most recent 30 from http://mathoverflow.net 2013-05-21T03:21:49Z http://mathoverflow.net/feeds/question/14264 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14264/an-integral-arising-in-statistics2 vilvarin 2010-02-05T13:41:42Z 2010-02-05T13:47:27Z

<p>The integral I am interested in is:<br /> $$t(x)=\int_{-K}^{K}\frac{\exp(ixy)}{1+y^{2q}}dy$$</p> <p>$K&lt;\infty$, q natural number</p> <p>For q=1 one can use contour integration. So for K>1 we have :</p> <p>$$\pi/2-\int_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy $$ Where Arc has radius $K$</p> <p>Is it correct that for K&lt;1 this integral is: $$-\int_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy ?$$</p> <p>What about K=1?</p>

http://mathoverflow.net/questions/14264/an-integral-arising-in-statistics2/14265#14265 Thorny 2010-02-05T13:47:27Z 2010-02-05T13:47:27Z

<p>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 $0$).</p>