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?



An integral arising  in statistics(2)
For $K=1$ your arc passes through the pole of the function $\frac{exp(ixy)}{1+y^2} = \frac{exp(ixy)}{2i}\left(\frac{1}{y-i}-\frac{1}{y+i}\right)$, so you don't get a sensible value (the discontinuity of the integrand is asymptotically $\frac{c}{t}$ for $t$ around $0$).


