As part of a proof of the admissibility of the complex Morlet wavelet, I am trying to show that the following integral is positive and finite $$ 0<\int_0^\infty{\frac{(e^{\sigma \omega}-1)^2e^{-\omega^2}}{\omega}}d\omega<\infty $$
Where $\sigma > 0$. Is anyone willing to give any guidance, or provide an explicit solution for the integral? Alternatively, I believe that one of the niceties of the Morlet wavelet is that it suffices to show that the following condition holds instead (for admissibility):
$$ \hat{\psi}(0)=0\implies\int^\infty_{-\infty}\psi(t)dt=0 $$
Where the Fourier transform $\hat{\psi}(\omega)=(e^{\sigma\omega}-1)e^{-\frac{1}{2}(\sigma^2+\omega^2)}$
If anyone is interested the actual admissibility criterion is that $0<C_\psi <\infty$, where:
$$ C_\psi = \int_{-\infty }^\infty{\frac{\left|\hat{\psi}(\omega)\right|^2}{|\omega |}}d\omega = \int_{-\infty }^\infty{\frac{\left|c_\sigma(e^{\sigma \omega}-1)e^{-\frac{1}{2}(\sigma^2 + \omega^2)}\right|^2}{|\omega |}}d\omega $$
The inverse Fourier transform of
$$\hat{\psi}(\omega)=(e^{\sigma\omega}-1)\, e^{-\frac{1}{2}(\sigma^2+\omega^2)}\tag{1}$$
is
$$\psi(t)=\mathcal{F}_{\omega}^{-1}\left[\hat{\psi}(\omega)]\right](t)=\frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{\infty} \left(e^{\sigma \omega}-1\right) e^{-\frac{1}{2} \left(\sigma^2+\omega^2\right)} e^{-i t \omega} \, d\omega\\=e^{-\frac{1}{2} t (t+2 i \sigma )}-e^{-\frac{\sigma^2}{2}-\frac{t^2}{2}}\tag{2}$$
and
$$\int\limits_{-\infty}^{\infty} \psi(t) \, dt=\int\limits_{-\infty}^{\infty} \left(e^{-\frac{1}{2} t (t+2 i \sigma)}-e^{-\frac{\sigma^2}{2}-\frac{t^2}{2}}\right) \, dt=0\tag{3}.$$
With respect to the admissibility criterion, Mathematica gives the result
$$C_\psi=\int\limits_{-\infty}^\infty{\frac{\left|\hat{\psi}(\omega)\right|^2}{|\omega|}}d\omega=\int\limits_{-\infty}^{\infty} \frac{\left|\left(e^{\sigma \omega}-1\right)\, e^{-\frac{1}{2} \left(\sigma^2+\omega^2\right)}\right|^2}{|\omega|} \, d\omega\\=-e^{-\sigma^2} \sigma^2 \left(\, _2F_2\left(1,1;\frac{3}{2},2;\frac{\sigma^2}{4}\right)-2 \, _2F_2\left(1,1;\frac{3}{2},2;\sigma^2\right)\right),\quad\sigma\in\mathbb{R}\tag{4}$$
and I believe $0<C_\psi<\infty$ for $0<|\sigma|\le\beta$ and $C_\psi=0$ for $\sigma=0\lor|\sigma|>\beta$ where $\beta\approx 27.3$.
The following plots illustrate formula (4) above for $C_\psi$ as a function of $\sigma$.