complex analysis - Show that $e^{ta}=frac1{2pi i}int_{gamma_alpha}frac{e^{lambda t}}{lambda-a},dlambda$

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?

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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.

Answer

I will use a strategy that it is used in the last chapter of the book.

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The parametrization $\gamma_\alpha:\Bbb R\to\Bbb C,\, t\mapsto \alpha+it$ defines a vertical ray at $\Re(\alpha)\in\Bbb R$. Without lose of generality we can assume that $\alpha=a+\ell$ for some $\ell\in\Bbb R\setminus\{0\}$. Then

$$f(h):=\frac1{2\pi i}\int_{\gamma_\alpha}\frac{e^{\lambda h}}{\lambda-a}\, d\lambda=\frac1{2\pi}\int_{-\infty}^\infty\frac{e^{h(a+\ell+it)}}{\ell+it}\, dt=e^{ha}\cdot\frac1{2\pi}\int_{-\infty}^\infty\frac{e^{h(\ell+it)}}{\ell+it}\, dt\\ =e^{ha}\cdot\frac{e^{h\ell}}{2\pi}\int_{-\infty}^\infty\frac{e^{hit}}{\ell+it}\, dt=e^{ha}\cdot\frac{e^{h\ell}}{2\pi}\int_{-\infty}^\infty\frac{e^{is}}{h\ell+is}\, ds=e^{ha}\cdot\frac1{2\pi i}\int_{\gamma_{h\ell}}\frac{e^\zeta}{\zeta}\,d\zeta\tag1$$
after the changes of variable $ht=s$ and $h\ell+is=\zeta$. Then setting $g(h):=e^{-ha}f(h)$ the question reduces to show that $g(h)=1$ when $h>0$. Also from the expansion of $\frac{e^{h(\ell+it)}}{\ell+it}$ it can be seen that the improper integral of Riemann $\int_{-\infty}^\infty\frac{e^{h(\ell+it)}}{\ell+it}\, dt$ converges conditionally.

We set $r:=h\ell$ and the paths defined by

$$\alpha:[0,\pi]\to\Bbb C,\quad t\mapsto r-iRe^{it}\\\beta:[0,\pi]\to\Bbb C,\quad t\mapsto r+iRe^{it}\\ \gamma:[-R,R]\to\Bbb C,\quad t\mapsto r+it\tag2$$
Then from the Cauchy integral theorem we knows that $\int_\gamma\frac{e^\zeta}{\zeta}d\zeta=\int_\alpha\frac{e^\zeta}{\zeta}d\zeta$, and for $R>r$ from the Cauchy integral formula that

$$\int_{\alpha+\beta}\frac{e^\zeta}{\zeta}d\zeta=2\pi i\implies\frac1{2\pi i}\int_\gamma\frac{e^\zeta}{\zeta}d\zeta=1-\frac1{2\pi i}\int_\beta\frac{e^\zeta}{\zeta}d\zeta\tag3$$
Thus it is enough to show that $\lim_{R\to\infty}\int_\beta\frac{e^\zeta}{\zeta}d\zeta=0$. Now observe that
$$\begin{align}\left|\int_\beta\frac{e^\zeta}{\zeta}d\zeta\right|&=\left|\int_0^\pi\frac{-Re^{it}e^{r+iRe^{it}}}{r+iRe^{it}}dt\right|\\&\le\left|\int_0^\epsilon\frac{-Re^{it}e^{r+iRe^{it}}}{r+iRe^{it}}dt\right|+\left|\int_{\pi-\epsilon}^\pi\frac{-Re^{it}e^{r+iRe^{it}}}{r+iRe^{it}}dt\right|+\left|\int_\epsilon^{\pi-\epsilon}\frac{-Re^{it}e^{r+iRe^{it}}}{r+iRe^{it}}dt\right|\\ &\le 2\epsilon\, e^r\max_{t\in[0,\epsilon]}\frac{R e^{-R\sin t}}{|r+iRe^{it}|}+(\pi-2\epsilon) e^r\max_{t\in[\epsilon,\pi-\epsilon]}\frac{R e^{-R\sin t}}{|r+iRe^{it}|}\end{align}\tag4$$
Also observe that $\sin t>0$ for $t\in[\epsilon,\pi-\epsilon]$ for any chosen $\epsilon\in(0,\pi/2)$. Also is easy to check that
$$\lim_{R\to\infty}\frac{R}{|r+iRe^{it}|}=\lim_{R\to\infty}\frac1{\sqrt{(r/R)^2+1-2(r/R)\sin t}}=1\tag5$$
Then putting all together we found that
$$\lim_{R\to\infty}\left|\int_\beta\frac{e^\zeta}{\zeta}d\zeta\right|\le 2\epsilon\, e^r,\quad\forall \epsilon\in(0,\pi/2)\implies\lim_{R\to\infty}\int_\beta\frac{e^\zeta}{\zeta}d\zeta=0\tag6$$
as desired.$\Box$