First observe that $\mathcal{W}(\mathcal{F}(z),\mathcal{U}(z))=\mathcal{F}^{\prime}(z)\mathcal{U}(z)-\mathcal{F}(z)\mathcal{U}^{\prime}(z)$ is a constant. In fact,

	$\displaystyle\frac{d}{dz}\mathcal{W}(\mathcal{F}(z),\mathcal{U}(z))$	$\displaystyle=\Big(\mathcal{F}^{\prime\prime}(z)\mathcal{U}(z)+\mathcal{F}^{\prime}(z)\mathcal{U}^{\prime}(z)\Big)-\Big(\mathcal{F}(z)\mathcal{U}^{\prime\prime}(z)+\mathcal{F}^{\prime}(z)\mathcal{U}^{\prime}(z)\Big)$
(4.49)			$\displaystyle=j(jz^{2}+2h+1)(\mathcal{F}(z)\mathcal{U}(z)-\mathcal{F}(z)\mathcal{U}(z))=0.$

Hence $\mathcal{W}(\mathcal{F}(z),\mathcal{U}(z))$ has to be a constant. Now we evaluate it at $z=0$, we get

$$\mathcal{W}(\mathcal{F},\mathcal{U})=2\mathcal{F}^{\prime}(0)\mathcal{U}(0).$$

Now

$$\mathcal{F}^{\prime}(0)=2\sqrt{j}\int_{0}^{\infty}e^{-t^{2}}t^{h+1}dt=\sqrt{j}\Gamma(\frac{h}{2}+1),$$

and similarly

$$\mathcal{U}(0)=\frac{1}{2}\Gamma(\frac{h}{2}+\frac{1}{2}).$$

Applying Legendre duplication formula

(4.50)

$$\frac{\Gamma(t)\Gamma(t+\frac{1}{2})}{\Gamma(2t)}=\frac{\sqrt{\pi}}{2^{2t-1}},\ t>0,$$

we have

$$\mathcal{W}(\mathcal{F},\mathcal{U})=2^{-h}\sqrt{j\pi}\Gamma(h+1)>0.$$

∎