# Inverse Laplace Transform of $\frac{1}{\lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs})}$.

I am trying to find the inverse Laplace transform of some function of the form:

$$\mathrm{F}\left(s\right) = \frac{1}{\lambda\cosh\left(\,\sqrt{\, as\,}\, + \,\sqrt{\, bs\, }\right) + \left(1 - \lambda\right)\cosh\left(\,\sqrt{\, as\,}\, -\,\sqrt{\, bs\,}\right)},$$

where $$\lambda \in \left[0,1\right]$$. I've tried to use the residue theorem, however, finding the poles ( roots of the denominator ) seems quite complicated, which I have not figured out, yet.

I've also tried to calculate the integral: $$\int_{\sigma - \mathrm{i}\infty}^{\sigma + \mathrm{i}\infty} \mathrm{F}\left(s\right)\mathrm{e}^{st}\,\mathrm{d}s,$$ by using some substitution of the form $$u = \mathrm{e}^{\,\sqrt{\, s\,}\,}$$, which, however, yields another complicated form from which I can proceed further.

Any ideas, hints, partial solutions, are appreciated!

• What does $\sigma+i\infty$ mean? Jun 2, 2019 at 22:44
• Cumbersome !!!. Jun 3, 2019 at 0:24
• @Chase Ryan Taylor. For a quick reference, see this staff.city.ac.uk/~george1/laplace_residue.pdf. Jun 3, 2019 at 8:23

Concerning the zeros for

$$\lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs}) = 0$$

making $$s = x + i y$$ and taking the real and the imaginary parts

$$R=\lambda \cos \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \cosh \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)-(\lambda -1) \cos \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \cosh \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)=0\\ I = \lambda \sin \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \sinh \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)-(\lambda -1) \sin \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \sinh \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)=0$$

and analyzing the root locus for some values of $$a,b,\lambda$$ we have good information.

For instance with $$a = 3, b = 2, \lambda = 0.9$$ no zeroes.

with $$a = 3, b = 2, \lambda = 0.1$$ the zeroes are at the intersections.

with $$a = 3, b = 2, \lambda = 0.5$$ no zeroes.

So we can visualize the zeroes position depending on the parameters. This can help with the residues method. Follows the MATHEMATICA script to make the plots

re0 = -(-1 + lambda) Cos[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Cosh[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]] + lambda Cos[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Cosh[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]]
im0 = -(-1 + lambda) Sin[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Sinh[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]] + lambda Sin[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Sin[1/2 Arg[x + I y]]] Sinh[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
Cos[1/2 Arg[x + I y]]]

a = 3;
b = 2;
lambda = 0.1;
r = 50;
gr1 = ContourPlot[re0 == 0, {x, -r, r}, {y, -r, r}, ContourStyle -> Red, PlotPoints -> 25];
gr2 = ContourPlot[im0 == 0, {x, -r, r}, {y, -r, r}, ContourStyle -> Blue, PlotPoints -> 25];
Show[gr1, gr2]

• Well, that's nice, thanks! I also had some values for which there were no zeroes, so, after that, I shall revisit the residues method. Jun 3, 2019 at 8:22
• @ΒασίληςΜάρκος Introduced the MATHEMATICA script in case it can be useful. Jun 3, 2019 at 9:44
• Thanks a lot, you saved me much time! Jun 3, 2019 at 11:06