First of all, by definition, for any $x \in \mathbb{R}_+$, $M_x \le 1$. Therefore, setting $\eta = (2 - \alpha)/4 \in (0, 1/4)$ and using (34),

$$ \left| \int_{0}^{|\omega|^{-\eta}} (1 - e^{\omega x}) M_x dx \right| \le \int_{0}^{|\omega|^{-\eta}} 2|\omega| x dx \le |\omega|^{1-2\eta} = |\omega|^{\alpha/2} $$

for $|\omega|$ small enough. As a consequence, $|\int_{0}^{|\omega|^{-\eta}} (1 - e^{\omega x}) M_x dx| = o(|\omega|^{\alpha-1} L(1/|\omega|))$ as $|\omega| \to 0$. Second term.

Fix $\epsilon > 0$. By the estimate (31), as $|\omega| \to 0$, uniformly for $x \ge |\omega|^{-\eta}$, $M_x \sim \frac{2-\alpha}{\alpha} x^{-\alpha} L(x)$. This allows us to write for any $\epsilon' > 0$, for $|\omega|$ small enough (depending on $\epsilon'$):

$$ \left| \int_{|\omega|^{-\eta}}^{+\infty} (1 - e^{\omega x}) M_x dx - \frac{2-\alpha}{\alpha} \int_{|\omega|^{-\eta}}^{+\infty} (1 - e^{\omega x}) x^{-\alpha} L(x) dx \right| \le \epsilon' \int_{|\omega|^{-\eta}}^{+\infty} |1 - e^{\omega x}| x^{-\alpha} L(x) dx. $$

In particular, mimicking the proof of Lemma 6.3 (ii), we bound the right-hand term and get, for $|\omega|$ small enough,

$$ \left| \int_{|\omega|^{-\eta}}^{+\infty} (1 - e^{\omega x}) M_x dx - \frac{2-\alpha}{\alpha} \int_{|\omega|^{-\eta}}^{+\infty} (1 - e^{\omega x}) x^{-\alpha} L(x) dx \right| \le \epsilon |\omega|^{\alpha-1} L\left(\frac{1}{|\omega|}\right). $$

The desired result then follows from the estimate of Lemma 6.3 (ii). □

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