I am trying to calculate the integral $${\int _{_{_{\left|z\right|=\frac{1}{2}}}}}\frac{e^{zt}}{z^2+1}dz$$ for all $t \in \mathbb{R}$
According to answers, it is zero because it is a closed curve.
I tried to find the solution by brute force and got $2\pi i sin(t)$
My way of solution:
$${\int _{_{_{\left|z\right|=\frac{1}{2}}}}}\frac{e^{zt}}{z^2+1}dz={\int _{_{_{\left|z\right|=\frac{1}{2}}}}}\frac{-e^{zt}}{\left(zi-1\right)\left(zi+1\right)}dz=\frac{1}{2}{\int _{_{_{\left|z\right|=\frac{1}{2}}}}}\frac{e^{zt}}{\left(zi+1\right)}dz-\frac{1}{2}{\int _{_{_{\left|z\right|=\frac{1}{2}}}}}\frac{e^{zt}}{\left(zi-1\right)}dz$$
which in fact is equal to:
$$\frac{\pi }{2i\pi }{\int _{_{_{\left|z\right|=\frac{1}{2}}}}}\frac{e^{zt}}{\left(z-i\right)}dz-\frac{\pi }{2\pi i}{\int _{_{_{\left|z\right|=\frac{1}{2}}}}}\frac{e^{zt}}{\left(z+i\right)}dz$$
Then, by using cauchy integral formula ( since $\gamma$ is a closed curve, derivative and $f(z)=e^{zt}$ is complete in complex ) I will receive: $$\pi f\left(i\right)-\pi f\left(-i\right)\:=\:\pi e^{it}-\pi e^{-it}=\pi \left(e^{it}-e^{-it}\right)=2\pi i\sin\left(t\right)\ne 0$$
Where is my mistake?
Actually, $$ \int_{|z|=1/2}\frac{e^{zt}}{z\pm i}\,dz=0 $$ since $\pm i$ is lies outside the disk centered at $0$ with radius $\frac12$.
Note that it is not true that the integral of an analytic function along a closed curve is always $0$. This is true however when the domain of the function contains the region of $\mathbb{C}$ bounded by the curve.