Find the Laplace Transform of $\sin\sqrt{t}$

Question

To find the Laplace Transform of $\sin\sqrt{t}$, I use the general formula $F(s)=\int_0^\infty e^{-st}f(t)\,dt$ and I get that: $$\mathcal{L}[\sin\sqrt{t}]=\int_0^\infty e^{-st}\sin\sqrt{t}\,dt=\int_0^\infty e^{-st}\sin\frac{\sqrt{st}}{\sqrt{s}}\,dt$$ Now I make the substitution $st=u \implies dt=\frac{du}{s}$ $$\mathcal{L}[\sin\sqrt{t}]=\int_0^\infty \frac{e^{-u}}{s} \sin\left(\sqrt{\frac{u}{s}}\right)\,du=\frac{1}{s}\int_0^\infty e^{-u}\sin\left(\sqrt{\frac{u}{s}}\right)\,du=\frac{1}{s}\int_0^\infty e^{-u}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\frac{\sqrt u \cdot u^n}{\sqrt s \cdot s^n}\,du=\frac{1}{s\sqrt s}\int_0^\infty e^{-u}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\frac{\sqrt u \cdot u^n}{s^n}\,du$$

How to keep solving this integral?

Answer 1

Using Euler's formula, $\mathcal{L}\{\sin\sqrt{t}\}(s)=\operatorname{Im}\int_0^\infty e^{i\sqrt{t}-st}\, dt$. If $I(s,t)=\int e^{i\sqrt{t}-st}\, dt$, then using the substitution $u=\sqrt{t}$ yields $$I(s,t)=\frac{1}{s}\int (2su-i)e^{iu-su^2}\, du+\frac{i}{s}\int e^{iu-su^2}\, du=\frac{1}{s}J+\frac{i}{s}K.$$ For $J$, substitute $v=iu-su^2.$ Therefore $$J=-\int e^v\, dv=-e^{i\sqrt{t}-st}+C_1.$$ For $K$, complete the square: $$K=\int e^{-\left(u\sqrt{s}-\frac{i}{2\sqrt{s}}\right)^2-\frac{1}{4s}}\, du.$$ Then substitute $u=\frac{2w\sqrt{s}+i}{2s}$ to get $$\begin{align}K&=\frac{1}{\sqrt{s}}\int e^{-w^2-\frac{1}{4s}}\, dw\\&=\frac{\sqrt{\pi}e^{-\frac{1}{4s}}}{2\sqrt{s}}\operatorname{erf}w+C_2\\&=\frac{\sqrt{\pi}e^{-\frac{1}{4s}}}{2\sqrt{s}}\operatorname{erf}\frac{2s\sqrt{t}-i}{2\sqrt{s}}+C_2.\end{align}$$ So we get $$I(s,t)=\frac{i\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}}\operatorname{erf}\frac{2s\sqrt{t}-i}{2\sqrt{s}}-\frac{e^{i\sqrt{t}-st}}{s}+C.$$ Now, evaluate the limits: $$\int_0^\infty e^{i\sqrt{t}-st}\, dt=\lim_{t\to\infty}I(s,t)-\lim_{t\to 0}I(s,t)$$ which is just $$\frac{i\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}}+\frac{i\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}}\operatorname{erf}\frac{i}{2\sqrt{s}}+\frac{1}{s}.$$ Since $\operatorname{erf}\frac{i}{2\sqrt{s}}\in i\mathbb{R}$ (under the assumption that $s\gt 0$), taking the imaginary part of the above expression gives $$\bbox[5px,border:2px solid red]{\mathcal{L}\{\sin\sqrt{t}\}(s)=\frac{\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}},\, s\gt 0.}$$

Answer 2

To find : $L(\sin (\sqrt{t}))=\int_0^{\infty} e^{-s t} \sin (\sqrt{t}) d t$

solution : $L(\sin (\sqrt{t}))=\int_0^{\infty} e^{-s t} \sin (\sqrt{t}) d t=f(s)$ $$ \begin{aligned} & \because \sin (\sqrt{t})=(\sqrt{t})-\frac{(\sqrt{t})^3}{3!}+\frac{(\sqrt{t})^5}{5!}-\cdots\\ & \Rightarrow f(s)=\int_0^{\infty} e^{-s t} t^{3 / 2-1} d t-\frac{1}{3!} \int_0^{\infty} e^{-s t} \cdot t^{5 / 2-1} d t+\cdots \\ & \because \int_0^{\infty} e^{-\lambda t} t^{n-1} d t=\frac{\Gamma(n)}{\lambda^n}, \Rightarrow \int_0^{\infty} e^{-\lambda t} t^n d t=\frac{\Gamma(n+1)}{\lambda^{n+1}} \text {. } \\ & \Rightarrow f(s)=\frac{\Gamma(3 / 2)}{1! s^{3 / 2}}-\frac{\Gamma(5 / 2)}{3! s^{5 / 2}}+\frac{\Gamma(7 / 2)}{5! s^{7 / 2}}-\frac{\Gamma(9 / 2)}{7! s^{9 / 2}}+\cdots \\ & \Rightarrow f(s)=\frac{1}{2} \sqrt{\pi} \cdot \frac{1}{2 s \sqrt{s}}-\frac{1}{1 \cdot 2 \cdot 3} \cdot \frac{1 \cdot 3}{2 \cdot 2} \frac{\sqrt{\pi}}{s \cdot(s \sqrt{s})}+\frac{1 \cdot 3 \cdot 5 \sqrt{\pi}}{(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5)(2 \cdot 2 \cdot 2) s^2(s \sqrt{s})}+\cdots \\ & f(s)=\frac{\sqrt{\pi}}{2 s \sqrt{s}}-\frac{1}{1} \cdot \frac{\sqrt{\pi}}{2 s \sqrt{s}} \cdot \frac{1}{(4 s)}+\frac{\sqrt{\pi}}{2 s \sqrt{s}} \cdot \frac{1}{2(4 s)^2}-\frac{\sqrt{\pi}}{2 s \sqrt{s}} \cdot \frac{1}{6(4 s)^s}+\cdots \\ & \Rightarrow f(s)=\frac{\sqrt{\pi}}{2 s \sqrt{s}}\left\{1-\frac{1}{1!}(1 / 4 s)+\frac{1}{2!}(1 / 4 s)^2-\frac{1}{3!}(1 / 4 s)^3+\cdots\right\} \\ & f(s)=\frac{\sqrt{\pi}}{2 s \sqrt{s}} e^{-1 / 4 s}=\frac{\sqrt{\pi} e^{-1 / 4 s}}{2 s \sqrt{s}} \\ & \because \Gamma\left(\frac{2 n+1}{2}\right)=\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2 n-1)}{2^n} \cdot \sqrt{\pi} \\ & \because e^{-x}=1-x+\frac{x^2}{2!}-\cdots\\ & \end{aligned} $$

Therefore : $L(\sin (\sqrt{t}))=\int_0^{\infty} e^{-s t} \sin (\sqrt{t}) d t=\frac{\sqrt{\pi} e^{-1 / 4 s}}{2 s \sqrt{s}}$