Given a wavelet family $\psi_{s, a}$ generated by translations and dilations of a mother wavelet $\psi$ $$ \psi_{s, a}(x)=\frac{1}{s} \psi\left(\frac{x-a}{s}\right) $$ we can show a wavelet decomposition of signal $f$ as $$ W_{f}(s, a)=\int_{-\infty}^{\infty} f(x) \psi^*_{s, a}(x) d x $$ where $^*$ denotes complex conjugation. In a paper, $\psi$ is defined in the Fourier domain as $$\psi_{s, a}(x)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \hat{\psi}(s \omega) e^{-j \omega a} e^{j \omega x} d \omega$$ I was wondering if someone can help me to see where this comes from. Thanks!
My try: An inverse Fourier transform is written as $$\psi_{s, a}(x)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \widehat{\psi_{s, a}}(\omega) e^{i \omega x} d \omega$$ On the other hand \begin{align} \widehat{\psi_{s, a}}(\omega) &= \int_{-\infty}^{\infty} \psi_{s, a}(x) e^{-i \omega x} d x\\ &= \int_{-\infty}^{\infty} \frac{1}{s} \psi\left(\frac{x-a}{s}\right) e^{-i \omega x} d x\\ \end{align}
If $\widehat{\psi}$ represents the Fourier transform of $\psi$ then
$$ \psi(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{\psi}(\omega) e^{j\omega x}{\rm d}\omega \tag{1} $$
Now consider the expression
\begin{eqnarray} \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{\psi}(s\omega) e^{-j\omega a}e^{j\omega x}{\rm d}\omega &=& \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{\psi}(s\omega) e^{j\omega(x-a)}{\rm d}\omega \\ &=& \frac{1}{2\pi s}\int_{-\infty}^{+\infty} \widehat{\psi}(\color{blue}{s\omega}) e^{j(\color{blue}{s\omega})(x-a)/s}{\rm d}(\color{blue}{s\omega}) \\ &\stackrel{\nu = \color{blue}{s\omega}}{=}& \frac{1}{2\pi} \int_{-\infty}^{+\infty} \widehat{\psi}(\nu) e^{j\nu(x - a)/s}{\rm d}\nu \\ &\stackrel{y = (x-a)/s}{=}& \frac{1}{s}\left\{\frac{1}{2\pi} \int_{-\infty}^{+\infty} \widehat{\psi}(\nu) e^{j\nu y}{\rm d}\nu\right\} \\ &\stackrel{(1)}{=}& \frac{1}{s}\psi(y) = \frac{1}{2}\psi\left(\frac{x - a}{s}\right) \\ &=& \psi_{s,a}(x) \tag{2} \end{eqnarray}