How to evaluate inverse Laplace transform of $e^{- \sqrt{s}} $ using series?

Question

I tried to find an inverse Laplace transform by series as follows $$ f(t)=L^{-1}_s\left(e^{-\sqrt{s}}\right)(t)=L^{-1}_s\left(\sum_{k=0}^{\infty}\frac{(-1)^k}{k!} s^{\frac{k}{2}}\right)(t)$$ and by separate even orders with odd orders and get $$ f(t)=L^{-1}_s\left(\sum_{k=0}^{\infty}\frac{(-1)^{2k}}{(2k)!} s^{k}+\sum_{k=0}^{\infty}\frac{(-1)^{2k+1}}{(2k+1)!} s^{k+\frac{1}{2}}\right)(t)$$ then $$ f(t)=L^{-1}_s\left(\sum_{k=0}^{\infty}\frac{s^k}{(2k)!}\right)-\sum_{k=0}^{\infty}\frac{1}{(2k+1)!} \frac{1}{\Gamma\left(-k-\frac{1}{2}\right)t^{k+\frac{3}{2}}} $$ now its easy to evaluate the series $$ f(t)=L^{-1}_s\left(\cosh\left(\sqrt{s}\right)\right)+\frac{t^{-\frac{3}{2}}}{2\sqrt{\pi}}e^{-\frac{1}{4t}} $$ now I know the result is only $$\frac{t^{-\frac{3}{2}}}{2\sqrt{\pi}}e^{-\frac{1}{4t}} $$ So is the Laplace transform of $\cosh\left(\sqrt{s}\right)$ is zero ? and what if we use this way $$ L^{-1}_s\left(\cosh\left(\sqrt{s}\right)\right)=L^{-1}_s\left(\sum_{k=0}^{\infty}\frac{s^k}{(2k)!}\right)=\sum_{k=0}^{\infty}\frac{1}{(2k)!\Gamma(-k)t^{k+1}} $$ now gamma for negative integers is infinity So we can say (as wolfram say) $$ \frac{1}{\Gamma(-k)}=0$$ finally $$ f(t)=L^{-1}_s\left(e^{-\sqrt{s}}\right)(t)=\frac{t^{-\frac{3}{2}}}{2\sqrt{\pi}}e^{-\frac{1}{4t}} $$ So is that using of series to get the result of inverse Laplace transform wrong ? or its true just here and how ?

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Answer 1

Let me first look at a simpler example, instead of the square root consider the inverse Laplace transform of $e^{-s}$. If you write the series expansion and invert term by term you obtain $$L^{-1}_s\left(e^{-{s}}\right)(t)=L^{-1}_s\left(\sum_{k=0}^{\infty}\frac{(-1)^k}{k!} s^k\right)(t)=\sum_{k=0}^\infty \frac{(-1)^k t^{-k-1}}{k! \Gamma (-k)}=0.$$ The correct answer is $$L^{-1}_s\left(e^{-{s}}\right)(t)=\delta(t-1),$$ so you miss the delta function.

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A similar contribution is missed in the square root case from the OP. The inverse Laplace transform of $\cosh\sqrt{s}$ is not zero but a series of derivatives of delta functions, $$L^{-1}_s\left(\cosh\left(\sqrt{s}\right)\right)=L^{-1}_s\left(\sum_{k=0}^{\infty}\frac{s^k}{(2k)!}\right)=\sum_{k=0}^\infty\frac{1}{2k!}\frac{d^k}{dt^k}\delta(t).$$