Interchange limit and (definite) integral I'm trying to calculate the following limit

\begin{equation}\label{eq}\large\lim_{R\to\infty}\,i\,\int_{-\pi/2}^{\pi/2}\frac{e^{-\alpha(R+i\,w)}e^{t\,e^{R+i\,w}}}{(R+i\,w)^{\beta}}dw\end{equation}

with $\alpha\geq0$, $\beta, t>0$ .
The problem is that (I think, I'm not sure...) I can't change limit by integral.
Then, I have 2 doubs:


*

*Can I change limit by integral?

*If the answer is not, another way to manipulate this limit is welcomed.
Update 1:
Using the generating function of Bell polynomials of first kind $B_n(t)$
$$\large e^{t(e^u-1)}=\sum_{n=0}^\infty\frac{B_n(t)}{n!}u^n$$
reemplacing $u \rightarrow R+i\,w$ we have
$$\lim_{R\to\infty}\,i\,\sum_{n=0}^\infty\frac{e^t\,B_n(t)}{n!}\int_{-\pi/2}^{\pi/2}e^{-\alpha(R+i\,w)}\,(R+i\,w)^{n-\beta}dw=$$

$$\large\lim_{R\to\infty}\sum_{n=0}^\infty\frac{e^t\,B_n(t)}{n!\,\alpha^{n-b-1}}\left[\,\Gamma(n-\beta-1,\alpha(R-i\,\pi/2))-\Gamma(n-\beta-1,\alpha(R+i\,\pi/2))\,\right]$$

Update 2:
Changing the variable $R+i\,w\rightarrow u$ we have

$$\large\lim_{R\to\infty}\,i\,\int_{-\pi/2}^{\pi/2}\frac{e^{-\alpha(R+i\,w)}e^{t\,e^{R+i\,w}}}{(R+i\,w)^{\beta}}dw=\lim_{R\to\infty}\,\int_{R-i\,\pi/2}^{R+i\,\pi/2}\frac{e^{-\alpha\,u}e^{t\,e^u}}{u^{\beta}}du$$ 

and maybe we can apply complex integration (Cauchy theorem,...) in the last one.
 A: 
Take $z = e^{R + i w}$. The integrand becomes $f(z) = e^{t z} z^{-\alpha - 1} \ln^{-\beta} z$, and
$$I(R) = \int_{\gamma_1}  f(z) \, dz = \int_{\gamma_2}  f(z) \, dz.$$
The integrals over the arcs of the left semicircle tend to zero, therefore
$$I = \lim_{R \to \infty} I(R) = \int_{\gamma(1)} f(z) \, dz.$$
If $\beta = 1$, then
$$I = 2 \pi i e^t + \int_{\gamma(0)} f(z) \, dz.$$
Alternatively, since $I$ can be converted into the Bromwich integral,
$$I = 2 \pi i \mathcal L^{-1}[z \mapsto z^{-\alpha - 1} \ln^{-\beta} z](t), \\
I \bigg\rvert_{(\alpha, \beta, t) = (0, 1, 1)} =
 2 \pi i \int_0^1 \int_0^\infty
 \frac {\tau^{u - 1}} {\Gamma(u)} \, du d\tau =
2 \pi i \int_0^\infty \frac {du} {\Gamma(u + 1)}.$$
A: We have $-\frac{\pi}{2}<w<\frac{\pi}{2}$ and
$$
\left|\frac{e^{-\alpha(R+iw)}\exp\left(te^{R+iw}\right)}{(R+iw)^\beta}\right|=\left|\frac{e^{-\alpha R}e^{-i\alpha w}\exp\left(te^R(\cos w+i\sin w)\right)}{(R+iw)^\beta}\right|=
$$
$$
=e^{-aR}\exp\left(te^{R}|\cos w|\right)\left|\exp\left(ite^R\sin w\right)\right|\frac{1}{(\sqrt{R^2+w^2})^{\beta}}=
$$
$$
=\frac{e^{-aR}\exp\left(te^{R}|\cos w|\right)}{(\sqrt{R^2+w^2})^{\beta}}=\frac{\exp\left(|\cos w|te^{R}-aR\right)}{(R^2+w^2)^{\beta/2}}>>\exp(t|\cos w|e^{R})\textrm{, }R\rightarrow\infty\tag 1
$$
Then also
$$
i\int^{\pi/2}_{-\pi/2}f(R,w)dw=i\int^{\pi/2}_{-\pi/2}|f(R,w)|e^{i\theta(R,w)}dw=
$$
$$
=-\int^{\pi/2}_{-\pi/2}|f(R,w)|\sin(\theta(R,w))dw+i\int^{\pi/2}_{-\pi/2}|f(R,w)|\cos(\theta(R,w))dw.
$$
Assume now we can interchange the limit and integral.
Since $\sin(\theta)\geq -1$, for all $\theta\in \textbf{R}$, we get
$$
Re\left(\lim_{R\rightarrow\infty}i\int^{\pi/2}_{-\pi/2}f(R,w)dw\right)=Re\left(\int^{\pi/2}_{-\pi/2}\lim_{R\rightarrow\infty}f(R,w)dw\right)\geq
$$
$$
\geq\int^{\pi/2}_{-\pi/2}\lim_{R\rightarrow\infty}|f(R,w)|dw=\infty\textrm{, from relation }(1).
$$
Hence we can not interchange limit and integral.
