# An integral arising in statistics

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

-
It would help attract people's attention if you could give your questions slightly more descriptive or specific titles, e.g. "An integral arising from a question in probability/statistics/wizardry". To say "difficult integral" is subjective and not very informative, in my opinion –  Yemon Choi Feb 1 '10 at 19:43

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.