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

  • 2
    $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Brevan Ellefsen May 29 at 17:27
  • $\begingroup$ This is fine, but how can i proceed with the integral in complex plane ? That wont be possible in the way it was done for sin(x)/x, i mean, converting it to e^(ix)/x $\endgroup$ – user157588 May 30 at 3:42
  • $\begingroup$ Shouldn't you have converted it into $\dfrac{ \sin z}z$? $\endgroup$ – Chris Custer May 30 at 3:54
  • $\begingroup$ Then also it will not have a residue, the dx term, in terms of new variable t=x^q, shall have fractional powers and hence residue theorem wont be applicable . $\endgroup$ – user157588 May 30 at 7:57
  • $\begingroup$ As pointed out, there are no fractional exponents. $\endgroup$ – Chris Custer May 30 at 14:50

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