$ as checked by Claude Leibovici with a CAS. It is well-known that the hyperbolic secant is more or less a fixed point of the Fourier transform.

Martin Sleziak
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Jack D'Aurizio
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  • I am trying to write a derivation for the above identity using the logarithmic derivative of Euler's infinite product formulas for the sine and cosine functions, but am unable to find a way to do so. It seems to depend on $\displaystyle\sum_{m=-\infty}^\infty\frac{(-1)^m}{m^2+a^2}=\frac\pi a~\text{csch }a\pi$. – Lucian Aug 23 '15 at 19:18
  • It is obviously related to $\displaystyle\sum_{m=-\infty}^\infty\frac1{m^2+a^2}=\frac\pi a~\coth a\pi$, but I'm not sure how exactly. – Lucian Aug 23 '15 at 19:25
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    @Lucian: $$\sum_{m\in\mathbb{Z}}\frac{(-1)^m}{m^2+a^2}+\sum_{m\in\mathbb{Z}}\frac{1}{m^2+a^2}=2\sum_{m\in\mathbb{Z}}\frac{1}{4m^2+a^2}=\frac{\pi}{a}\coth\frac{\pi a}{2},$$ hence: $$\sum_{m\in\mathbb{Z}}\frac{(-1)^m}{m^2+a^2}=\frac{\pi}{a}\coth\frac{\pi a}{2}-\frac{\pi}{a}\coth\frac{\pi a}{1}.$$ – Jack D'Aurizio Aug 23 '15 at 19:44
  • I thought about something similar along those lines, but wasn't able to spot the exact details, for some reason. Thanks! – Lucian Aug 23 '15 at 20:01
  • @Lucian: you're welcome, as always. – Jack D'Aurizio Aug 23 '15 at 20:04