<s>[INST] Laplace transform of $x^{3/2}$ Solve the Laplace transform of $x^{3/2}$
$L[x^{3/2}]=L[x (x^{1/2})]=-\frac{d}{dp}L[x^{1/2}]=-\frac{d}{dp}L[x(x^{-1/2})]=-\frac{d}{dp}(-\frac{d}{dp})\sqrt{\frac{\pi}{p}}=\frac{1}{2p}\frac{1}{2p}\sqrt{\frac{\pi}{p}}=\frac{1}{4p^2}\sqrt{\frac{\pi}{p}}$ 
The correct answer is $\frac{3}{4p^2}\sqrt{\frac{\pi}{p}}$, could someone tell me where I missed the $3$ please?
 [/INST] Here's the correct way to do the integral.
\begin{align*}
\int_0 ^\infty e^{-st} t^{3/2} \ dt &= \int_0 ^\infty e^{-u} \left(\frac{u}{s} \right)^{3/2} \ \frac{du}{s} \\
&= \frac{1}{s^{5/2}} \int_0 ^\infty e^{-u} u^{\frac{5}{2}-1} \ du \\
&= \frac{1}{s^{5/2}} \Gamma \left( \frac{5}{2} \right) \\
&= \frac{1}{s^{5/2}} \cdot \frac{3\sqrt{\pi}}{4}
\end{align*}
The change of variables $u = st$ was used, and tables were used to evaluate the particular value of the gamma function.
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