# Fractional pole residue calculation

According to m.l.boas, one can solve integrations involving a simple pole using residue theorem and using those principles, I can solve integrals like

$$\int_{-\infty}^\infty \frac{\sin{x}}x dx$$

Converting to $$\frac{e^{iz}}z$$ and calculating residue at origin .

But what to do when we encounter something like a fractional pole .

If I am to do

$$\int_{-\infty}^{\infty} \frac{\sin{x^q}}{x^q}dx$$

Then we get a laurent series at origin with fractional power by converting to a complex integral in a similar way . How to proceed with such integrals ?

Apart from way of making use of residue theorem, I am looking for a way of solving this integral, too .

q may be a fraction

• there is no such a thing as a fractional pole - one of the fundamental properties of holomorphic and meromorphic functions is that their special set (zeroes/poles) is discrete with all points of integral order (positive for zeroes, negative for poles); truly fractional powers (e.g square root) have non-discrete singularities in the plane and are not holomorphic/meromorphic on punctured discs but only on cut discs by say a ray (or analytic arc) through the origin – Conrad May 29 at 18:47

Converting to $$\dfrac {\sin z^q}{z^q}$$ and using the Taylor series for sin, we get: $$\dfrac1{z^q}\sum_{n=0}^\infty (-1)^n\dfrac{(z^q)^{2n+1}}{n!}=\sum_{n=0}^\infty (-1)^n\dfrac{z^{2nq}}{n!}$$. Thus the function is analytic.
• Indeed. @OP - note this is a straightforward generalization of $\lim_{z \to 0} \frac{\sin z}{z} = 1$ by a simple change of variables $z^q \to z$ (since both go to zero simultaneously and the integrand is clearly analytic elsewhere). We have a removable singularity at the origin. – Brevan Ellefsen May 29 at 17:27
• Shouldn't you have converted it into $\dfrac{ \sin z}z$? – Chris Custer May 30 at 3:54