I'm trying to compute this integral as part of a larger problem I'm working on. I'm trying to solve the integral $\int_0^\infty \frac{\sin(x)}{x}dx$ and to do it I'm using the method where you integrate the function $f(z) = \frac{e^{iz}}{z}$ around the indented semicircle contour.
The part that I'm working on right now is that I need to do the integral for a small half circle of radius $\epsilon$.
My path is $z(t) = \epsilon e^{it}$ for $t \in [-\pi,0]$. Thus $z'(t) = i\epsilon e^{it}$
Then the integral along the contour is
$$\int_{-\pi}^0 \frac{e^{i\epsilon e^{it}}}{\epsilon e^{it}}i\epsilon e^{it}dt = i \int_{-\pi}^0 e^{i \epsilon e^{it}}dt$$
I know the answer I'm supposed to get for this piece is $-\pi i$, and it seems like all I have to do is take the limit as $\epsilon \to 0$, but I'm not sure if I can do the following
$$\lim_{\epsilon \to 0} i \int_{-\pi}^0 e^{i \epsilon e^{it}}dt = i \int_{-\pi}^0 \lim_{\epsilon \to 0} e^{i \epsilon e^{it}}dt$$