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
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.