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$$ A:=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1} $$

$$=\lim_{\beta\to\infty}\int_{0}^{\beta}\left(x\cdot\sin^{}\left(x\right)\right)\cdot\exp\left(-sx\right)\,dx$$

$$ = \lim_{\beta\to\infty}\int_{0}^{\beta}\left(x\right)\cdot \underbrace{\left( \sin^{}\left(x\right) \exp\left(-sx\right) \right)}_{\text{This part is to be integrated} } \,dx $$

$$ = \lim_{ \beta \to \infty} \left\{ \left[ x \cdot \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \right]_{0}^{\beta} - \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx \right\} $$

$$ \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx = - \frac{ s \cdot \exp\left(-sx\right) }{ \left( s^2+1 \right) } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right) \right) +\text{const} \tag{2} $$

Should I have written the derivation of the above equation?

$$ A= \lim_{ \beta \to \infty} \left\{ \left[ -\frac{ x \cdot s \cdot \exp\left(-sx\right) }{ \left( s^2+1 \right) } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right) \right) \right]_{0}^{\beta} - \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx \right\} $$

$$ = \lim_{ \beta \to \infty} \left\{ \underbrace{\left[ -\frac{ x \cdot s }{ \left( s^2+1 \right) e^{sx} } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right) \right) \right]_{0}^{\beta}}_{\text{About}~\beta~ \text{,it converges to }~0 } - \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx \right\} $$

$$ = -\lim_{ \beta \to \infty} \underbrace{\int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx}_\text{What can be done at here?} $$

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  • $\begingroup$ Wait , I think I can solve it in my own just greedily proceeding calculations. $\endgroup$ Nov 1, 2021 at 4:52
  • $\begingroup$ Solved. I am now writing a way to the answer. $\endgroup$ Nov 1, 2021 at 5:06

1 Answer 1

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Just proceed calculations using equations of laplace transformations of $~ \sin^{}\left(x_{}\right) ~~,~~ \cos^{}\left(x\right) ~$

$$ A=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1} $$

$$ = -\lim_{ \beta \to \infty} \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right) \,dx \right) \,dx $$

$$ = -\lim_{ \beta \to \infty} \int_{0 }^{\beta } \left(- \frac{ s }{ s^2+1 } \right) \frac{ 1 }{ e^{sx} } \left( \sin^{}\left(x\right) + \frac{1}{ s } \cos^{}\left(x\right) \right) \,dx $$

$$ = \frac{ s }{ s^2+1 } \lim_{ \beta \to \infty} \int_{ 0}^{ \beta} \frac{1}{ e^{sx} } \left( \sin^{}\left(x\right) + \frac{1}{ s } \cos^{}\left(x\right) \right) \,dx $$

$$ = \frac{ s }{ s^2+1 } \lim_{ \beta \to \infty} \int_{ 0}^{\beta } \left( \sin^{}\left(x\right) \exp\left(-sx\right) + \frac{1}{ s } \cos^{}\left(x\right) \exp\left(-sx\right) \right) \,dx $$

$$ = \frac{ s }{ s^2+1 } \left( \mathcal{L}\left[\sin^{}\left(x\right) \right]\left(s\right) + \frac{1}{ s } \mathcal{L}\left[\cos^{}\left(x\right) \right]\left(s\right)\right) $$

$$ = \frac{ s }{ s^2+ 1 } \left( \frac{ 1 }{ s^2+1 } + \frac{1}{ s } \frac{ s }{ s^2+1 } \right) $$

$$ = \frac{ s }{ s^2+ 1 } \left( \frac{ 2 }{ s^{2}+1 } \right) $$

$$ = \frac{ 2s }{ \left( s^{2}+1 \right) ^2 } $$

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  • $\begingroup$ I think I finally derived the correct answer. $\endgroup$ Nov 1, 2021 at 5:22
  • $\begingroup$ It is definitely correct. $\endgroup$ Nov 1, 2021 at 14:46

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