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

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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
up vote 1 down vote accepted

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.

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.

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I am sorry, but I don't understand how the limits of integration will change. Will I have an exponential integral over arc? – vilvarin Feb 5 '10 at 15:03
If the integrand is analytic, the integral over any curve is the difference of the antiderivative evaluated at the end points. If the integrand has poles, then integration curves not deformable into each other without hitting one of the poles will require you to use different branches of the antiderivative. The exponential integrals give you the antiderivative. The integration curve Arc tells you which branch to use, in case there is any ambiguity. – Igor Khavkine Feb 6 '10 at 23:33

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