Use Residues to find the inverse Laplace transform $F(s)=\frac{2s^3}{(s^2-4)}$

Question

Use Residues to find the inverse Laplace transform $F(s)=\frac{2s^3}{(s^2-4)}$.

The answer from the text book is $f(t)=\cosh^2(t)+\cos^2(t)$.

But my result is $2\cos^2(t)\cdot \cosh^2(t)$. Which is correct?

Answer 1

The Laplace of $\cos^2(t)$ can be derived from the famous half angle formula. What you get is $$\mathcal{L}\{\cos^2(t)\}(s) = \frac{1}{2s} + \frac{s}{2(s^2+4)}$$ On the other hand, Laplace of $\cosh^2(t)$ is found by employing its definition in terms of exponentials. In this case, you get $$\mathcal{L}\{\cosh^2(t)\}(s) = \frac{1}{4} \frac{1}{s-2} + \frac{1}{2s} + \frac{1}{4}\frac{1}{s+2}$$ So the Laplace of the book's answer is $$F_{\mathrm{book}}(s) = \frac{1}{s} + \frac{s}{2(s^2+4)} + \frac{1}{4} [\frac{1}{s-2} + \frac{1}{s+2}]$$ By Mathematica, put this all together to see that the function with the inverse laplace of the book's answer is $$F_{\mathrm{book}}(s) = \frac{2s^4-16}{s(s^4-16)}$$ This is simply not equal to the Laplace transform you stated. By computer algebra, the answer you gave does not equal the desired Laplace transform either.

Neither seems correct. So let's work through it from the beginning.

Note (by long division and then a little partial fractions) $$\frac{2s^3}{s^2-4} = 2s + \frac{4}{s-2} + \frac{4}{s+2}$$ Integrating over a Bromwich contour with real part greater than 2, we need consider the inegral $$\mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{2\pi i } \int_{B} [2s + \frac{4}{s-2} + \frac{4}{s+2}] e^{st} ds$$ For a moment ignore the first term in the integrand. You are then left with a simple pole at both $s=2$ and $s=-2$. Hence the inverse transform gives $$f(t) \approx 4e^{-2t} + 4e^{2t} = 8(\frac{e^{-2t} + e^{2t}}{2}) = 8\cosh(2t)$$ The integral from the first term is $$2 \frac{1}{2\pi i }\int_{B} s e^{st} ds$$ You can recognize this as the derivative of the dirac delta function in its Fourier-type expression. Hence a full answer is $$f(t) = 2 \delta'(t) + 8\cosh(2t)$$ But if you assume $t \geq 0$, that first term in the integrand is holomorphic over the contour and hence disappears giving the answer with only the hyperbolic function.

Hope this helps.