Small trouble in finding Laplace Transform of $\sqrt{t}$ without using Gamma function

Question

We know that :

$$\mathcal{L} \left \{ t^{\frac{1}{2}} \right \}=\frac{\Gamma (\frac{1}{2}+1)}{S^{\frac{1}{2}+1}}=\frac{\sqrt{\pi }/2}{S^{\frac{3}{2}}}$$

I'm trying to prove that without using the Gamma function .

$$\mathcal{L} \left \{ \sqrt{t} \right \} = \mathcal{L} \left \{ t\frac{1}{\sqrt{t}} \right \} = -\frac{d}{ds}(\mathcal{L} \left \{ \frac{1}{\sqrt{t}} \right \})\Rightarrow (1)$$

Now I find $\mathcal{L} \left \{ \frac{1}{\sqrt{t}} \right \}$ using the property :

$$\mathcal{L} \left \{ {f}'(t) \right \} = s \mathcal{L} \left \{ f(t) \right \}-f(0)$$

by putting $$f(t)=\sqrt{t} \Rightarrow {f}'(t)=\frac{1/2}{\sqrt{t}}$$

$$\mathcal{L} \left \{ \frac{1/2}{\sqrt{t}} \right \} = s \mathcal{L} \left \{ \sqrt{t} \right \}- \sqrt{0} \Rightarrow 2\mathcal{L} \left \{ \frac{1/2}{\sqrt{t}} \right \} = 2s \mathcal{L} \left \{ \sqrt{t} \right \}$$

$$\mathcal{L} \left \{ \frac{1}{\sqrt{t}} \right \} = 2s\mathcal{L} \left \{ \sqrt{t} \right \} \Rightarrow (2)$$ , substituting from (2) in (1) :

$$\mathcal{L} \left \{ \sqrt{t} \right \} = -\frac{d}{ds}(2S\mathcal{L} \left \{ \sqrt{t} \right \} )$$

Let : $$\mathcal{L} \left \{ \sqrt{t} \right \}= h(s)$$ , then :

$$h = -\frac{d}{ds}(2Sh ) \Rightarrow h = -2h -2S {h}'$$

$$\Rightarrow 2S{h}'+3h=0 \Rightarrow {h}'+\frac{3}{2S}h=0$$

The integrating factor is :

$$e^{\int \frac{3}{2S}ds} = e^{\frac{3}{2}ln(S)}=S^{\frac{3}{2}}$$

$${(S^{\frac{3}{2}}h)}'=0 \Rightarrow S^{\frac{3}{2}}h = C \Rightarrow h(s)=\frac{C}{S^{\frac{3}{2}}}$$

Finally : back substitute

$$\mathcal{L} \left \{ \sqrt{t} \right \} = h(s) \Rightarrow \mathcal{L} \left \{ \sqrt{t} \right \} = h(s) \Rightarrow \mathcal{L} \left \{ \sqrt{t} \right \} = \frac{C}{S^{\frac{3}{2}}}$$

Here is the trouble how can I prove that $C = \sqrt{\pi }/2$ , or what is the wrong in the steps above ?

Answer 1

Note that we have

$$\begin{align} \int_0^\infty \sqrt t e^{-st}\,dt&\overbrace{=}^{t\mapsto t^2}2\int_0^\infty t^2e^{-st^2}\,dt\\\\ &=-2\frac{d}{ds}\int_0^\infty e^{-st^2}\,dt\\\\ &=-2\frac{d}{ds}\left(\frac1{\sqrt s}\frac{\sqrt\pi}{2}\right)\\\\ &=\frac{\sqrt\pi}{2s^{3/2}} \end{align}$$