complex-analysislaplace transformresidue-calculus
Doing some computations on a unknown but desirable function $f:[0,\infty) \to \Bbb R $ has lead me to that its Laplace transform can be known:
$$\hat{f} (s) = \frac{1}{s(\cosh{(\sqrt{2s})}-1)}, \quad s>0$$
I read that getting back the original function $f$ can be achieved using Mellin's inverse formula in combination with the residue theorem. Unfortunately, due to lack of experience with those theorems or complex integrals/analysis I am stuck at this point.
Any help is appreciated.
The way to attack a problem like this is via Cauchy's theorem on a properly distorted Bromwich contour. Here, we want our contour to avoid the branch point at $z=0$. This, we consider
$$\oint_C dz \, z^{-a-1} e^{-z^a} e^{z t} $$
where $a \in (0,1)$ and $C$ is the following contour:
We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.
$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.
$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.
$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.
$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.
$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.
$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.
When $t \gt 0$, the integral over the contours $C_2$ and $C_6$ vanish in the limit as $R \to \infty$.
The contour integral is thus equal to, in this limit,
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} + e^{-i \pi a} \int_{\infty}^{\epsilon} dx \, x^{-a-1} e^{-e^{i \pi a} x^a} e^{-x t} \\ + i \epsilon^{-a} \int_{\pi}^{-\pi} d\phi \, e^{-i a \phi} e^{-\epsilon^a e^{i a \phi}} e^{\epsilon t e^{i \phi}} + e^{i a \pi} \int_{\epsilon}^{\infty} dx \, x^{-a-1} e^{-e^{-i \pi a} x^a} e^{-x t}$$
Note that there is an apparent singularity at $\epsilon = 0$; however, the divergences cancel in the limit as $\epsilon \to 0$.
In this limit, the third integral has the following leading behavior:
$$-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi$$
Rescaling and combining the second and fourth integrals, we get for the contour integral:
$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} -i 2 t^a \operatorname{Im}{\left [e^{-i \pi a}\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} e^{-e^{i \pi a} t^{-a} u^a} \right]}-i \frac{2}{a} \epsilon^{-a} \sin{\pi a} +i 2 \pi $$
We may Taylor expand the second exponential in the integrand because it is subdominant to the first exponential (at least for the first $n$ terms, where $n$ is that largest integer such that $\lfloor n a \rfloor = 0$). We need only expand to the first two terms to treat the limit as $\epsilon \to 0$. Note that
$$\int_{\epsilon t}^{\infty} \frac{du}{u^{1+a}} e^{-u} = \frac{t^{-a}}{a} \epsilon^{-a} + \Gamma(-a) + O \left ( \epsilon^a \right ) $$
The second term produces
$$-e^{i \pi a} t^{-a}\int_{\epsilon t}^{\infty} \frac{du}{u} e^{-u} $$
Because the exponentials outside the integral cancel, the imaginary part of the term is zero. Thus, we now take the limit as $\epsilon \to 0$. Because the contour integral is zero by Cauchy's theorem, we get for the ILT,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-a-1} e^{-s^a} e^{s t} = \frac{t^a}{\pi} \operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} - 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} + \frac{t^a}{\Gamma(1+a)} - 1$$
ADDENDUM
Even though the above result is suitable for numerical calculation, we can illustrate the above result with an analytical example. Consider the case $a=1/2$. Subbing $u=x^2$ and taking the imaginary part of the integral, we end up with
$$\operatorname{Im}{\left [e^{-i \pi a}\int_{0}^{\infty} \frac{du}{u^{1+a}} e^{-u} \left ( e^{-e^{i \pi a} t^{-a} u^a} - 1 + e^{i \pi a} t^{-a} u^a \right ) \right]} = 2 \int_{-\infty}^{\infty} dx \, e^{-x^2} \frac{\sin^2{\beta x}}{x^2}$$
where $\beta = 1/(2 \sqrt{t})$.
The latter integral may be evaluated using Parseval's theorem, because the individual factors of the integrand are inverse Fourier transforms of simple functions. For example,
$$\int_{-\infty}^{\infty} dx \, e^{-x^2} e^{i k x} = \sqrt{\pi} e^{-k^2/4} $$ $$\int_{-\infty}^{\infty} dx \, \frac{\sin^2{\beta x}}{x^2} e^{i k x} =\begin{cases} \pi \beta \left ( 1-\frac{|k|}{2 \beta} \right ) & |k| \lt 2 \beta \\ 0 & |k| \gt 2 \beta \end{cases}$$
The integral is then equal to
$$2 \frac1{2 \pi} \sqrt{\pi} \pi \beta \int_{-2 \beta}^{2 \beta} dk \, \left ( 1-\frac{|k|}{2 \beta} \right ) e^{-k^2/4} = \sqrt{\pi} \beta \int_0^{2 \beta} dk \, \left ( 1-\frac{k}{2 \beta} \right ) e^{-k^2/4}$$
The evaluation is fairly straightforward using the definition of the error function. The result is, for the integral,
$$2 \pi \beta \operatorname{erf}{\beta} - 2 \sqrt{\pi} \left (1-e^{-\beta^2}\right ) $$
Now plugging this back into the main result above and using $\beta = 1/(2 \sqrt{t})$, we get that
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{-s^{1/2}} e^{s t} &= \operatorname{erf}{\left ( \frac1{2 \sqrt{t}} \right )} + \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - 1 \\ &= \frac{2}{\sqrt{\pi}} \sqrt{t} e^{-\frac1{4 t}} - \operatorname{erfc}{\left ( \frac1{2 \sqrt{t}} \right )}\end{align}$$
Best Answer
$$I:=\mathcal L^{-1}\left\{ \frac{1}{s(\cosh{(\sqrt{2s})}-1)} \right\}=\frac1{2\pi i}\int^{\gamma+i\infty}_{\gamma-i\infty}\underbrace{\frac{e^{st}}{s(\cosh{(\sqrt{2s})}-1)}}_{f(s)}ds$$ with $\gamma,t>0$.
Consider the integral $$\oint_C f(s)ds$$ where $C$ is the original integral path $\gamma-i\infty\to\gamma+i\infty$ plus a semicircle on the left.
Let's denote $a_n=-2n^2\pi^2$.
By estimation lemma, the arc integral vanishes. Thus, by residue theorem, $$I=\sum_{n=0}^\infty \operatorname{Res} (f(s),a_n)$$ (note that $f(s)$ is a meromorphic function on $\mathbb C$.)
Due to the Laurent/Taylor expansions at $s=a_n,n\ne 0$: $$\frac1{\cosh\sqrt{2s}-1}=\frac{4a_n}{(s-a_n)^{2}}+\frac2{s-a_n}+\cdots$$ $$\frac{e^{st}}{s}=\frac{e^{a_n t}}{a_n}+\frac{e^{a_n t}}{a_n^2}(a_nt-1)\cdot(s-a_n)+\cdots$$ we have $$\operatorname*{Res}_{s=a_n}f(s)=\left(4t-\frac{2}{a_n}\right)e^{a_n t}\qquad{n\ne 0}$$
Moreover, since $$\frac1{s(\cosh\sqrt{2s}-1)}=s^{-2}-\frac16 s^{-1}+\cdots$$ $$e^{st}=1+ts+\cdots$$ , we have $$\operatorname*{Res}_{s=0}f(s)=-\frac16 +t$$
Hence, $$I(t)=-\frac16+t+\sum^\infty_{n=1}\left(4t+\frac1{n^2\pi^2}\right)e^{-2\pi^2 n^2t}$$
A special value is $I(0)=0$.
On the other hand, for $t<0$, we can choose $C$ to be the semicircle on the right and use residue theorem to show $I(t)\equiv 0$.
Thoughts: By merely claiming $I(t)$ to be continuous at $0$, the above result leads to another proof of $\sum_1 \frac1{n^2}=\frac{\pi^2}6$.