Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The integral I am interested in is:

$K<\infty$, 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?

share|cite|improve this question

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 $0$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.