An integral arising in statistics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:06:34Z http://mathoverflow.net/feeds/question/13665 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13665/an-integral-arising-in-statistics An integral arising in statistics vilvarin 2010-02-01T12:10:40Z 2010-02-05T13:31:47Z <p>The integral I need: $$t(x)=\int_{-K}^{K}\frac{\exp(ixy)}{1+y^{2q}}dy$$</p> <p>$K&lt;\infty$, q natural number</p> <p>For q=1 this integral is $$\pi/2-\int_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy$$ Where Arc has radius $K$</p> <p>Upper bound is $$K\pi/(K^2-1)^2$$</p> <p>Can I obtain a better expression for the integral?</p> <p>One more question about this integral. For K&lt;1 this integral is just $$-\int_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy?$$</p> http://mathoverflow.net/questions/13665/an-integral-arising-in-statistics/13673#13673 Answer by Igor Khavkine for An integral arising in statistics Igor Khavkine 2010-02-01T13:43:52Z 2010-02-01T13:43:52Z <p>Are you asking if this integral can be expressed in terms of elementary functions?</p> <p>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 <a href="http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/07/01/01/" rel="nofollow">exponential integral</a>.</p> <p>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.</p>