Laplace Transforms

Question

Show that ${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\frac{(2n-1)!!}{2^n}\frac{1}{s^n} \sqrt{\frac{\pi}{s}}$

My Attempt:

${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\int_{0}^{\infty}\mathcal e^{-st}t^{n-\frac{1}{2}}dt$

Integration by parts gives:

${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\frac{(2n-1)}{2s}{\mathcal L} {\lbrace t^{n-\frac{3}{2}}\rbrace}$

${\mathcal L} {\lbrace t^{n-\frac{3}{2}}\rbrace}=\frac{(2n-3)}{2s}{\mathcal L} {\lbrace t^{n-\frac{5}{2}}\rbrace}$

${\mathcal L} {\lbrace t^{n-\frac{5}{2}}\rbrace}=\frac{(2n-5)}{2s}{\mathcal L} {\lbrace t^{n-\frac{7}{2}}\rbrace}$

I start to see the $\frac{(2n-1)!!}{2^n}\frac{1}{s^n}$, but not the $\sqrt{\frac{\pi}{s}}$

Thanks, in advance.

Answer 1

The Laplace Transform of $t^{n-1/2}$ is given by

$$\mathscr{L}\left(t^{n-1/2}\right)(s)=\int_0^\infty t^{n-1/2}\,e^{-st}\,dt$$

Now, enforce the substitution $st\to t$ reveals

$$\begin{align} \mathscr{L}\left(t^{n-1/2}\right)(s)&=\left(\frac{1}{s}\right)^{n+1/2}\int_0^\infty t^{n-1/2}\,e^{-t}\,dt\\\\ &=\left(\frac{1}{s}\right)^{n+1/2}\Gamma(n+1/2)\tag 1 \end{align}$$

where $\Gamma(z)\equiv \int_0^\infty t^{z-1}\,e^{-t}\,dt$ is the Gamma Function.

The Gamma function satisfies the Functional Relationship $\Gamma(z+1)=z\Gamma (z)$. Applying this functional relationship repeatedly to the right-hand side of $(1)$ yields

$$\begin{align} \mathscr{L}\left(t^{n-1/2}\right)(s)&=\left(\frac{1}{s}\right)^{n+1/2}\left(\frac12\,\cdot \frac32\,\cdot\frac52\,\cdots\frac{2n-1}{2}\right)\Gamma(1/2)\\\\ &=\frac{(2n-1)!!}{2^n}\left(\frac{1}{s}\right)^{n}\sqrt{\frac{\pi}{s}} \end{align}$$

where we used $\Gamma(1/2)=\int_0^\infty t^{-1/2}e^{-t}\,dt=\sqrt{\pi}$. Note that this integral can be evaluated by enforcing the substitution $t\to t^2$, which results in the well-know Gaussian integral.

Answer 2

By the definition of the $\Gamma$ function, $$\begin{eqnarray*}\mathcal{L}\left(t^{n-\frac{1}{2}}\right)&=&\int_{0}^{+\infty} t^{n-\frac{1}{2}}e^{-st}\,dt = \frac{1}{s^{n+\frac{1}{2}}}\int_{0}^{+\infty}u^{n-\frac{1}{2}}e^{-u}\,du\\&=&\frac{\Gamma\left(n+\frac{1}{2}\right)}{s^{n+\frac{1}{2}}}=\frac{\left(n-\frac{1}{2}\right)\Gamma\left(n-\frac{1}{2}\right)}{s^{n+\frac{1}{2}}}\\&=&\ldots = \frac{(2n-1)!!\cdot \Gamma\left(\frac{1}{2}\right)}{2^n s^{n+\frac{1}{2}}}\end{eqnarray*}$$ and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$.