calculate the integral Compute $$I=\int_C\frac{e^{zt}}{1+z^2}dz$$ where $t>0$, a real number, and $C$ is the line $\{z\mid \mathrm{Re}(z)=1\}$ with direction of increasing imaginary part.

I tried to integral along the boundary of $\{z| -1<\mathrm{Re}(z)<1, -R<\mathrm{Im}(z)<R\}$ with $R\to\infty$. The integral along the top and bottom line goes to zero, but computation becomes ugly when calculate along the left line.
 A: Consider the rectangular contour $$-R-iT \to 1-iT \to 1 + iT \to -R+iT \to -R-iT$$
By Cauchy's theorem, the integral is
$$2 \pi i\left( \dfrac{e^{it}}{2i} + \dfrac{e^{-it}}{-2i}\right) = 2i \pi \sin(t)$$
Now let $T \to \infty$. We will then get that
$$\int_{C_1} \dfrac{e^{tz}}{z^2+1}dz - \int_{C_{-R}} \dfrac{e^{tz}}{z^2+1}dz = 2i \pi \sin(t)$$
where $C_1$ is the line $\text{Re}(z) = 1$ and $C_{-R}$ is the line $\text{Re}(z) = -R$.
Now let $R \to \infty$, then $e^{-tz} \to 0$. Hence
$$\lim_{R \to \infty} \int_{C_{-R}} \dfrac{e^{tz}}{z^2+1}dz = 0$$
Hence,$$\int_{C_1} \dfrac{e^{tz}}{z^2+1}dz =  2i \pi \sin(t)$$
A: For $R>\sqrt{2}$ denote by $D_R$ the bounded domain of $\mathbb{C}$ with boundary 
$$
[1-iR,1+iR]\cup\gamma_R([0,\pi]),
$$
where 
$$
\gamma_R(\theta)=1+iRe^{i\theta} \quad \forall \theta \in [0,\pi].
$$
Since the function $f_t: z \mapsto \frac{e^{tz}}{1+z^2}$ is holomorphic on $D_R\setminus\{\pm i\}$, it follows from the Residue Theorem that
$$
\int_{\partial D_R}f_t(z)\,dz=2\pi i(\text{Res}(f_t,-i)+\text{Res}(f_t,i))=2\pi i\Big(-\frac{e^{-it}}{2i}+\frac{e^{it}}{2i}\Big)=2\pi i\sin t.
$$
We have
$$
\Big|\int_{\gamma_R}\frac{e^{tz}}{1+z^2}\,dz\Big|=e^t\Big|\int_0^\pi\frac{-Re^{i(\theta+tR\cos\theta)}e^{-tR\sin\theta}}{1+(1+iRe^{i\theta})^2}\,d\theta\Big|\le Re^t\int_0^\pi\frac{d\theta}{(R-\sqrt{2})^2}
=\frac{\pi Re^t}{(R-\sqrt{2})^2},
$$
and therefore
$$
\lim_{R\to \infty}\int_{\gamma_R}f_t(z)\,dz=0.
$$
Hence
$$
\int_Cf_t(z)\,dz=\lim_{R \to \infty}\int_{\partial D_R}f_t(z)\,dz=2\pi i\sin t.
$$
