An inverse Laplace transform and its asymptotics

Question

I am wondering if there is an analytic expression for the inverse Laplace transform $f(t):=\mathcal L^{-1}[F](t)$ for $F(s):=\frac1{s(\cosh{\sqrt{2s}}-1)}$. If there is no such analytic expression, what would be an asymptotics of $f(t)$ for $t\to\infty$?

Answer 1

The function has second order poles along the negative real axis at $s = -2\pi^2k^2$ for $k\in\Bbb{Z}$. By the Final Value Theorem we have that

$$\lim_{s\to0^+}s^2F(s) = 1 = \lim_{t\to\infty}f'(t)$$

and

$$\lim_{s\to0^+}s\left(F(s)-\frac{1}{s^2}\right) = \lim_{s\to0^+}\frac{1}{s}\left(\frac{1}{1+\frac{1}{6}s+\frac{1}{90}s^2+\cdots}-1\right) = -\frac{1}{6} = \lim_{t\to\infty}f(t)-t$$

The inverse Laplace transform of each of the other second order poles is

$$\mathcal{L}^{-1}\left\{F(s)+\frac{1}{6s}-\frac{1}{s^2}\right\} = \sum_{k=1}^\infty\left(4t+\frac{1}{\pi^2k^2}\right)e^{-2\pi^2k^2t}$$

by the Residue theorem

$$\frac{1}{2\pi i}\int_{-i\infty}^{+i\infty}e^{st}\left(F(s)+\frac{1}{6s}-\frac{1}{s^2}\right)ds = \sum_{k=1}^\infty\operatorname{Res}\left(e^{st}\left(F(s)+\frac{1}{6s}-\frac{1}{s^2}\right),s=-2\pi^2k^2\right)$$

$$ = \sum_{k=1}^\infty-\frac{e^{-2\pi^2k^2t}}{2\pi^2k^2}\left(t+\frac{1}{2\pi^2k^2}\right)\cdot\lim_{s\to-2\pi^2k^2}\frac{(s+2\pi^2k^2)^2}{\cosh\sqrt{2s}-1}$$

$$-\frac{e^{-2\pi^2k^2t}}{2\pi^2k^2}\cdot\lim_{s\to-2\pi^2k^2}\left(\frac{(s+2\pi^2k^2)^2}{\cosh\sqrt{2s}-1}\right)'$$

which can be evaluated by using

$$\lim_{s\to-2\pi^2k^2}\frac{(s+2\pi^2k^2)^2}{\cosh\sqrt{2s}-1} = -8\pi^2k^2$$

and

$$\lim_{s\to-2\pi^2k^2}\left(\frac{(s+2\pi^2k^2)^2}{\cosh\sqrt{2s}-1}\right)' = \frac{8\pi^2k^2}{4\pi^2k^2}=2$$

This gives us a final answer for the inverse Laplace transform of $F$ itself as

$$\mathcal{L}^{-1}\left\{\frac{1}{s(\cosh\sqrt{2s}-1)}\right\} = t - \frac{1}{6} + \sum_{k=1}^\infty\left(4t+\frac{1}{\pi^2k^2}\right)e^{-2\pi^2k^2t}$$

Answer 2

Note that \begin{align*} F(s) &= \frac{1}{{s(\cosh (\sqrt {2s} ) - 1)}} = \frac{1}{{2s}}{\mathop{\rm csch}\nolimits} ^2\! \left( {\sqrt {\frac{s}{2}} } \right) \\ &= \frac{1}{{s^2 }} + \frac{2}{s}\sum\limits_{n = 1}^\infty {\frac{1}{{s + 2\pi ^2 n^2 }}} - \frac{{8\pi ^2 }}{s}\sum\limits_{n = 1}^\infty {\frac{{n^2 }}{{(s + 2\pi ^2 n^2 )^2 }}} . \end{align*} (This is a re-written form of $(4.36.4)$.) Hence, $$ f'(t) = \mathscr{L}^{ - 1} (sF(s)) = 1 + 2\sum\limits_{n = 1}^\infty {{\rm e}^{ - 2\pi ^2 n^2 t} } - 8\pi ^2 t\sum\limits_{n = 1}^\infty {n^2 {\rm e}^{ - 2\pi ^2 n^2 t} } , $$ and therefore $$ f(t) = t - \frac{1}{6} + \frac{1}{{\pi ^2 }}\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}{\rm e}^{ - 2\pi ^2 n^2 t} } + 4t\sum\limits_{n = 1}^\infty {{\rm e}^{ - 2\pi ^2 n^2 t} } , $$ provided $\operatorname{Re}(t)>0$. The constant $-\frac{1}{6}$ follows from $$ \mathop {\lim }\limits_{s \to 0^ + } s\!\left( {F(s) - \frac{1}{{s^2 }}} \right) = \mathop {\lim }\limits_{t \to + \infty } (f(t) - t). $$ This is a consequence of the final value theorem.