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I would like to see how to compute the inverse Laplace transform $${\mathscr L}^{-1}\left(\frac{\coth (s/(2a))}{2s}\right).$$ I am interested in the proof, not just the answer. I am reading the book "The Laplace Transform" by Joel Schiff, where the inverse Laplace transform $${\mathscr L}^{-1}\left(\frac{\coth(\sqrt{s})}{\sqrt{s}}\right)=1+2\sum_{n=1}^\infty e^{-n^2\pi^2 t}, t>0$$ is computed using a parabolic contour of integration, so I wonder if something similar would work. Thanks for any suggestions.

adriaanJ
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rpembroke
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1 Answers1

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There are several posts on this website that dealt with similar problems. Your function has poles at $s=0$ and at $s=2\mathrm{i}\,a\pi n$, with $n=-2,-1,0,1,2$, for which $\sinh(s/(2a))=0$. Calculate the residues like here, or here.

The answer can be easily found with this identity (see here and here) \begin{align} \coth (X)=\frac{1}{X}+2\sum_{n=1}^{\infty}\frac{X}{n^2\pi^2+X^2} \end{align} hence \begin{align} \frac{1}{2s}\coth \frac{s}{2a}=\frac{a}{s^2}+\sum_{n=1}^{\infty}\frac{2a}{(2a n\pi)^2+s^2}\,. \end{align} Applying inverse Laplace transformations on each term, and using $\mathcal{L}^{-1}\left\{\frac{1}{k^2+s^2}\right\}=\frac{1}{k}\sin(kt)$ gives \begin{align} \mathcal{L}^{-1}\left\{\frac{1}{2s}\coth \frac{s}{2a}\right\}=at+\sum_{n=1}^{\infty}\frac{\sin(2a n \pi t)}{\pi n} \end{align} The code and figure below confirm this result for the case $a=2$.

enter image description here

import numpy as np
import matplotlib.pyplot as plt
from mpmath import *

def f(t,a):
    f=0
    for n in range(1,100):
        f+=np.sin(2anp.pint)/n
    return a*t+f/np.pi


def F(s,a):
    return mp.coth(s/(2a))/(2s)


mp.dps = 30
def fnum(t,a):
    ft = lambda s: F(s,a)
    return invertlaplace(ft,t,method='dehoog')


t  = np.linspace(0,3,100)
t2 = np.linspace(0.1,3,30)


fig, ax = plt.subplots()
ax.plot(t,[f(i,2) for i in t],  'g',lw=2,label='analytical')
ax.plot(t2,[fnum(i,2) for i in t2],'g',lw=0,marker='o',label='numerical')
ax.set(xlabel=r'$t$', ylabel=r'$f(t)$') 
ax.legend()
adriaanJ
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