&= v\int_{-R^{-1}}^{R^{-1}} \frac{u p(\frac{1}{u})}{u-v}\frac{1}{u^2}du \\  
&= v\int_{C_R} \frac{w p(\frac{1}{w})}{w-v}\frac{1}{w^2}dw,    
\end{split}$$ where $C_R$ is the semicircle $C_R =\{R^{-1}e^{i \theta}; - \pi \leq \theta \leq 0 \}$ with the counterclockwise direction. The last expression is also valid for $|v| < R^{-1}$ as well as for $v \in \operatorname*{\mathbb{C}_{+}}$. The Taylor expansion of it becomes $$\sum_{n=0}^\infty v^{n+1}\int_{C_R}\frac{p(\frac{1}{w})}{w^{n+2}}dw.$$

The tensor convolution (the usual convolution) is characterized by the multiplication of Fourier transforms. Therefore, we in turn consider analytic continuation of Fourier transforms.

\[prop2\] Let $\lambda$ be a positive measure on $(-\infty,-R] \cup [R, \infty)$ which satisfies (A2). Then $\mathcal{F}_\lambda$ defined for $z > 0$ is the restriction of an entire function, which we denote by $\widetilde{\mathcal{F}}_\lambda$. Therefore the Taylor series $$\widetilde{\mathcal{F}}_\lambda(z) = \sum_{n=0}^\infty \frac{c_n}{n!}(iz)^n$$ is convergent for all $z \in \operatorname*{\mathbb{C}}$.

We have for $z > 0$ $$\label{eq12}
\begin{split}
\mathcal{F}_\lambda (z) &= \int_{|x|>R} e^{ixz}p(x)dx \\
   &= \int_{-R^{-1}}^{R^{-1}}e^{i\frac{z}{u}}
p\Big(\frac{1}{u}\Big)\frac{1}{u^2}du.  
\end{split}$$ Since ${\text{\normalfont Re~}}\!\big( i\frac{z}{u}\big) \leq 0$ for ${\text{\normalfont Im~}}u \leq 0$, $u \neq 0$, we have $$\lim_{\varepsilon \searrow 0} \int_{C_{\varepsilon^{-1}}}e^{i\frac{z}{u}}
p\Big(\frac{1}{u}\Big)\frac{1}{u^2}du = 0.$$ Then the path in the integral (\[eq12\]) can be changed as $$\begin{split}
\int_{-R^{-1}}^{R^{-1}}e^{i\frac{z}{u}}
p\Big(\frac{1}{u}\Big)\frac{1}{u^2}du &= \lim_{\varepsilon \searrow 0} \Big( \int_{-R^{-1}}^{-\varepsilon} + \int^{R^{-1}}_{\varepsilon}\Big) \Big(e^{i\frac{z}{u}}
p\Big(\frac{1}{u}\Big)\frac{1}{u^2}du\Big)\\ 
&= \lim_{\varepsilon \searrow 0} \int_{\Gamma_{R, \varepsilon}}e^{i\frac{z}{u}}
p\Big(\frac{1}{u}\Big)\frac{1}{u^2}du \\
  