In my textbook for wavelets I found an example which shows that the Haar-Wavelet $H$ is indeed a wavelet.
So I need to show $0 < 2\pi \int_\mathbb{R} \frac{|\widehat{H}(\omega)|^2}{|\omega|} d\omega < \infty $.
The Fourier transform is $\widehat{H} = \frac{i}{\omega \sqrt{2\pi}} e^{-\frac{-\omega}{2}}4sin^2(\frac{\omega}{4})$. Then:
$\begin{split} \frac{|\widehat{H}(\omega)|^2}{|\omega|} &= \frac{1}{|\omega |} {\left| \frac{i}{\omega \sqrt{2\pi}} e^{-\frac{i\omega}{2}} 4 sin^2(\frac{\omega}{4}) \right|}^2 \\ &{\leq \; \;} \frac{1}{{|\omega |}^3} {\left| e^{-\frac{i\omega}{2}} sin^2(\frac{\omega}{4}) \right| }^2 \\ &= \left| \frac{sin^4(\frac{\omega}{4})}{{\omega}^3} \right|. \end{split}$
But my textbook says that it can be estimated by $\frac{|\widehat{H}(\omega)|^2}{|\omega|} \leq \left|\frac{{sin(\omega)}^4}{{\omega}^3} \right|$. Did I miss something?
When I type this integral into WolframAlpha i get $0.69314718055994...$ as value for $[0, \infty)$. But is there a more elegant way to show that $0 < \int_{\mathbb{R}} \left|\frac{{sin(\omega)}^4}{{\omega}^3} \right| d\omega <\infty$?