The inverse Laplace transform of the function $${\text{F}}\left( {\text{s}} \right) = \frac{1}{{{\text{s}}\left( {{\text{s}} + 1} \right)}}$$   is given by

The Laplace transform F(s) of the exponential function. f(t) = e^at when t ≥ 0, where a is a constant and (s - a) > 0, is

Evaluate $$\int\limits_0^\infty {\frac{{\sin {\text{t}}}}{{\text{t}}}{\text{dt}}} $$

If the Laplace transform of $${{\text{e}}^{\omega {\text{t}}}}$$  is $$\frac{1}{{{\text{s}} - \omega }},$$  the Laplace transform of tcosh t is

The function f(t) satisfies the differential equation $$\frac{{{{\text{d}}^2}{\text{f}}}}{{{\text{d}}{{\text{t}}^2}}} + {\text{f}} = 0$$   and the auxiliary conditions, f(0) = 0, $$\frac{{{\text{df}}}}{{{\text{dt}}}}\left( 0 \right) = 4.$$  The Laplace transform of f(t) is given by