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

$K<\infty$, q natural number

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

Upper bound is $$K\pi/(K^2-1)^2$$

Can I obtain a better expression for the integral?

One more question about this integral. For K<1 this integral is just $$-\int_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy?$$