Evaluating improper integrals of odd functions with hyperbolic and circular elements

Question

The integral $$ \int_{0}^{\infty} {\sin\left(\omega t\right) \over \cosh^{2}\left(t/\sqrt{2\,}\,\right)}\,{\rm d}t $$ with $ \omega >0 $ is an odd function in variable t.

This precludes any contour type integration method for the same. The function for some chosen $\omega$ in the domain gives a nice curve that simply oscillates in a very small band about the t axis as it goes to infinity.

This implies that the integral exists (Riemann Integral, of course), but neither Maple or Mathematica is able to evaluate it.

I have tried a series type expression for the function and attempted to evaluate the integral as the sum of integrals of individual terms of the infinite series which also does not give any values for $\omega >3.9.$

Are there any known methods or possibilities to evaluate this type of integral- possibly by complex ananlysis?

Answer 1

Using a CAS, I have been able to first compute the antiderivative of the integrand (I suppose that Mathematica could give it); it is quite ugly.

Starting from there, I computed the integral as $$\frac{1}{2} \omega \left(-\psi ^{(0)}\left(\frac{1}{2}-\frac{i \omega }{2 \sqrt{2}}\right)+\psi ^{(0)}\left(1-\frac{i \omega }{2 \sqrt{2}}\right)-\psi ^{(0)}\left(\frac{i \omega }{2 \sqrt{2}}+\frac{1}{2}\right)+\psi ^{(0)}\left(\frac{i \omega }{2 \sqrt{2}}+1\right)\right)$$ which is real valued function of $\omega$ I have not been able to simplify further.

Plotting the result as a function of $\omega$ does not show any problem; just a nice curve starting at $0$, going through a maximimum value of $0.807934$ for $\omega=1.06508$ and decreasing asymptotically to zero.

For small values of $\omega$, the integral varies as
$$\omega \log (4)-\frac{3 \omega ^3 \zeta (3)}{4}+\frac{15 \omega ^5 \zeta (5)}{32}+O\left(\omega ^6\right)$$

For large values of $\omega$, the integral varies as
$$\frac{1}{\omega }+\left(\frac{1}{\omega }\right)^3+\frac{4}{\omega ^5}+O\left(\left(\frac{1}{\omega }\right)^6\right)$$ This is not much but I hope and wish it could help.

Answer 2

Just to show that evaluation is possible with complex analysis, I will outline the following solution. I warn, though, that I merely get an alternative ugly expression for this integral. Claude's expression is about as good as we likely will get for this one.

Consider the contour integral

$$\oint_C dz \frac{e^{i \omega z}}{\sinh^2{a z}}$$

where $C$ is the rectangle having vertices $\pm i \pi/(2 a)$ and $R \pm i \pi/(2 a)$, and $a=1/\sqrt{2}$. Note that we will need to deform $C$ at the origin by a semicircle of radius $\epsilon$ to avoid a pole there. Then by respective parametrization of each segment of $C$, the contour integral is

$$\int_0^R dx \frac{e^{i \omega x} e^{\pi \omega/(2 a)}}{-\cosh^2{a x}} + i \int_{-\pi/(2 a)}^{\pi/(2 a)} dy \frac{e^{i \omega R} e^{-\omega y}}{\sinh^2{(R+i y)}}\\ + \int_R^0 dx \frac{e^{i \omega x} e^{-\pi \omega/(2 a)}}{-\cosh^2{a x}} + i \int_{\pi/(2 a)}^{\epsilon} dy \frac{e^{-\omega y}}{-\sin^2{a y}} \\+ i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{e^{i \omega \epsilon e^{i \phi}}}{\sinh^2{(a \epsilon e^{i \phi})}}+ i \int_{-\epsilon}^{-\pi/(2 a)} dy \frac{e^{-\omega y}}{-\sin^2{a y}}$$

As $R \to \infty$, the second integral vanishes. As $\epsilon \to 0$, the fifth integral has a limit

$$\frac{i \epsilon}{a^2} \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{1+i \epsilon \omega e^{i \phi}}{\epsilon^2 e^{i 2 \phi}} \sim -i \frac{2}{a^2} \frac{1}{\epsilon} +\frac{\pi \omega}{a^2}$$

In this limit, the fourth and sixth integrals, and the $1/\epsilon$ term, combine to form a Cauchy principal value. By Cauchy's theorem, the contour integral is zero because there are no poles of the integrand inside $C$. Therefore,

$$2 \sinh{\frac{\pi \omega}{2 a}} \int_0^{\infty} dx \frac{e^{i \omega x}}{\cosh^2{a x}} = \frac{\pi \omega}{a^2} + i \, PV \int_{-\pi/(2 a)}^{\pi/(2 a)} dy \frac{e^{-\omega y}}{\sin^2{a y}}$$

Again, note that by $PV$, I mean

$$\lim_{\epsilon \to 0} \left [\int_{-\pi/(2 a)}^{-\epsilon} dy \frac{e^{-\omega y}}{\sin^2{a y}} + \int_{\epsilon}^{\pi/(2 a)} dy \frac{e^{-\omega y}}{\sin^2{a y}} - \frac{2}{a^2} \frac{1}{\epsilon}\right ]$$

Taking imaginary parts, I get that

$$2 \sinh{\frac{\pi \omega}{2 a}} \int_0^{\infty} dx \frac{\sin{\omega x}}{\cosh^2{a x}} = \, PV \int_{-\pi/(2 a)}^{\pi/(2 a)} dy \frac{e^{-\omega y}}{\sin^2{a y}}$$

which really doesn't help matters much. The integrand has an antiderivative involving hypergeometric functions; if you went this route, you would have to take the limit defined above and that seems a lot to ask. Ugly all around. Then again, I hope this at least illustrates how complex analysis may be used to derive an alternative expression for the integral.

Note that, when we take real parts, we get a lovely integral for free:

$$\int_0^{\infty} dx \frac{\cos{\omega x}}{\cosh^2{a x}} = \frac{\pi \omega}{2 a^2} \operatorname*{csch}{\frac{\pi \omega}{2 a}}$$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's $\ds{\mu \equiv {1 \over \root{2}}}$: \begin{align} &\color{#00f}{\large\int_{0}^{\infty} {\sin\left(\omega t\right) \over \cosh^{2}\left(\mu t\right)}\,{\rm d}t} =4\Im\int_{0}^{\infty} {\expo{\ic\omega t}\expo{-2\mu t} \over \pars{1 + \expo{-2\mu t}}^{2}}\,{\rm d}t \\[3mm]&=4\Im\int_{0}^{\infty}\expo{\ic\omega t}\expo{-2\mu t} \sum_{n = 0}^{\infty}\pars{-1}^{n}\pars{n + 1}\expo{-2n\mu t}\,{\rm d}t \\[3mm]&=4\sum_{n = 0}^{\infty}\pars{-1}^{n}\pars{n + 1} \Im\int_{0}^{\infty}\exp\pars{\bracks{\ic\omega - 2\pars{n + 1}\mu}t}\,\dd t \\[3mm]&=4\sum_{n = 0}^{\infty}\pars{-1}^{n}\pars{n + 1} \,\Im\pars{-1 \over \ic\omega - 2\pars{n + 1}\mu} ={2 \over \mu}\sum_{n = 0}^{\infty}\pars{-1}^{n} \,\Im\pars{n + 1 \over n + 1 - \ic\omega/2\mu} \\[3mm]&={\omega \over \mu^{2}}\Re\sum_{n = 0}^{\infty} {\pars{-1}^{n} \over n + 1 - \ic\omega/2\mu} ={\omega \over \mu^{2}}\Re\sum_{n = 0}^{\infty}\pars{% {{1 \over 2n + 1 - \ic\omega/2\mu} - {1 \over 2n + 2 - \ic\omega/2\mu}}}\ \\[3mm]&={\omega \over 4\mu^{2}}\Re\sum_{n = 0}^{\infty} {1 \over \pars{n + 1/2 - \ic\omega/4\mu}\pars{n + 1 - \ic\omega/4\mu}} \\[3mm]&=\color{#00f}{\large{\omega \over 2\mu^{2}}\bracks{% \Re\Psi\pars{1 -\ic\,{\omega \over 4\mu}} -\Re\Psi\pars{\half -\ic\,{\omega \over 4\mu}}}}\,,\qquad \mu = {1 \over \root{2}} = {\root{2} \over 2} \end{align}