Show that $e^{ta}=\frac1{2\pi i}\int_{\gamma_\alpha}\frac{e^{\lambda t}}{\lambda-a}\,d\lambda$ This is the exercise 18 on page 359 of Analysis II of Amann and Escher. I'm stuck in this exercise

Suppose $a\in\Bbb C$ and $\alpha\neq\Re(a)$. Show that for $\gamma_\alpha:\Bbb R\to\Bbb C,\,s\mapsto \alpha+is$ we have $$e^{ta}=\frac1{2\pi i}\int_{\gamma_\alpha}\frac{e^{\lambda t}}{\lambda-a}\,d\lambda\quad\text{for }t>0\tag1$$ (HINT: the Cauchy integral formula gives $$e^{ta}=\frac1{2\pi i}\int_{\partial\Bbb D(a,r)}\frac{e^{\lambda t}}{\lambda-a}\,d\lambda\quad\text{for }t\in\Bbb R\text{ and }r>0\tag2$$ Now apply the Cauchy integral theorem.)

Trying to follow the hint I tried to create a family of closed paths $\Gamma_r=[\gamma_r]+[\delta_r]$, such that $a$ doesn't belong to it bounded regions, and then use the Cauchy integral theorem, that is
$$\int_{\Gamma_r}g(\lambda)\, d\lambda=0\quad r>0$$
for $g(\lambda):=\frac{e^{\lambda t}}{\lambda-a}$, and exploit some kind of symmetry to relate the integration on the paths $[\gamma_r]$ or $[\delta_r]$ to the Cauchy integral formula. By example, without lose of generality suppose that $\alpha>\Re(a)$, then I could define
$$\gamma_r:[-r,r]\to\Bbb C,\quad s\mapsto \alpha+is\\\delta_r:[r,r+\pi]\to\Bbb C,\quad s\mapsto re^{-i(s-r-\pi/2)}\tag3$$
and try to relate the integration on the half circle defined by $\delta_r$ with the integral in the complete circle, where for suitable enough big $r$ I can use the Cauchy integral formula. However this is not easy to deal with, because it don't show a symmetry to exploit.
Maybe I'm over-complicating and the exercise want to do something different. Can someone help me?

EDIT:
It can be shown that $(1)$, as an improper integral of Riemann, converges conditionally, and after some changes of variables the question reduces to show that
$$\frac1{2\pi i}\int_{\gamma_r}\frac{e^\zeta}{\zeta}\,d\zeta=1$$
where $\gamma_r:\Bbb R\to\Bbb C,\, t\mapsto r+it$
for any chosen $r\in\Bbb R\setminus\{0\}$. Graphing the integrand we can see that it describes two non-rectifiable spirals (symmetric respect to the real axis) that converges to some point on the real axis.
 A: I'm also confused by this problem.
But I suspect that the problem is wrong. Please see the following.
$$
\begin{aligned}
e^{ta}&=\frac{1}{2\pi i}\int_{\gamma_\alpha} e^{\lambda t}(\lambda -a)^{-1}  d \lambda\\
\iff 1&=\frac{1}{2\pi i}\int_{\gamma_\alpha} \frac{e^{(\lambda-a) t}}{(\lambda -a)} d\lambda \qquad &&\eta=t(\lambda-a)\\
\iff 1&=\frac{1}{2\pi i}\int_{\gamma_\beta} \frac{e^{\eta}}{\eta} d\eta \qquad &&\gamma_\beta=\phi(\gamma_\alpha), \qquad \phi:x\mapsto t(x-a).
\end{aligned}$$
Here note that the line $\gamma_\alpha=\alpha+ i \mathbb R$ and $\gamma_\beta=t(\alpha-a)+ i \mathbb R$.
If $\gamma_\beta$ is right to $0$, then we only need to consider the left half circle $$\gamma_1: [0,\pi]\to \mathbb C,\quad t\mapsto \beta +i Re^{it}.$$
Here $\beta$ is the center of the circle on the line $\gamma_\beta$.
$$\begin{aligned}
&\int_{\gamma_1} \frac{e^{\eta}}{\eta} d\eta\\
=&\int_0^\pi\frac{e^{\beta+i Re^{it}}\cdot (-R e^{it})}{\beta+i Re^{it}}dt \to 0,\quad R\to \infty.
\end{aligned}$$
Just as @Masacroso answered.
This part is checked by computer.

But the rest is also checked by computer to be wrong.
If $\gamma$ is left to $0$, that is $Re( \alpha -a)<0$, then we need to consider the right half circle
$$\gamma_2: [0,\pi]\to \mathbb C,\quad t\mapsto \beta -i Re^{it}.$$
Here
$$\begin{aligned}
&\int_{\gamma_1} \frac{e^{\eta}}{\eta} d\eta\\
=&\int_0^\pi\frac{e^{\beta-i Re^{it}}\cdot (R e^{it})}{\beta-i Re^{it}}dt\to 2\pi i,\quad R\to \infty
\end{aligned}$$
It is also checked by computer directly that $e^{ta}\neq \frac{1}{2\pi i}\int_{\gamma_\alpha} e^{\lambda t}(\lambda -a)^{-1}  d \lambda$ when $Re(\alpha -a)<0$.
