Any good math calculators online to find inverse laplace transforms? Specifically $s\log\frac{s+4}{s-4}$

Question

I think I have found the final answer $\left(4 \left[\frac{2t \cosh(4t) - \sinh(4t)}{t^2}\right]\right)$ but need to verify it.

Answer 1

Let $$F(s) = s\ln\frac{s+4}{s-4}$$ The first property to use is $$\mathcal L[g'] = s\mathcal L[g] - g(0^-)$$ So we consider $\mathcal L[g] = \ln\frac{s+4}{s-4}$. Another property is $$\mathcal L[\frac{h(t)}{t}] = \int_s^\infty H(\lambda)d\lambda$$ Where $\mathcal L[h] = H(s)$. Note that $$\int_s^\infty (\frac{1}{\lambda - 4} - \frac{1}{\lambda + 4})d\lambda = \ln\frac{s+4}{s-4}$$ So $H(s) = \frac{1}{s - 4} - \frac{1}{s + 4}$ and that implies $$h(t) = e^{4t} - e^{-4t}$$ Therefore $g(t) = \frac{e^{4t} - e^{-4t}}{t}$. Then $$\lim_{t \to 0} g(t) = \lim_{t \to 0} \frac{e^{4t} - e^{-4t}}{t} = \lim_{t \to 0} \frac{4e^{4t} +4e^{-4t}}{1} = 8$$ Note that $$\mathcal L[\delta] = 1$$ Finally $$f(t) = g'(t) + 8\delta(t) = \frac{t(4e^{4t} + 4e^{-4t}) - (e^{4t} - e^{-4t})}{t^2} + 8\delta(t)$$