# Find the Laplace transform F of the following functions and give the domain of F. f(x)=cos(2x)

The Laplace transform of a continuous function over the interval $$\displaystyle{\left[{0},∞\right)}{\left[{0},∞\right)}$$ is defined by $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{x}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}.$$
This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of F is the set of all real numbers s such that the improper integral converges.
Find the Laplace transform F of the following functions and give the domain of F. $$\displaystyle{f{{\left({x}\right)}}}={\cos{{\left({2}{x}\right)}}}$$

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yunitsiL
Since the Laplace transform of a continious function over the interval $$\displaystyle{\left[{0},∞\right)}$$ is defined by
$$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{x}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{\infty}}{\cos{{\left({2}{x}\right)}}}{e}^{{{s}{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\lim_{{{t}\to\infty}}{\int_{{0}}^{{t}}}{\cos{{\left({2}{x}\right)}}}{e}^{{{s}{x}}}{\left.{d}{x}\right.}$$
Using integration by parts we get
$$\displaystyle\int{\cos{{\left({2}{x}\right)}}}{e}^{{{s}{x}}}{\left.{d}{x}\right.}=\frac{{1}}{{{s}^{{2}}+{4}}}{e}^{{-{s}{r}}}{\left({2}{\sin{{\left({2}{x}\right)}}}-{s}{\cos{{2}}}{x}\right)}$$
Hence
$$\displaystyle{F}{\left({s}\right)}=\lim_{{{t}\to\infty}}{\left(\frac{{1}}{{{s}^{{2}}+{4}}}{e}^{{-{s}{r}}}{\left({2}{\sin{{\left({2}{x}\right)}}}-{s}{\cos{{2}}}{x}\right)}\right)}$$
$$\displaystyle=\lim_{{{t}\to\infty}}{\left(\frac{{s}}{{{s}^{{2}}+{4}}}-\frac{{1}}{{{s}^{{2}}+{4}}}{e}^{{-{s}{t}}}{\left({2}{\sin{{\left({2}{t}\right)}}}-{t}{\cos{{\left({2}{t}\right)}}}\right)}\right)}$$
$$\displaystyle=\frac{{s}}{{{s}^{{2}}+{4}}}$$
Domain of F is $$\displaystyle{\left[{0},∞\right)}$$
$$\displaystyle{F}{\left({s}\right)}=\frac{{1}}{{s}}$$
Domain of F is $$\displaystyle{\left[{0},∞\right)}$$