Cauchy Principal Value for Poles stronger than 1 While calculating the Cauchy Principal Value of $\int_{-\infty}^{+\infty}\frac{\sin x}{x}dx$ we take an indented semicircle (in the upper half plane) of radius $\delta$ centered at the simple pole of $\frac{e^{iz}}{z}$ at $z=0$. Then we Laurent expand and by integrating and taking $\delta\rightarrow0$ we obtain over the small semicircle (in $\pi\rightarrow0$ clockwise sense): $$\int\frac{e^{iz}}{z}dz=-i\pi\qquad\text{Residue at}\,z=0=-i\pi$$ (As done in Arfken $7^{th}$ edition,  $11.75$).
But this method fails when we have poles stronger than of order $1$, e.g., $\frac{e^{iz}}{z^{2}}$. How do I approach such cases?
 A: $\oint z^n dz =0$ except for $n=-1$. So expand $e^{iz}$ as a series
$$\oint \frac{e^{iz}}{z^2} dz =\oint \frac{1+iz-z^2+\dots}{z^2} dz =i\oint \frac{1}{z} dz=-2\pi$$
More generally, if $f(z)$ is regular at $z=0$
$$\oint \frac{f(z)}{z^2} dz =2\pi i f'(0)$$

As pointed out in the comments, the question is actually about a semicircle. Let me show this is still a (partially) valid way to calculate it in that case for the function $f(z)=e^{iz}$
$$\oint \frac{e^{iz}}{z^2} dz =\int_+ \frac{e^{iz}}{z^2} dz + \int_- \frac{e^{iz}}{z^2} dz$$
The plus and minus refer to upper and lower semicircles. We can bring the lower semicircle integration to an upper semicircle integration by changing variables from $z\rightarrow -z$
$$\oint \frac{e^{iz}}{z^2} dz =\int_+ \frac{e^{iz}-e^{-iz}}{z^2} dz =\int_+ \frac{2i \sin z}{z^2} dz$$
So it is true that
$$\oint \frac{e^{iz}}{z^2} dz =\oint \frac{i\sin z}{z^2} dz=2\int_+ \frac{i \sin z}{z^2} dz$$
In principle though there is a part coming from $\int_+ \frac{ \cos z}{z^2} dz$ which cancels on the full closed integral (since it comes with opposite sign). This part is logarithmically divergent as the radius of semicircle integration goes to zero, so without specifying the regularization of the integral further you can't proceed.
