Quadratic function and the Gaussian integral

Question

I'm trying to find the CWT (Continuous Wavelet Transform) for the function $$f(t) = \alpha t^2 + \beta t + c$$ and the wavelet $$\psi(t) = (1-\frac{t^2}{\sigma^2})\exp(-\frac{t^2}{2\sigma^2})$$Note that CWT is defined as $$WT(a,b) = \int_{-\infty}^{+\infty}f(t)\frac{1}{\sqrt{a}}\psi(\frac{t-b}{a})dt$$So we should evaluate the $$WT(a,b) = \int_{-\infty}^{+\infty}(\alpha t^2 + \beta t + c)\frac{1}{\sqrt{a}}(1-\frac{(\frac{t-b}{a})^2}{\sigma^2})\exp(-\frac{(\frac{t-b}{a})^2}{2\sigma^2})dt$$ The integral which contains $c$ is $0$ because $\Psi(0) = 0$. Note that $\Psi(\omega)$ is the Fourier transform of $\psi(t)$. So we have $$WT(a,b) = \int_{-\infty}^{+\infty}(\alpha t^2 + \beta t)\frac{1}{\sqrt{a}}(1-\frac{(\frac{t-b}{a})^2}{\sigma^2})\exp(-\frac{(\frac{t-b}{a})^2}{2\sigma^2})dt$$ It seems that the integral doesn't simplify further but surprisingly, it can be simplified to (Using Mathematica) $$WT(a,b) = -\frac{1}{\sqrt{a}}\times 2a^4\sigma^4\alpha \sqrt{2\pi}\frac{1}{a\sigma} \tag{1}$$ It shows that in this case $WT(a,b)$ doesn't depend on $b$ and also $\beta$. I tried to derive $(1)$ by hand but it didn't work. For example I tried the changing of the variable $$u = \frac{t-b}{a\sigma}$$ but after that many terms appear when we expand the multiplications. Maybe differentiation under the integral is useful but I don't see how that applies here.

Answer 1

Let $$t=b+\sqrt{2}\, a \,\sigma \, u$$ $$\text{WT}(a,b)= \sqrt{2a}\int P(u)\,e^{-u^2}\,du$$ $$P(u)=\sum_{n=0}^4 A_n\,u^n$$ $$\left( \begin{array}{cc} n & A_n \\ 0 & \sigma (b (\alpha b+\beta )+c) \\ 1 & \sqrt{2} a \sigma ^2 (2 \alpha b+\beta ) \\ 2 & -2 \sigma \left(-a^2 \alpha \sigma ^2+b (\alpha b+\beta )+c\right) \\ 3 & -2 \sqrt{2} a \sigma ^2 (2 \alpha b+\beta ) \\ 4 & -4 a^2 \alpha \sigma ^3 \\ \end{array} \right)$$ The required integrals $$\int_{-\infty}^{+\infty} \,e^{-u^2}\,u^n\,du=\frac{1}{2} \left(1+(-1)^n\right) \Gamma \left(\frac{n+1}{2}\right)$$ So, we do not care about the odd powers and finally

$$\text{WT}(a,b)= \sqrt{2\pi a}\left(A_0+\frac{1}{2}A_2+\frac{3 }{4}A_4\right)$$ In other words $$\color{red}{\large{\text{WT}(a,b)=-2 \,\sqrt{2 \pi }\, a^{5/2}\, \alpha\, \sigma ^3}}$$