The integral I am interested in is:

$$t(x)=\int_{-K}^{K}\frac{\exp(ixy)}{1+y^{2q}}dy$$

$K<\infty$, q natural number

For q=1 one can use contour integration. So for K>1 we have :

$$\pi/2-\int_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy $$ Where Arc has radius $K$

Is it correct that for K<1 this integral is: $$-\int_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy ?$$

What about K=1?