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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?

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1 Answer 1

up vote 1 down vote accepted

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 $0$).

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