# I have a Laplace Transform problem which I'm having trouble

I have a Laplace Transform problem which I'm having trouble with:
$$\displaystyle{L}{\left\lbrace{t}{\sin{{h}}}{\left({4}{t}\right)}\right\rbrace}=?$$

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MoxboasteBots5h
The definition of the Laplace transform is $$\displaystyle{L}{\left({f}\right)}{\left({s}\right)}\:={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}{f{{\left({t}\right)}}}{\left.{d}{t}\right.}{a}{s}{k}{\quad\text{and}\quad}{f}{\quad\text{or}\quad}{\text{sinh}{{t}}}{h}{e}{f}{o}{l}{l}{o}{w}\in{g}{h}{o}{l}{d}{s}:{P}{S}{K}{\sin{{h}}}{x}={\frac{{{1}}}{{{2}}}}{\left({e}^{{x}}-{e}^{{-{x}}}\right)}$$. Now you put these two things together and compute
$$\displaystyle{\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{\infty}}}{x}{e}^{{{4}{x}}}{e}^{{-{x}{s}}}{\left.{d}{x}\right.}-{\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{\infty}}}{x}{e}^{{-{4}{x}}}{e}^{{-{x}{s}}}{\left.{d}{x}\right.}$$
Hope this helps
###### Not exactly what you’re looking for?
levurdondishav4
You got $$\displaystyle{t}{\sin{{h}}}{4}{t}={t}{\frac{{{e}^{{{4}{t}}}-{e}^{{-{4}{t}}}}}{{{2}}}}$$ ,so:
$$\displaystyle{L}{\left\lbrace{t}{\sin{{h}}}{4}{t}\right\rbrace}={\frac{{{1}}}{{{2}}}}{\left({L}{\left\lbrace{t}{e}^{{{4}{t}}}\right\rbrace}+{L}{\left\lbrace{t}{e}^{{-{4}{t}}}\right\rbrace}\right)}$$
We know that $$\displaystyle{L}{\left\lbrace{e}^{{{a}{t}}}{f{{\left({t}\right)}}}\right\rbrace}={F}{\left({s}-{a}\right)}$$. Furthermore $$\displaystyle{L}{\left\lbrace{t}\right\rbrace}={\frac{{{1}}}{{{s}^{{2}}}}}$$.So in our case $$\displaystyle{L}{\left\lbrace{t}{e}^{{{4}{t}}}\right\rbrace}={F}{\left({s}-{4}\right)}={\frac{{{1}}}{{{\left({s}-{4}\right)}^{{2}}}}}$$. Finally:
$$\displaystyle{L}{\left\lbrace{t}{\sin{{h}}}{4}{t}\right\rbrace}={\frac{{{0.5}}}{{{\left({s}-{4}\right)}^{{2}}}}}+{\frac{{{0.5}}}{{{\left({s}+{4}\right)}^{{2}}}}}$$