Bessel function to $\sin(kr)$ $J_{\frac{1}{2}}(kr)=\frac{\sqrt{\frac{2}{\pi }} \text{Sin}[\text{kr}]}{\sqrt{\text{kr}}})$ This can be easily obtained by Mathematica,
How to do the details?
 A: One way is to use the series definition of the Bessel function together with the duplication formula for the Gamma function.
For $x \geq 0$ we have
$$
\begin{align}
J_{1/2}(x) &= \sqrt{\frac{x}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+1/2)} \left(\frac{x}{2}\right)^{2n} \\
&= \sqrt{\frac{2}{x}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+1/2)} \left(\frac{x}{2}\right)^{2n+1}.
\end{align}
$$
Now
$$
\Gamma(n+1)\Gamma(n+1+1/2) = 2^{-1-2n} \sqrt{\pi} \,\Gamma(2n+2)
$$
by the duplication formula, so
$$
J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma(2n+2)} x^{2n+1} = \sqrt{\frac{2}{\pi x}} \sin x.
$$
A: Here's the quick and dirty way to do it for $kr\ge0$. The case of $kr<0$ involves a little trick with Bessel functions of negative integers. I will show this below. The following is often taken as the definition for $J_{\nu}$.
$$J_{\nu}(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\nu+1)}\left(\frac{x}{2}\right)^{2n+\nu}$$
Letting $\nu = \frac{1}{2}$ we have
$$J_{\frac{1}{2}} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\frac{1}{2}+1)}\left(\frac{x}{2}\right)^{2n+\frac{1}{2}} = \left(\frac{2}{x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\Gamma(n+\frac{1}{2}+1)}\left(\frac{x}{2}\right)^{2n+1}$$
Here I've just multiplied by a clever form of $1$ to get something we're used to seeing in the power series for $\sin$ (this is where your term in the denominator comes from). Now all that is really left is to evaluate $\Gamma(m+\frac{1}{2})$ for $m$ a natural number (plus 0).
From the properties of the gamma function, we have that $\Gamma(m+\frac{1}{2}) = \frac{(2n)!\sqrt{\pi}}{4^nn!}$. Substituting this into our expression we have
$$J_{\frac{1}{2}} = \left(\frac{2}{x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!\frac{(2(n+1))!\sqrt{\pi}}{4^{n+1}(n+1)!}}\left(\frac{x}{2}\right)^{2n+1}$$
This quickly reduces to $\sin$ after a little manipulation:
$$J_{\frac{1}{2}} = \left(\frac{2}{x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n4^{n+1}(n+1)}{(2n+2)!\sqrt{\pi}}\left(\frac{x}{2}\right)^{2n+1} = \left(\frac{2}{\pi x}\right)^\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n+2}(n+1)}{2(n+1)(2n+1)!}\left(\frac{x}{2}\right)^{2n+1}$$
$$J_{\frac{1}{2}} = \left(\frac{2}{\pi x}\right)^\frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} = \sqrt{\frac{2}{\pi x}}\sin x$$
The case of $x<0$ is pretty simple once you have this done. We have for $x>0$ that $J_{\nu}(-x) = e^{\nu\pi i}J_{\nu}(x)$. In our case $\nu = \frac{1}{2}$ and as we know, $e^{i\pi/2} = i$ so $J_{\nu}(-x) = i J_{\nu}(x)$. This is verified with Wolfram Alpha.
