The questions asks to show that for every real number $\alpha>0$ the function $x \mapsto e^{-\alpha x}\big(\frac{\sin x}{x} \big)^3$ is Lebesgue integrable over $(0,\infty)$ and that the function $\alpha \mapsto \int_{0}^{\infty} e^{-\alpha x}\big(\frac{\sin x}{x} \big)^3 $ is continuous on $(0,\infty)$.
For every $\alpha,x>0$ we have $$0\leq e^{-\alpha x}\bigg(\frac{\sin x}{x} \bigg)^3 \leq \bigg(\frac{\sin x}{x} \bigg)^3$$
so both claims will follow if I can show that $\big(\frac{\sin x}{x} \big)^3$ is Lebesgue integrable on $(0,\infty)$ (the continuity claim follows because of the the DCT). For every $k\in \mathbb{N}$ we have
$$\int_{k\pi}^{(k+1)\pi}\frac{\sin^3 x}{x^3} dx=\int_{0}^{\pi}\frac{\sin^3 t}{(t+k\pi)^3} dt \leq \frac{1}{k^3\pi^3}\int_{0}^{\pi}\sin^3 t \hspace{0.1cm} dt=\frac{4}{3k^3\pi^3}$$
Since $\sum_{k=1}^n 1_{[k\pi,(k+1)\pi)}\frac{\sin^3 x}{x^3}$ increase to $1_{[\pi,\infty)}\frac{\sin^3 x}{x^3}$ on $(0,\infty)$ we get by the MCT
$$\int_{\pi}^{\infty}\frac{\sin^3 x}{x^3}=\sum_{k=1}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{\sin^3 x}{x^3}\leq \frac{4}{3\pi^3} \sum_{k=1}^{\infty}\frac{1}{k^3}<\infty $$
Also $\int_{0}^{\pi}\frac{\sin^3 x}{x^3}<\infty$ since $\frac{\sin^3 x}{x^3}$ is Riemann integrable between $0$ and $\pi$. Putting both pieces together gives the result.
Am I missing something?
Let $f(\alpha)$ be the integral. By the mean value theorem, one has \begin{equation} |e^{-\alpha x} - e^{-\beta x}|\le |\alpha - \beta|\,x\, e^{-\min(\alpha, \beta)x} \end{equation} as $|\frac{sin x}{x}|\le 1$, one deduces \begin{equation} |f(\alpha) - f(\beta)|\le \int_0^{+\infty} |\alpha - \beta|\,x\, e^{-\min(\alpha, \beta)x}\, d x = \frac{|\alpha-\beta|}{\min(\alpha,\beta)^2} \end{equation} which proves continuity.