Prove the boundedness of $R(z; D)v(x) = -i \int_x^{\infty} e^{i z(x-y)} v(y) dy, \quad x \in \mathbb{R}_+ \ a.e.$ in $L_1 (\mathbb{R}_+)$

Question

Also, $Im z < 0$

That's what I've got:

$$\DeclareMathOperator{\Dd}{\operatorname{d}\!} \begin{split} \|R(z; D)v(x) \|_{L_1} &= \int\limits_0^{+\infty} \Bigg|\int\limits_x^{+\infty} -i e^{iz(x-y)} v(y) \Dd y\Bigg|\Dd x \\ &\le \int\limits_0^{+\infty} \Bigg(\int\limits_x^{+\infty} |-i e^{iz(x-y)} v(y)| \Dd y \Bigg)\Dd x \\ &= \int\limits_0^{+\infty} \Bigg(\int\limits_x^{+\infty} |-i| |e^{iz(x-y)}| |v(y)| \Dd y \Bigg)\Dd x \\ & = \int\limits_0^{+\infty} \Bigg(\int\limits_x^{+\infty} e^{-(\mathrm{Im} z )(x-y)}|v(y)| \Dd y \Bigg)\Dd x \end{split} $$

I don't know what to do further