Consider the following two signals
$$\begin{aligned} x(t) &= e^{-kt} u(t)\\\\ y(t) &= x(t) \,\frac{\sin(\theta t)}{\pi t}\end{aligned}$$
Determine the relation between $k$ and $\theta$ so that the energy of $y$ is half of the energy of $x$.
What I did was the following.
$$\mathcal{E}_x = \int_{-\infty}^{\infty}\vert x(t)\vert^2 = \frac{1}{2k}$$
$$\mathcal{E}_y = \int_{-\infty}^{\infty}\vert X(f)P_F(f) \vert^2 = \frac{\theta/\pi}{k^2+\theta^2}$$
with
$$P_F(f) = \begin{cases} 1 && -\frac F2 <f< \frac F2\\ 0 && \text{otherwise}\end{cases}$$
where $F=\frac{\theta}{\pi}$. A classmate suggested to use the duality property of the $\mathscr{F}$ transform, meaning $$\mathscr{F}\Bigl[\mathscr{F}\left[x(t)\right]\Bigr]=x(-t)$$ but I can't see where or how.