Refined decay rates of $C_0$-semigroups on Banach spaces

Abstract

We study rates of decay for $C_0$-semigroups on Banach spaces under the assumption that the norm of the resolvent of the semigroup generator grows with $|s|^{\beta }\log (|s|)^b$, $\beta , b \ge 0$, as $|s|\rightarrow \infty$, and with $|s|^{-\alpha }\log (1/|s|)^a$, $\alpha , a \ge 0$, as $|s|\rightarrow 0$. Our results do not suppose that the semigroup is bounded. In particular, for $a=b=0$, our results improve the rates involving Fourier types obtained by Rozendaal and Veraar (J Funct Anal 275(10):2845–2894, 2018).

Appendices

Appendix

Proof of Proposition 4.3

Item (a). Let $\zeta >1$ and set $\tilde{c}:=\zeta +a$.

$\bullet$ Case 1: $\alpha =1$.

Case 1(a): $\tilde{c} \in (1,2]$. Note that in this case, $a\in [0,1)$. Set $h_{\alpha ,\zeta }(\lambda )=\lambda ^{\alpha }(2\pi -i\log (\lambda ))^{\zeta }$, with $\lambda \in i{\mathbb {R}}\setminus \{0\}$, and define the operator $L_{\nu ,\tilde{c}}(A):=(1+A)^{-\nu }(2\pi -i\log (A))^{-\tilde{c}}\in {\mathcal {L}}(X)$. Since $(\lambda +A)^{-1}$ commutes with $L_{\nu ,\tilde{c}}(A)$, it follows from the moment inequality that

$$\begin{aligned}{} & {} \Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,\tilde{c}}(A)\Vert _{{\mathcal {L}}(X)} \lesssim \Vert h_{1,1-a}(\lambda )(\lambda +A)^{-1}L_{\nu ,1}(A)\Vert ^{2-\tilde{c}}_{{\mathcal {L}}(X)}\nonumber \\{} & {} \quad \Vert h_{1,2-a}(\lambda )(\lambda +A)^{-1}L_{\nu ,2}(A)\Vert ^{\tilde{c}-1}_{{\mathcal {L}}(X)}. \end{aligned}$$

(A.1)

Let $\varepsilon >0$, set $A_\varepsilon :=(A+\varepsilon )(1+\varepsilon A)^{-1}$ and note that $A^{-1}_\varepsilon \in {\mathcal {L}}(X)$. For each $\lambda \in i{\mathbb {R}}{\setminus }\{0\}$, let $r\in (0,|\lambda |/2]$ and $R\ge 2|\lambda |+2$ be such that $\sigma (A_\varepsilon )\subset \{z\in {\mathbb {C}}\mid r<|z|<R\}$, let $\theta \in (\pi /2,\pi )$ and set $\gamma _{+}=\{se^{i\theta }\mid s\in [r,R]\}$, $\gamma _{-}=\{te^{-i\theta }\mid t\in [r,R]\}$, $\gamma _{r}=\{re^{is}\mid s\in [-\theta ,\theta ]\}$, $\gamma _{R}=\{Re^{is}\mid s\in [-\theta ,\theta ]\}$ and $\gamma :=\gamma _{+}\cup \gamma _{-}\cup \gamma _{r}\cup \gamma _{R}$. Then, by the Riesz–Dunford functional calculus (see (2.6)), for each $x\in X$ (here, $y:=(1+A)^{-\nu }x$),

$$\begin{aligned}{} & {} h_{1,1-a}(\lambda )(\lambda +A_\varepsilon )^{-1}(2\pi -i\log (A_\varepsilon ))^{-1}y\\{} & {} \quad = \frac{h_{1,1-a}(\lambda )}{2\pi i} \int _{\gamma } \frac{1}{(2\pi -i\log (z))}R(z,A_\varepsilon ) (\lambda +A_\varepsilon )^{-1}ydz\\{} & {} \quad = \frac{h_{1,1-a}(\lambda )}{2\pi i} \int _{\gamma } \frac{1}{(2\pi -i\log (z))(\lambda +z)}dz(\lambda +A_\varepsilon )^{-1} y\\{} & {} \qquad + \frac{h_{1,1-a}(\lambda )}{2\pi i} \int _{\gamma } \frac{1}{(2\pi -i\log (z))(\lambda +z)}R(z,A_\varepsilon ) y dz\\{} & {} \quad =\frac{h_{1,1-a}(\lambda )(\lambda +A_\varepsilon )^{-1}y}{2\pi -i\log (-\lambda )} +\frac{1}{2\pi i} \int _{r}^{R}\frac{h_{1,1-a}(\lambda )e^{-i\theta }R(te^{-i\theta },A_\varepsilon )y}{(2\pi -\theta -i\log (t))(\lambda +te^{-i\theta })} dt\\{} & {} \qquad - \frac{h_{1,1-a}(\lambda )}{2\pi i} \int _{r}^{R}\frac{e^{i\theta }}{(2\pi +\theta -i\log (t))(\lambda +te^{i\theta })}R(te^{i\theta },A_\varepsilon )y dt\\{} & {} \qquad + \frac{h_{1,1-a}(\lambda )}{2\pi i} \int _{-\theta }^{\theta }\frac{iRe^{is}}{(2\pi -s+i\log (R))(\lambda +Re^{is})}R(Re^{is},A_\varepsilon )y ds\\{} & {} \qquad - \frac{h_{1,1-a}(\lambda )}{2\pi i} \int _{-\theta }^{\theta }\frac{ire^{is}}{(2\pi -s+i\log (r))(\lambda +re^{is})}R(re^{is},A_\varepsilon )yds, \end{aligned}$$

where we have used the residue theorem in the third identity. By taking the limit $\theta \rightarrow \pi$ on both sides of the identity above, one gets

$$\begin{aligned}{} & {} h_{1,1-a}(\lambda )(\lambda +A_\varepsilon )^{-1}(2\pi -i\log (A_\varepsilon ))^{-1}y= \frac{h_{1,1-a}(\lambda )}{2\pi -i\log (-\lambda )}(\lambda +A_\varepsilon )^{-1}y\\{} & {} \quad +\frac{1}{2\pi i} \int _{r}^{R}\frac{h_{1,1-a}(\lambda )}{(\pi -i\log (t))(\lambda -t)}(t+A_\varepsilon )^{-1}y dt\\{} & {} \quad - \frac{1}{2\pi i} \int _{r}^{R}\frac{h_{1,1-a}(\lambda )(t+A_\varepsilon )^{-1}y}{(3\pi -i\log (t))(\lambda -t)} dt\\{} & {} \quad + \frac{1}{2\pi i} \int _{-\pi }^{\pi }\frac{ih_{1,1-a}(\lambda )Re^{is} R(Re^{is},A_\varepsilon )y}{(2\pi -s-i\log (R))(\lambda +Re^{is})} ds\\{} & {} \quad - \frac{h_{1,1-a}(\lambda )}{2\pi i} \int _{-\pi }^{\pi }\frac{ire^{is}}{(2\pi +s-i\log (r))(\lambda +re^{is})}R(re^{is},A_\varepsilon )y ds \end{aligned}$$

Now, by taking the limits $r\rightarrow 0$ and $R\rightarrow \infty$ on both sides of the last identity, one gets for each $x\in X$,

$$\begin{aligned}{} & {} h_{1,1-a}(\lambda )(\lambda +A_\varepsilon )^{-1}(2\pi -i\log (A_\varepsilon ))^{-1}y\\{} & {} \quad = \frac{h_{1,1-a}(\lambda )(\lambda +A_\varepsilon )^{-1}y}{2\pi -i\log (-\lambda )}\\{} & {} \qquad + \int _{0}^{\infty } \frac{ih_{1,1-a}(\lambda )}{(3\pi ^2-4\pi i\log (t)-\log (t)^2)(\lambda -t)}(t+A_\varepsilon )^{-1}y\,dt. \end{aligned}$$

Finally, by taking the limit $\varepsilon \rightarrow 0^{+}$ on both hands of the identity above, one gets

$$\begin{aligned}{} & {} h_{1,1-a}(\lambda )(\lambda +A)^{-1}(2\pi -i\log (A))^{-1}y\nonumber \\{} & {} \quad =\frac{h_{1,1-a}(\lambda )(\lambda +A)^{-1}y}{2\pi -i\log (-\lambda )}\nonumber \\{} & {} \qquad +\int _{0}^{\infty } \frac{ih_{1,1-a}(\lambda )(t+A)^{-1}y}{(3\pi ^2-4\pi i\log (t)-\log (t)^2)(\lambda -t)}dt, \end{aligned}$$

(A.2)

where we have used on the left-hand side that $(\lambda +A_\varepsilon )^{-1}\rightarrow (\lambda +A)^{-1}$ uniformly (by Lemma 2.10), $(2\pi -i\log (A_\varepsilon ))^{-1}\rightarrow (2\pi -i\log (A))^{-1}$ strongly (see the proof of Lemma 3.5.1 [20]), and on the right-hand side dominated convergence.

Then, by (A.2), one gets

$$\begin{aligned}{} & {} |h_{1,1-a}(\lambda )|\left\| (\lambda +A)^{-1}(2\pi -i\log (A))^{-1}(1+A)^{-\nu }\right\| _{{\mathcal {L}}(X)}\nonumber \\{} & {} \quad \lesssim \left\| \frac{h_{1,1-a}(\lambda )(\lambda +A)^{-1}}{2\pi -i\log (-\lambda )}\right\| _{{\mathcal {L}}(X)}+ \int _{0}^{\infty } \frac{|h_{1,1-a}(\lambda )| }{(\pi ^2+\log (t)^{2})|\lambda -t|}\Vert (t+A)^{-1}\Vert _{{\mathcal {L}}(X)} dt\nonumber \\{} & {} \quad \lesssim \left\| \frac{h_{1,1-a}(\lambda )(\lambda +A)^{-1}}{2\pi -i\log (-\lambda )}\right\| _{{\mathcal {L}}(X)}+\int _{0}^{\infty } \frac{|h_{1,1-a}(\lambda )| }{t(\pi ^2+\log (t)^{2})|(\lambda |+t)} dt\nonumber \\{} & {} \quad \lesssim \left\| \frac{h_{1,1-a}(\lambda )(\lambda +A)^{-1}}{2\pi -i\log (-\lambda )}\right\| _{{\mathcal {L}}(X)} +\int _{0}^{\infty }\frac{|h_{1,1-a}(\lambda )|(t+1)}{t(\pi ^2+\log (t)^2)(|\lambda |+t)}dt\nonumber \\{} & {} \quad =\left\| \frac{h_{1,1-a}(\lambda )(\lambda +A)^{-1}}{2\pi -i\log (-\lambda )}\right\| _{{\mathcal {L}}(X)}+ \frac{|h_{1,1-a}(\lambda )|(|\lambda |-1)}{|\lambda |\log (|\lambda |)}, \end{aligned}$$

(A.3)

where we have used relation (2.3) in the last identity.

Note that for each $\lambda \in i{\mathbb {R}}\setminus \{0\}$ with $|\lambda |\le 1$, it follows from (4.5) that

$$\begin{aligned} \left\| \frac{h_{1,1-a}(\lambda )(\lambda +A)^{-1}}{2\pi -i\log (-\lambda )}\right\| _{{\mathcal {L}}(X)}\lesssim 1, \end{aligned}$$

and since for each $\eta >0$, $\displaystyle {\lim _{|\lambda |\rightarrow 0^{+}} |\lambda |\log (|\lambda |)^{\eta }=0}$, one gets

$$\begin{aligned} \frac{|h_{1,1-a}(\lambda )|(|\lambda |-1)}{|\lambda |\log (|\lambda |)}\le \frac{(2\pi +|\log (|\lambda |)|)^{1-a}(|\lambda |-1)}{\log (|\lambda |)} \lesssim |\lambda ||\log (|\lambda |)|^{-a} {\mathop {\longrightarrow }\limits ^{|\lambda |\rightarrow 0^+}} 0 \end{aligned}$$

and $|h_{1,1-a}(\lambda )|\rightarrow 0$ as $|\lambda |\rightarrow 0^{+}$. Hence, one concludes that

$$\begin{aligned}{} & {} \sup \left\{ \left\| h_{1,1-a}(\lambda )(\lambda +A)^{-1}(1+A)^{-\nu }(2\pi -i\log (A))^{-1}\right\| _{{\mathcal {L}}(X)} \mid \lambda \in i{\mathbb {R}}\setminus \{0\}, |\lambda |\le 1\right\} \nonumber \\{} & {} \quad <\infty . \end{aligned}$$

(A.4)

Now, by using the same ideas as before, one has for each $\varepsilon >0$ and each $x\in X$,

$$\begin{aligned}{} & {} h_{1,2-a}(\lambda )(\lambda +A_\varepsilon )^{-1}(2\pi -i\log (A_\varepsilon ))^{-2}(1+A)^{-\nu }x\\{} & {} \quad = \frac{h_{1,2-a}(\lambda )(\lambda +A_\varepsilon )^{-1}(1+A)^{-\nu }x}{(2\pi -i\log (\lambda ))^2}\\{} & {} \qquad -\int _{0}^{\infty } \frac{2ih_{1,2-a}(\lambda )(2\pi -i\log (t))(t+A_\varepsilon )^{-1}(1+A)^{-\nu }x}{(3\pi ^2-4\pi i\log (t)-\log (t)^2)^2(\lambda -t)} dt. \end{aligned}$$

So, by taking the limit $\varepsilon \rightarrow 0^+$ on both sides of the identity, one gets

$$\begin{aligned}{} & {} h_{1,2-a}(\lambda )(\lambda +A)^{-1}(2\pi -i\log (A))^{-2}(1+A)^{-\nu }x = \frac{h_{1,2-a}(\lambda )(\lambda +A)^{-1}(1+A)^{-\nu }x}{(2\pi -i\log (\lambda ))^2}\\{} & {} \quad -\int _{0}^{\infty } \frac{2ih_{1,2-a}(\lambda )(2\pi -i\log (t))(t+A)^{-1}(1+A)^{-\nu }x}{(3\pi ^2-4\pi i\log (t)-\log (t)^2)^2(\lambda -t)} dt. \end{aligned}$$

Then,

$$\begin{aligned}{} & {} \left\| h_{1,2-a}(\lambda )(\lambda +A)^{-1}(2\pi -i\log (A))^{-2}\right\| _{{\mathcal {L}}(X)}\lesssim \left\| \frac{h_{1,2-a}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (\lambda ))^2}\right\| _{{\mathcal {L}}(X)}\\{} & {} \qquad + \int _{0}^{e^{-2\pi }} \frac{|h_{1,2-a}(\lambda )||\log (t)|}{t(\pi ^{2}+\log (t)^2)^2(|\lambda |+t)}dt+\int _{e^{-2\pi }}^{e^{2\pi }} \frac{|h_{1,2-a}(\lambda )|(|\log (t)|+2\pi )}{|(3\pi ^2-4\pi i\log (t)-\log (t)^2)^2|}\frac{dt}{t}\\{} & {} \qquad +\int _{e^{2\pi }}^{\infty } \frac{|h_{1,2-a}(\lambda )| |\log (t)|}{t(\pi ^{2}+\log (t)^2)^2(|\lambda |+t)}dt \lesssim \left\| \frac{h_{1,2-a}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (\lambda ))^2}\right\| _{{\mathcal {L}}(X)}\\{} & {} \qquad +|h_{1,2-a}(\lambda ) |\int _{0}^{\infty } \frac{\pi ^2-2(1+1/t)\log (t)+\log (t)^2}{(\pi ^2+(\log (t))^{2})^2(|\lambda |+t)} dt +|h_{1,2-a}(\lambda )|\\{} & {} \quad = \left\| \frac{h_{1,2-a}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (\lambda ))^2}\right\| _{{\mathcal {L}}(X)}+ \frac{|h_{1,2-a}(\lambda )|(|\lambda |\log (|\lambda |)-|\lambda |+1 +|\lambda |\log (|\lambda |)^2)}{|\lambda |\log (|\lambda |)^2}, \end{aligned}$$

where we have used relation (2.4) in the last identity.

By using the same reasoning as before, one concludes that

$$\begin{aligned}{} & {} \sup \left\{ \left\| h_{1,2-a}(\lambda )(\lambda +A)^{-1}(2\pi -i\log (A))^{-2}(1+A)^{-\nu }\right\| _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}\setminus \{0\}, |\lambda |\le 1\right\} \nonumber \\{} & {} \quad <\infty . \end{aligned}$$

(A.5)

Finally, by combining (A.1), (A.4) and (A.5), it follows that

$$\begin{aligned} \sup \{\Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,\tilde{c}}(A)\Vert _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}\setminus \{0\}, |\lambda |\le 1\}<\infty . \end{aligned}$$

Case 1(b): $\tilde{c}\in (2,3]$. In this case, $a\in [1,2)$; then, by the moment inequality, one gets for each $\lambda \in i{\mathbb {R}}{\setminus }\{0\}$,

$$\begin{aligned} \Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,\tilde{c}}(A)\Vert _{{\mathcal {L}}(X)}\lesssim & {} \Vert h_{1,2-a}(\lambda )(\lambda +A)^{-1}L_{\nu ,2}(A)\Vert ^{2-\tilde{c}}_{{\mathcal {L}}(X)}\\{} & {} \Vert h_{1,3-a}(\lambda )(\lambda +A)^{-1}L_{\nu ,3}(A)\Vert ^{\tilde{c}-1}_{{\mathcal {L}}(X)}, \end{aligned}$$

and it remains to estimate $\Vert h_{1,3-a}(\lambda )(\lambda +A)^{-1}L_{\nu ,3}(A)\Vert ^{\tilde{c}-1}_{{\mathcal {L}}(X)}$. Note that for each $\lambda \in i{\mathbb {R}}\setminus \{0\}$, $\varepsilon >0$ and each $x\in X$, one has (here, $y=(1+A)^{-\nu }x$)

$$\begin{aligned}{} & {} h_{1,3-a}(\lambda )(\lambda +A_\varepsilon )^{-1}(2\pi -i\log (A_\varepsilon ))^{-3}y = \frac{h_{1,3-a}(\lambda )(\lambda +A_\varepsilon )^{-1}y}{(2\pi -i\log (-\lambda ))^{3}}\\{} & {} \quad + ih_{1,3-a}(\lambda )\int _{0}^{\infty } \frac{(26\pi ^3-24\pi ^2 i \log (t)+6\pi \log (t))(t+A_\varepsilon )^{-1}y}{(3\pi ^2-4\pi i\log (t)-\log (t)^2)^{3}(\lambda -t)} dt, \end{aligned}$$

and then, by taking the limit $\varepsilon \rightarrow 0^+$ on both sides of the last identity, one gets

$$\begin{aligned}{} & {} h_{1,3-a}(\lambda )(\lambda +A)^{-1}L_{\nu ,3}(A)x=\frac{h_{1,3-a}(\lambda )(\lambda +A)^{-1}y}{(2\pi -i\log (-\lambda ))^{3}}+ \\{} & {} \quad + \int _{0}^{\infty } \frac{ih_{1,3-a}(\lambda )(26\pi ^3-24\pi ^2 i \log (t)+6\pi \log (t)}{(3\pi ^2-4\pi i\log (t)-\log (t)^2)(\lambda -t)}(t+A)^{-1}y\,dt. \end{aligned}$$

Thus, by relation (2.5),

$$\begin{aligned}{} & {} \left\| h_{1,3-a}(\lambda )(\lambda +A)^{-1}L_{\nu ,3}(A)\right\| _{{\mathcal {L}}(X)} \lesssim \left\| \frac{h_{1,3-a}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (-\lambda ))^{3}}\right\| _{{\mathcal {L}}(X)}\\{} & {} \qquad +\int _{0}^{e^{-\sqrt{3}\pi }} \frac{|h_{1,3-a}(\lambda )| (26\pi ^3+24\pi ^2|\log (t)| +6\pi \log (t)^2)\Vert (t+A)^{-1}\Vert _{{\mathcal {L}}(X)}}{(\pi ^2+\log (t)^{2})^3|\lambda -t|} dt\\{} & {} \qquad + \int _{e^{-\sqrt{3}\pi }}^{e^{\sqrt{3}\pi }} \frac{|h_{1,3-a}(\lambda )|}{|3\pi ^2-4\pi i\log (t)-\log (t)^2|||\lambda -t|}\Vert (t+A)^{-1}\Vert _{{\mathcal {L}}(X)} dt\\{} & {} \qquad +|h_{1,3-a}(\lambda )|\int _{e^{\sqrt{3}\pi }}^{\infty } \frac{26\pi ^3+24\pi ^2|\log (t)|+6\pi \log (t)^2}{(\pi ^2+\log (t)^{2})^{3}|\lambda -t|}\Vert (t+A)^{-1}\Vert _{{\mathcal {L}}(X)} dt\\{} & {} \quad \lesssim \left\| \frac{h_{1,3-a}(\lambda )}{(2\pi -i\log (-\lambda ))^{3}}(\lambda +A)^{-1}\right\| _{{\mathcal {L}}(X)}+ \int _{0}^{\infty } \frac{|h_{1,3-a}(\lambda )|f(t)}{t(\pi ^2+\log (t)^{2})^{3}(|\lambda |+t)} dt\\{} & {} \qquad + \int _{e^{-\sqrt{3}\pi }}^{e^{\sqrt{3}\pi }} \frac{|h_{1,3-a}(\lambda )|}{t^2|3\pi ^2-4\pi i\log (t)-\log (t)^2|}dt\\{} & {} \qquad \frac{|h_{1,3-c}(\lambda )|f(t)}{t(\pi ^2+\log (t)^{2})(|\lambda |+t)} dt \lesssim \left\| \frac{h_{1,3-a}(\lambda )}{(2\pi -i\log (-\lambda ))^3} (\lambda +A)^{-1}\right\| _{{\mathcal {L}}(X)}\\{} & {} \qquad +|h_{1,3-a}(\lambda )| \frac{(|\lambda |\log (|\lambda |)^2-2(|\lambda |\log (|\lambda |)-|\lambda |+1)}{|\lambda |\log (|\lambda |)^3}\\{} & {} \qquad + |h_{1,3-a}(\lambda )|, \end{aligned}$$

where for each $t>0$,

$$\begin{aligned} f(t)= & {} \pi ^2((-2+\pi ^2)t-2)+t\log (t)^4-4t\log (t)^3\\{} & {} +2((3+\pi ^2)t+3)\log (t)^2-4\pi ^2t\log (t). \end{aligned}$$

By proceeding as in Case 1(a), one concludes that

$$\begin{aligned} \sup \{\Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,\tilde{c}}(A)\Vert _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}\setminus \{0\}, |\lambda |\le 1\}<\infty . \end{aligned}$$

Case 1(c): $\tilde{c}>3$. In this case, $a\ge 2$. Let $\zeta =\zeta _1+\zeta _2$, with $\zeta _2\in (1,2)$. Again, by applying the moment inequality over $\zeta _2$, one gets

$$\begin{aligned}{} & {} \Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,a+\zeta _1+\zeta _2}(A)\Vert _{{\mathcal {L}}(X)} \lesssim \Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,a+\zeta _2}(A)\Vert _{{\mathcal {L}}(X)}\\{} & {} \quad \lesssim \Vert h_{1,1}(\lambda )(\lambda +A)^{-1}L_{\nu ,1+a}(A)\Vert ^{2-\zeta _2}_{{\mathcal {L}}(X)} \Vert h_{1,2}(\lambda )(\lambda +A)^{-1}L_{\nu ,2+a}(A)\Vert ^{\zeta _2-1}_{{\mathcal {L}}(X)}. \end{aligned}$$

Let $\gamma$ be the same path as presented in Case 1(a). Then, for each $\varepsilon >0$ and each $x\in X$,

$$\begin{aligned}{} & {} h_{1,1}(\lambda )(\lambda +A_\varepsilon )^{-1}(2\pi -i\log (A_\varepsilon ))^{-(1+a)}x\\{} & {} \quad = \frac{h_{1,1}(\lambda )}{2\pi i} \int _{\gamma } \frac{1}{(2\pi -i\log (z))^{1+a}}R(z,A_\varepsilon ) (\lambda +A_\varepsilon )^{-1} x dz\\{} & {} \quad {\mathop {\longrightarrow }\limits ^{\theta \rightarrow \pi }} \frac{h_{1,1}(\lambda ) (\lambda +A_\varepsilon )^{-1}x}{(2\pi -i\log (-\lambda ))^{1+a}}+\frac{1}{2\pi i} \int _{r}^{R}\frac{h_{1,1}(\lambda )(t+A_\varepsilon )^{-1}}{(2\pi -i\log (t))^{1+a}(\lambda -t)} x dt\\{} & {} \quad - \frac{1}{2\pi i} \int _{r}^{R}\frac{h_{1,1}(\lambda )}{(3\pi -i\log (t))^{1+a}(\lambda -t)} (t+A_\varepsilon )^{-1} x dt\\{} & {} \quad + \frac{1}{2\pi i} \int _{-\pi }^{\pi }\frac{ih_{1,1}(\lambda )Re^{is}}{(2\pi -s-i\log (R))^{1+a}(\lambda +Re^{is})}R(Re^{is},A_\varepsilon )x ds\\{} & {} \quad - \frac{h_{1,1}(\lambda )}{2\pi i} \int _{-\pi }^{\pi }\frac{ire^{is}}{(2\pi +s-i\log (r))^{1+a}(\lambda +re^{is})}R(re^{is},A_\varepsilon ) x ds\\{} & {} \quad {\mathop {\longrightarrow }\limits ^{r\rightarrow 0,R\rightarrow \infty }}\frac{h_{1,1}(\lambda )}{(2\pi -i\log (-\lambda ))^{1+a}}(\lambda +A_\varepsilon )^{-1}x+\\{} & {} \quad \frac{1}{2\pi i} \int _{0}^{\infty }\frac{h_{1,1}(\lambda )(t+A_\varepsilon )^{-1}}{(\pi -i\log (t))^{1+a}(\lambda -t)}xdt\\{} & {} \quad - \frac{1}{2\pi i} \int _{0}^{\infty }\frac{h_{1,1}(\lambda )(t+A_\varepsilon )^{-1}}{(3\pi -i\log (t))^{1+a}(\lambda -t)}x dt. \end{aligned}$$

Now, it follows from dominated convergence that for each $x\in X$,

$$\begin{aligned}{} & {} h_{1,1}(\lambda )(\lambda +A)^{-1}(2\pi -i\log (A))^{-(1+a)}x = \frac{h_{1,1}(\lambda )(\lambda +A)^{-1}x}{(2\pi -i\log (-\lambda ))^{1+a}}\\{} & {} \quad + \frac{1}{2\pi i} \int _{0}^{\infty }\frac{h_{1,1}(\lambda )}{(\pi -i\log (t))^{1+a}(\lambda -t)}(t+A)^{-1} xdt\\{} & {} \quad - \frac{1}{2\pi i} \int _{0}^{\infty }\frac{h_{1,1}(\lambda )(t+A)^{-1}}{(3\pi -i\log (t))^{1+a}(\lambda -t)}x dt. \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} \Vert h_{1,1}(\lambda )(\lambda +A)^{-1}(2\pi -i\log (A))^{-(1+a)}\Vert _{{\mathcal {L}}(X)}\le \left\| \frac{h_{1,1}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (-\lambda ))^{1+a}}\right\| _{{\mathcal {L}}(X)}\\{} & {} \qquad + \frac{1}{2\pi } \int _{0}^{\infty }\frac{|h_{1,1}(\lambda )|\Vert (t+A)^{-1}\Vert _{{\mathcal {L}}(X)}}{(\pi ^2+\log (t)^2)(|\lambda |+t)} dt+ \frac{1}{2\pi } \int _{0}^{\infty }\frac{|h_{1,1}(\lambda )|\Vert (t+A)^{-1}\Vert _{{\mathcal {L}}(X)}}{(\pi ^2+\log (t)^2)(|\lambda |+t)} dt\\{} & {} \quad = \left\| \frac{h_{1,1}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (-\lambda ))^{1+a}}\right\| _{{\mathcal {L}}(X)}+2\frac{|h_{1,1}(\lambda )|(|\lambda |-1)}{|\lambda |\log (|\lambda |)}. \end{aligned}$$

Now, by the same reasoning as before, one gets

$$\begin{aligned}{} & {} h_{1,2}(\lambda )(\lambda +A)^{-1}(2\pi -i\log (A))^{-(2+a)} = \frac{h_{1,2}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (-\lambda ))^{2+a}}\\{} & {} \quad + \frac{1}{2\pi i} \int _{0}^{\infty }\frac{h_{1,2}(\lambda )}{(\pi -i\log (t))^{2+a}(\lambda -t)}(t+A)^{-1} dt\\{} & {} \quad - \frac{1}{2\pi i} \int _{0}^{\infty }\frac{h_{1,2}(\lambda )(t+A)^{-1}}{(3\pi -i\log (t))^{2+a}(\lambda -t)} dt, \end{aligned}$$

so

$$\begin{aligned}{} & {} \Vert h_{1,2}(\lambda )(\lambda +A)^{-1}L_{\nu ,2+a}(A)\Vert _{{\mathcal {L}}(X)}\\{} & {} \quad \lesssim \left\| \frac{h_{1,2}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (-\lambda ))^{2+a}}\right\| _{{\mathcal {L}}(X)}+ \\{} & {} \qquad + \frac{|h_{1,2}(\lambda )|}{\pi } \int _{0}^{\infty }\frac{(\pi ^2-2(1+1/t)\log (t)+\log (t)^2)}{(\pi ^2+\log (t)^2)^2(|\lambda |+t)}dt\\{} & {} \quad =\left\| \frac{h_{1,2}(\lambda )(\lambda +A)^{-1}}{(2\pi -i\log (-\lambda ))^{2+a}} \right\| _{{\mathcal {L}}(X)}\\{} & {} \qquad +\frac{|h_{1,2}(\lambda )|(|\lambda |\log (|\lambda |)-|\lambda |+1)}{\pi |\lambda |\log (|\lambda |)^2}. \end{aligned}$$

Again, by proceeding as in Case 1(a), one concludes that

$$\begin{aligned} \sup \{\Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,a+\zeta _1+\zeta _2}(A)\Vert _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}\setminus \{0\}, |\lambda |\le 1\}<\infty . \end{aligned}$$

$\bullet$ Case 2: $\alpha \ge 2$. By using the functional calculus for $H_0^\infty$ functions (see Remark 2.13), one gets for each $x\in X$,

$$\begin{aligned}{} & {} h_{1,\zeta }(\lambda )(\lambda +A)^{-1}A^{\alpha -1}(1+A)^{-(\alpha -1)}(2\pi -i\log (A))^{-\tilde{c}} x\\{} & {} \quad = \frac{h_{1,\zeta }(\lambda )}{2\pi i}\int _{\Gamma } \frac{z^{\alpha -1}(\lambda +A)^{-1}}{(1+z)^{\alpha -1}h_{0,\tilde{c}}(z)}R(z,A)xdz\\{} & {} \quad = \frac{h_{1,\zeta }(\lambda )(-\lambda )^{\alpha -1}}{(1-\lambda )^{\alpha -1}(2\pi -i\log (-\lambda ))^{\tilde{c}}}(\lambda +A)^{-1}x\\{} & {} \qquad + h_{1,\zeta }(\lambda )S^{''}_{\lambda }x, \end{aligned}$$

where

$$\begin{aligned} S^{''}_{\lambda }:=\frac{1}{2\pi i}\int _{\Gamma }\frac{z^{\alpha -1}}{(1+z)^{\alpha -1}h_{0,\tilde{c}}(z)(z+\lambda )}R(z,A)dz. \end{aligned}$$

The function $z\mapsto (2\pi -i\log (z))^{-\tilde{c}}R(z,A)$ is integrable on $\Gamma$ and by Lemma 5.9 in [34], for $z\in \Gamma$ and $|\lambda |\le 1$, one has

$$\begin{aligned} \left| \frac{z^{\alpha -1} h_{1,\zeta }(\lambda )}{(1+z)^{\alpha -1}(z+\lambda )}\right| \le \frac{C}{|1-\lambda |}\le C; \end{aligned}$$

hence, $\sup \{\Vert h_{1,\zeta }(\lambda )S^{''}_{\lambda } \Vert _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}{\setminus }\{0\}, |\lambda |\le 1\}<\infty$, and since $\left\| \dfrac{h_{1,\zeta }(\lambda )(-\lambda )^{\alpha -1}(\lambda +A)^{-1}}{(1-\lambda )^ {\alpha -1}(2\pi -i\log (-\lambda ))^{\tilde{c}}}\right\| _{{\mathcal {L}}(X)}$ is also bounded (by hypothesis), then

$$\begin{aligned} \sup \{\Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}A^{\alpha -1}(1+A)^{-(\alpha -1)} (2\pi -i\log (A))^{-\tilde{c}}\Vert _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}\setminus \{0\}, |\lambda |\le 1\}<\infty . \end{aligned}$$

$\bullet$ Case 3: $\alpha \in (1,2)$. By the moment inequality (applied over $\alpha -1\in (0,1)$), one gets

$$\begin{aligned}{} & {} \Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}(A(1+A)^{-1})^{\alpha -1}L_{\nu ,\tilde{c}}(A)\Vert _{{\mathcal {L}}(X)}\\{} & {} \quad \lesssim \Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}L_{\nu ,\tilde{c}}(A)\Vert ^{2-\alpha }_{{\mathcal {L}}(X)} \Vert h_{1,\zeta }(\lambda )(\lambda +A)^{-1}A(1+A)^{-1}L_{\nu ,\tilde{c}}(A)\Vert ^{\alpha -1}_{{\mathcal {L}}(X)}. \end{aligned}$$

The first factor is treated as in Case 1, and the second factor is treated as in Case 2.

Item (b)

$\bullet$ Case 1: $\alpha =1.$ Let $\zeta >1$ and set $\tilde{c}:=\zeta +a>1$.

Given that the operator $(\log (2+A))^{\tilde{c}}(2\pi -i\log (A))^{-\tilde{c}}$ is closed, it follows from the Closed Graph Theorem that it is bounded; hence,

$$\begin{aligned} \Vert (\lambda +A)^{-1}(1+A)^{-1}(2\pi -i\log (A))^{-\tilde{c}}\Vert _{{\mathcal {L}}(X)}\lesssim \Vert (\lambda +A)^{-1}(1+A)^{-1}\log (A+2)^{-\tilde{c}}\Vert _{{\mathcal {L}}(X)}. \end{aligned}$$

Now, by Proposition 3.3, one gets

$$\begin{aligned}{} & {} \sup \left\{ \frac{|\lambda |}{(1+|\lambda |)^{1-\beta _0}} |(2\pi -\log (\lambda ))|^{\zeta }\Vert (\lambda +A)^{-1}(1+A)^{-1}\right. \\{} & {} \quad \left. \log (A+2)^{-\tilde{c}}\Vert _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}, |\lambda |\ge 1\right\} <\infty . \end{aligned}$$

$\bullet$ Case 2: $\alpha \ge 2$. Let $g_{\alpha ,\zeta }(\lambda )=\dfrac{\lambda ^{\alpha }}{(1-\lambda )^{1-\beta _0}}(2\pi -i\log (\lambda ))^\zeta$, with $\lambda \in i{\mathbb {R}}\setminus \{0\}$; then, by the functional calculus for $H_0^\infty$ functions (see Remark 2.13), for each $x\in X$, one has

$$\begin{aligned}{} & {} g_{1,\zeta }(\lambda )(\lambda +A)^{-1}A^{\alpha -1}(1+A)^{-(\alpha +\beta +\beta _0-1)}(2\pi -i\log (A))^{-\tilde{c}}x\\{} & {} \quad =\frac{g_{1,\zeta }(\lambda )}{2\pi i}\int _{\Gamma }\frac{z^{\alpha -1}(\lambda +A)^{-1}}{(1+z)^{\alpha +\beta +\beta _0-1}(2\pi -i\log (z))^{\tilde{c}}}R(z,A)xdz\\{} & {} \quad = \frac{g_{1,\zeta }(\lambda )(-\lambda )^{\alpha -1}}{(1-\lambda )^{\alpha +\beta +\beta _0-1} (2\pi -i\log (-\lambda ))^{\tilde{c}}}(\lambda +A)^{-1}x+ g_{1,\zeta }(\lambda )T^{''}_{\lambda }x, \end{aligned}$$

where

$$\begin{aligned} T^{''}_{\lambda }:=\frac{1}{2\pi i}\int _{\Gamma }\frac{z^{\alpha -1}}{(1+z)^{\alpha +\beta +\beta _0-1}(2\pi -i\log (z))^{\tilde{c}}(z+\lambda )}R(z,A)dz, \end{aligned}$$

with $\Gamma$ the path defined in the proof of Proposition 3.3. The function $z\mapsto (2\pi -i\log (z))^{-\tilde{c}}R(z,A)$ is integrable on $\Gamma$ and by Lemma 5.9 in [34], for $z\in \Gamma$ and $|\lambda |\ge 1$, one has

$$\begin{aligned} \left| \frac{z^{\alpha -1} g_{1,\zeta }(\lambda )}{(1+z)^{\alpha +\beta +\beta _0-1}(z+\lambda )}\right| \lesssim \frac{ |g_{1,\zeta }(\lambda )|}{|1-\lambda |}\le C; \end{aligned}$$

thus, $\sup \{\Vert g_{1,\zeta }(\lambda )T^{''}_{\lambda }\Vert _{{\mathcal {L}}(X)}\mid \lambda \in i{\mathbb {R}},\;|\lambda |\ge 1\}<\infty$, and since

$$\begin{aligned} \sup \left\{ \left\| \frac{g_{1,\zeta }(\lambda )(-\lambda )^{\alpha -1}}{(1-\lambda )^{\alpha +\beta +\beta _0-1} (2\pi -i\log (-\lambda ))^{\tilde{c}}}(\lambda +A)^{-1}\right\| _{{\mathcal {L}}(X)}\mid \lambda \in i{\mathbb {R}}, |\lambda |\ge 1\right\} <\infty , \end{aligned}$$

by hypothesis, it follows that

$$\begin{aligned}{} & {} \sup \left\{ |g_{\alpha ,\zeta }(\lambda )|\Vert (\lambda +A)^{-1}A^{\alpha -1}(1+A)^ {-\beta -\beta _0-\alpha +1}\right. \\{} & {} \quad \left. (2\pi -i\log (A))^{-\tilde{c}}\Vert _{{\mathcal {L}}(X)}\mid \lambda \in i {\mathbb {R}}, |\lambda |\ge 1\right\} <\infty . \end{aligned}$$

$\bullet$ Case 3: $\alpha \in (1,2)$. It follows from the moment inequality (applied to $\alpha -1\in (0,1)$) that

$$\begin{aligned}{} & {} \Vert g_{1,\tilde{c}}(\lambda )(\lambda +A)^{-1}(A(1+A)^{-1})^{\alpha -1}L_{\beta +\beta _0,\tilde{c}}(A)\Vert _{{\mathcal {L}}(X)}\\{} & {} \quad \lesssim \Vert g_{1,\tilde{c}}(\lambda )(\lambda +A)^{-1}L_{\beta +\beta _0,\tilde{c}}(A)\Vert ^{2-\alpha }_{{\mathcal {L}}(X)} \Vert g_{1,\tilde{c}}(\lambda )(\lambda +A)^{-1}A(1+A)^{-1}L_{\beta +\beta _0,\tilde{c}}(A)\Vert ^{\alpha -1}_{{\mathcal {L}}(X)}. \end{aligned}$$

The first factor must be treated as in Case 1, and the second one as in Case 2.

Estimates

Lemma B.1

Let $\mu ,\zeta \ge 0$ and $\nu \ge 1$; then, for each $t\ge 0$,

• 1. $\displaystyle \int _{i\infty }^{-i\infty }e^{-\lambda t} \dfrac{1}{(1+\lambda )^{\nu }(\log (2+\lambda ))^{\zeta }}d\lambda =0$.

• 2. $\displaystyle \int _{i\infty }^{-i\infty }e^{-\lambda t} \dfrac{\lambda ^{\mu }}{(1+\lambda )^{\nu +\mu }(2\pi -i\log (\lambda ))^{\zeta }}d\lambda =0$.

Proof

We just present the proof of the first equality, since the proof of the other one is analogous. Let us first show the following statement.

Claim:

$$\begin{aligned}{} & {} \frac{1}{2\pi i}\int _{i\infty }^{-i\infty }e^{-\lambda t} \frac{1}{(1+\lambda )^{\nu }(\log (2+\lambda ))^{\zeta }}d\lambda \nonumber \\{} & {} \quad = \frac{1}{2\pi i}\int _{\Gamma _\varphi }e^{-\lambda t} \frac{1}{(1+\lambda )^{\nu }(\log (2+\lambda ))^{\zeta }}d\lambda , \end{aligned}$$

(B.6)

where $\Gamma _\varphi =\{re^{i\varphi }\mid r\in [0,\infty )\}\cup \{re^{-i\varphi }\mid r\in [0,\infty )\}$ and $0<\varphi <\frac{\pi }{2}$.

Namely, for $t\ge 0$, set $i{\mathbb {R}}\ni \lambda \mapsto h_t(\lambda ):=e^{-\lambda t} \dfrac{1}{(1+\lambda )^{\nu }(\log (2+\lambda ))^{\zeta }}$, and for each $R,r>0$ and each $\eta \in [\varphi ,\pi /2]$, set $\Gamma ^{+}_{R,\varphi }=\{Re^{i\theta }\mid \theta \in (\varphi ,\frac{\pi }{2})\}$, $\Gamma ^{+}_{r,\varphi }=\{re^{i\theta }\mid \theta \in (\varphi ,\frac{\pi }{2})\}$, $\Gamma ^{-}_{R,\varphi }=\{Re^{-i\theta }\mid \theta \in (\varphi ,\frac{\pi }{2})\}$, $\Gamma ^{-}_{r,\varphi }=\{re^{-i\theta }\mid \theta \in (\varphi ,\frac{\pi }{2})\}$, $\gamma ^{+}_{\eta }=\{se^{i\eta }\mid s\in [r,R]\}$ and $\gamma ^{-}_{\eta }=\{se^{-i\eta }\mid s\in [r,R]\}$. By Cauchy’s Integral Theorem,

$$\begin{aligned} -\int _{\Gamma ^{+}_{R,\varphi }}h_t(\lambda )d\lambda +\int _{\gamma ^{+}_{\frac{\pi }{2}}}h_ t(\lambda )d\lambda +\int _{\Gamma ^{+}_{r,\varphi }}h_t(\lambda )d\lambda -\int _{\gamma ^{+}_{\varphi }}h_t(\lambda )d\lambda =0, \end{aligned}$$

(B.7)

and

$$\begin{aligned} \int _{\Gamma ^{-}_{R,\varphi }}h_t(\lambda )d\lambda -\int _{\gamma ^{-}_{\frac{\pi }{2}}}h_t(\lambda ) d\lambda -\int _{\Gamma ^{-}_{r,\varphi }}h_t(\lambda )d\lambda +\int _{\gamma ^{-}_{\varphi }}h_t(\lambda )d\lambda =0. \end{aligned}$$

(B.8)

Note that, by Lemma 5.2.2 in [16],

$$\begin{aligned} \left| \int _{\Gamma ^{\pm }_{R,\varphi }}h_t(\lambda )d\lambda \right|\le & {} \int _{\varphi }^{\frac{\pi }{2}}\frac{R e^{-t\cos \theta }}{|(1+Re^{\pm i\theta })|^{\nu }|\log (2+Re^{\pm i\theta })|^{\zeta }}d\theta \\\le & {} 2^{\nu /2} \int _{\varphi }^{\frac{\pi }{2}}\frac{Re^{-t\cos \theta }}{(1+R)^{\nu }(1+\cos (\theta ))^{\nu /2}\left( \log (2+R)+\frac{1}{2}\log \left( \frac{1+\cos (\theta )}{2}\right) \right) ^{\zeta }}d\theta \\\lesssim & {} \frac{R^{1-\nu }}{\log (2+R)^{\zeta }} \end{aligned}$$

and

$$\begin{aligned} \left| \int _{\Gamma ^{\pm }_{r,\varphi }}h_t(\lambda )d\lambda \right| \lesssim r \end{aligned}$$

By adding Eqs. (B.7) and (B.8), and by taking the limits $R\rightarrow \infty$, $r\rightarrow 0$, one gets (B.6).

By Claim, it suffices to prove that

$$\begin{aligned} \frac{1}{2\pi i}\int _{\Gamma _\varphi }e^{-\lambda t} \frac{1}{(1+\lambda )^{\nu }\log (2+\lambda )^{\zeta }}d\lambda =0. \end{aligned}$$

It follows from Cauchy’s Integral Theorem that for each $0<r<R$,

$$\begin{aligned} \frac{1}{2\pi i}\int _{\Gamma _\varphi }h_t(\lambda )d\lambda +\frac{1}{2\pi i}\int _{\gamma _{R,\varphi }}h_t(\lambda )d\lambda +\frac{1}{2\pi i}\int _{\gamma _{r,\varphi }}h_t(\lambda )d\lambda =0, \end{aligned}$$

(B.9)

with $\gamma _{R,\varphi }:=\{Re^{i\theta }\mid \theta \in [-\varphi ,\varphi ]\}$ and $\gamma _{r,\varphi }:=\{r e^{-i\theta }\mid \theta \in [-\varphi ,\varphi ]\}$.

Note that for each sufficiently large R,

$$\begin{aligned} \left| \int _{\gamma _{R,\varphi }} e^{-\lambda t} \frac{1}{(1+\lambda )^{\nu }\log (2+\lambda )^{\zeta }}d\lambda \right| \lesssim \frac{R^{1-\nu }}{\log (2+R)^\zeta }, \end{aligned}$$

and for each sufficiently small r,

$$\begin{aligned} \left| \int _{\gamma _{r,\varphi }} e^{-\lambda t} \frac{1}{(1+\lambda )^{\nu }\log (2+\lambda )^{\zeta }}d\lambda \right| \lesssim r. \end{aligned}$$

The result follows by taking the limits $r\rightarrow 0$ and $R\rightarrow \infty$ in relation (B.9). $\square$

Lemma B.2

Let $\varphi \in (0,\frac{\pi }{2}]$ and $\theta \in (\pi -\varphi ,\pi )$. Set $\Omega := \overline{{\mathbb {C}}_{+}}{\setminus } (S_\varphi \cup \{0\})$ and let $\Gamma :=\{re^{i\theta }\mid r \in [0,\infty )\}\cup \{re^{i\theta }\mid r \in [0,\infty )\}$ be oriented from $\infty e^{i\theta }$ to $\infty e^{-i\theta }$. Then, for each $\alpha \in [0,\infty )$, $\beta \in (0,\infty )$, $\eta \in (0, 1]$ and each $\lambda \in \Omega$, one has

• (a) $\displaystyle \int _{\Gamma } \dfrac{1}{(\eta +z)^{\beta }(\log (1+\eta +z))^{\zeta }(z+\lambda +\eta -1)} dz= \dfrac{1}{(1-\lambda )^{\beta }(\log (2-\lambda ))^{\zeta }}$.

• (b) $\displaystyle \int _{\Gamma } \dfrac{z^{\alpha }}{(\eta +z)^{\alpha +\beta }(2\pi -i\log (-1+\eta +z))^{\zeta }(z+\lambda +\eta -1)} dz= \dfrac{(1-\lambda -\eta )^{\alpha }}{(1-\lambda )(2\pi -i\log (-\lambda ))^{\zeta }}$.

Proof

We just present the proof of item a). Let $\lambda \in \Omega$. For each $r \in (0, \text {Im}(\lambda )/2]$ and each $R\ge 2|\lambda | + 2$, set $\gamma _{+}:=\{se^{i\theta }\mid s\in [r,R] \}$, $\gamma _{-}:=\{se^{-i\theta }\mid s\in [r,R]\}$, $\gamma _{r}:=\{re^{i\nu }\mid \nu \in [-\theta ,\theta ]\}$, $\gamma _{R}:=\{Re^{i\nu }\mid \nu \in [-\theta ,\theta ]\}$ and $\gamma _{r,R}:=(-\gamma _{+})\cup \gamma _{-}\cup (-\gamma _{r})\cup \gamma _{R}$. Let $f_{\beta ,\zeta ,\lambda }:\overline{{\mathbb {C}}_{+}}\rightarrow {\mathbb {C}}$ be given by the law $f_{\beta ,\zeta ,\lambda }(z)=\dfrac{1}{(\eta +z)^{\beta }(\log (1+\eta +z))^{\zeta }(z+\lambda +\eta -1)}$; then,

$$\begin{aligned} \left| \int _{\gamma _R} f_{\beta ,\zeta ,\lambda }(z) dz\right|\le & {} \int _{-\theta }^{\theta } \frac{R}{|\eta +Re^{i\nu }|^{\beta }\log (|1+\eta +Re^{i\nu }|)^{\zeta }|Re^{i\nu }+\lambda +\eta -1|} d\nu \\\lesssim & {} \frac{R^{-\beta }}{\log (1+R)^{\zeta }}, \end{aligned}$$

which goes to zero as $R\rightarrow \infty$. Similarly, one can show that

$$\begin{aligned} \lim _{r\rightarrow 0} \left| \int _{\gamma _r} f_{\beta ,\zeta ,\lambda }(z) dz\right| =0. \end{aligned}$$

On the other hand, by the Residue Theorem, one has

$$\begin{aligned} \int _{\gamma _{r,R}} \frac{1}{(\eta +z)^{\beta }(\log (1+\eta +z))^{\zeta }(z+\lambda +\eta -1)} dz =\frac{1}{(1-\lambda )^{\beta }\log (2-\lambda )^{\zeta }}. \end{aligned}$$

Thus, it follows that

$$\begin{aligned}{} & {} \int _{\Gamma } \frac{1}{(\eta +z)^{\beta }(\log (1+\eta +z))^{\zeta }(z+\lambda +\eta -1)} dz\\{} & {} \quad =\lim _{r\rightarrow 0, R\rightarrow \infty } \int _{\gamma _{r,R}} \frac{1}{(\eta +z)^{\beta }(\log (1+\eta +z))^{\zeta }(z+\lambda +\eta -1)} dz=\\{} & {} \quad \frac{1}{(1-\lambda )^{\beta }\log (2-\lambda )^{\zeta }}. \end{aligned}$$

$\square$