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What is the Laplace transform for function:

\begin{equation} f(x) = \sin(a \sqrt{x})? \end{equation}

I tried a lot of variations, but still, not sure how to solve it with the a-parameter. Is it maybe possible to solve without the Gamma function?

amWhy
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abc
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    What have you tried? This would be easier to do with cosine instead of sine. – Ninad Munshi Jan 20 '24 at 21:59
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jan 20 '24 at 22:04
  • I have to solve it with sinus. I tried using integral from the definition but it is really hard to get anything... – abc Jan 20 '24 at 22:05
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    You need to prove $\int_0^\infty\sin(a\sqrt{x})e^{-sx}\mathrm{d}x=\frac{\sqrt{\pi}}{2}as^{-3/2}e^{-a^2/(4s)}$ for $a,,s>0$. Which techniques do you know for computing Laplace transforms? Could you edit your question to show what happened when you tried them? – J.G. Jan 20 '24 at 22:22
  • Yes, I know the result, but I got stuck while doing integral by parts – abc Jan 20 '24 at 22:28
  • Do you want to use some property or only calculate the integral? – daniel Jan 20 '24 at 22:30
  • I just want to know how to do this transform (if it is possible- without Gamma function) – abc Jan 20 '24 at 22:31
  • you can use this posts https://math.stackexchange.com/questions/3530529/laplace-transformations-of-left-frac-cos-sqrt-t-sqrt-t-right-and-lef?noredirect=1&lq=1 ,https://math.stackexchange.com/questions/1434918/compute-laplace-transform-of-frac-cos-sqrt-t-sqrt-t?rq=1 – daniel Jan 20 '24 at 23:09
  • but how to do it with parameter a? – abc Jan 20 '24 at 23:13

1 Answers1

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Using the posts that I suggest we have $$( \sin ( a\sqrt t))'=\frac a 2\frac {\cos ( a\sqrt t)}{ \sqrt t}$$ And $$\mathscr {L} \{f'(t)\}(s)=s\,\mathscr {L} \{f(t)\}(s)-f(0)$$ Therefore: $$\frac{a}{2s} \mathscr {L} \left\{\frac {\cos ( a\sqrt t)}{ \sqrt t} \right\}(s)=\,\mathscr {L} \{\sin ( a\sqrt t)\}(s)$$ now \begin{align*} \mathscr {L} \left\{\frac {\cos ( a\sqrt t)}{ \sqrt t} \right\}(s) & = \int_{0}^{\infty} \frac {\cos ( a\sqrt t)}{ \sqrt t} e^{-st} dt \\ & x = a \sqrt{t} \\ & = \frac{2}{a} \int_{0}^{\infty} \cos(x) e^{-sx^2/a^2} dx \end{align*} now let $$ I(b) = \int_{0}^{\infty} \cos(bx) e^{-sx^2/a^2} dx $$ using Feynman trick \begin{align*} I(b)' & = -\int_{0}^{\infty} x \sin(bx) e^{-sx^2/a^2} dx \\ & \text{use i.b.p} \\ & = 0 - \frac{a^2}{2s} b \int_{0}^{\infty} \cos(bx) e^{-sx^2/a^2} dx \\ & = -b\frac{a^2}{2s} I(b) \\ I(b)' & = -b\frac{a^2}{2s} I(b) \\ \frac{I(b)'}{I(b)} & = -b\frac{a^2}{2s} \\ \vdots & = \vdots \\ I(b) & = \dfrac{\sqrt{{\pi}}\,a}{2\sqrt{s}} e^{-b^2 a^2 / 4s} \\ I(1) & = \dfrac{\sqrt{{\pi}}\,a}{2\sqrt{s}} e^{- a^2 / 4s} \\ \end{align*} we replace what we were looking for \begin{align*} \mathscr {L} \left\{\frac {\cos ( a\sqrt t)}{ \sqrt t} \right\}(s) & =\frac{2}{a} I(1) \\ & = \frac{2}{a} \dfrac{\sqrt{{\pi}}\,a}{2\sqrt{s}} e^{- a^2 / 4s} \\ & = \dfrac{\sqrt{{\pi}}\,}{\sqrt{s}} e^{- a^2 / 4s} \end{align*} Finally we have to \begin{align*} \frac{a}{2s} \mathscr {L} \left\{\frac {\cos ( a\sqrt t)}{ \sqrt t} \right\}(s) & = \frac{a}{2s} \dfrac{\sqrt{{\pi}}\,}{\sqrt{s}} e^{- a^2 / 4s} \\ & = \dfrac{a\sqrt{{\pi}}\,}{2s^{3/2}} e^{- a^2 / 4s} \end{align*}

Edit

First we use the following property $$ \mathscr {L} \{f'(t)\}(s)=s\,\mathscr {L} \{f(t)\}(s)-f(0)$$ Then, since the integral is difficult to calculate, we use the Feynman trick, you can see this video ( very similar) to understand how it works https://www.youtube.com/watch?v=YO38MCdj-GM&t=538s. Now $I(b)$ is $$ I(b) = \int_{0}^{\infty} cos(bx) e^{-sx^2/a^2} dx $$ and $$ I(1) = \int_{0}^{\infty} cos( x) e^{-sx^2/a^2} dx $$ How does the $\pi$ number appear? Let's see, we have the following differential equation $$ I(b)' = -b \frac{a^2}{2s}I(b) $$ can be solved intuitively (is a function of $b$) \begin{align*} I(b)' & = -b \frac{a^2}{2s} I(b)\\ \frac{I(b)'}{I(b)} & = -b \frac{a^2}{2s}\\ \int \frac{I(b)'}{I(b)} db & =\int -\frac{b a^2}{s} db \\ ln(I(b)) & = -\frac{b^2 a^2}{4s} + C_1 \\ I(b) & = e^{-\frac{b^2 a^2}{4s}} C_2 \end{align*} To find the value of $C_2$ we do the following \begin{align*} I(0) & = e^{-\frac{0^2 a^2}{4s}} C_2 \\ I(0) & = C_2 \\ \int_{0}^{\infty} cos(0 \cdot x) e^{-sx^2/a^2} dx & = C_2\\ \int_{0}^{\infty} e^{-sx^2/a^2} dx & = C_2 \\ \int_{0}^{\infty} e^{- \left( \sqrt{s} x/ a \right)^2} dx & = C_2 \\ \frac{a}{\sqrt{s}} \int_{0}^{\infty} e^{- u^2} du & = C_2 \\ \frac{a}{\sqrt{s}} \frac{\sqrt{\pi}}{2} & = C_2 \\ \end{align*} Finally we have $$ I(b) = e^{-\frac{b^2 a^2}{4s}} \frac{a}{\sqrt{s}} \frac{\sqrt{\pi}}{2} $$

daniel
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