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
$$-\int\_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy?$$



An integral arising  in statistics
Are you asking if this integral can be expressed in terms of elementary functions?


Most likely no. The reason is that there's a fairly straight forward way of expressing it using exponential integrals, which are not elementary functions. To do that, expand the rational part $1/(1+y^{2q})$ in partial fractions. Each term should have a simple pole. Shift the pole to zero and use the definition of the [exponential integral](http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/07/01/01/).


Or are you interested in some asymptotic expression for the integral in the limit of large/small K or x? The answer would then depend on the limits you are interested in.


