For some reals $a$ and $b$, we have $h(t)=p\left(\cfrac{t-a}{b}\right)$. Also, $p(s)$ is real and even function of $s$. By the transformation rules, how can I determine the continuous Fourier transform $H$ of $h$ using the Fourier transform $P$ to $p$.
I am thinking I need to use $\displaystyle p(f)=\int^{\infty}_{\infty} p(t)\cdot\exp(2\pi i f t)\ \text dt$ as well as the time shifting and scaling.
Note that, as $p$ is even, we have for $b < 0$: $\def\abs#1{\left|#1\right|}$ $$ h(t) = p\left(\frac{t-a}b\right) = p\left(-\frac{t-a}{b}\right) = p\left(\frac{t-a}{\abs b}\right) $$ (for $b > 0$ we obviously also have $h(t) = p\bigl((t-a)/\abs b\bigr)$. Now \begin{align*} H(f) &= \int_{-\infty}^\infty h(t) \exp(2\pi i ft)\,dt\\ &= \int_{-\infty}^\infty p\left(\frac{t-a}{\abs b}\right)\exp(2\pi i ft)\,dt\\ &= \int_{-\infty}^\infty p(\tau)\exp\bigl(2\pi i f(\abs b \tau + a)\bigr)\,\abs b \,d\tau & \text{we substituted $\tau = \frac{t-a}{\abs b}$)}\\ &= \abs b \exp(2\pi i fa)\int_{-\infty}^\infty p(\tau)\exp(2\pi i f\abs b\tau)\, d\tau\\ &= \abs b \exp(2\pi i fa)\cdot P(f\abs b) \end{align*}
