# Can Cauchy principal values of functions with nonsimple poles be evaluated using complex contour integration methods?

Can Cauchy principal values of functions with nonsimple poles be evaluated using complex contour integration methods?

In all of the examples I have seen, poles are simple and this helps to avoid those singular points since integrals around them converge to a finite value (such as $i\pi$) when the radius of the circular path around them goes to zero. (In nonsimple poles I guess the integral does not converge around them, on the circle with infinitesimal path.)

For example, for the integral $$\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx$$ Wolfram alpha (correctly)gives $\pi$ as the principal value, but says $$\int_{-\infty}^{\infty}\frac{\sin(x)}{x^3}dx$$

or

$$\int_{-\infty}^{\infty}\frac{\sin(x)}{x^4}dx$$ do not converge.

• Sure you can, if the pole is "odd". i.e near a pole $z_0$, the function behaves like $$\sum_{k=0}^n \frac{\alpha_k}{(z-z_0)^{2k+1}} + \text{sth. regular at } z_0$$ Jan 8, 2015 at 20:05
• @achillehui Thanks, but how is the divergence of the example I added to the question explained? The pole at the origin is odd for sure. Jan 8, 2015 at 21:09
• $\sin(x)/x$ has a removable singularity at zero, and an essential singularity at infinity. So what does this have to do with "nonsimple poles"? Perhaps you should say what "principal value" you want to compute. Jan 8, 2015 at 21:16
• @user215721 $\frac{1}{z^{2k+1}}$ has anti-derivative $\frac{-1}{2k z^{2k}}$. When you integrate $\frac{1}{z^{2k+1}}$ around a semi-circle, $\frac{-1}{2k z^{2k}}$ is the same at the two end-points. As a result, the contour integral over the semi-circle vanishes. That's why the principle-value of $\int \frac{\sin x}{x^4} dx$ remains finite (because the integrand is odd under $x \to -x$). Jan 8, 2015 at 21:56

The first one is an ordinary improper integral, and Maple says

$$\int_{-\infty}^\infty \sin(x)/x$$

is $$\pi$$. The others are, indeed, poles.

This one has a pole of order $$1$$, and Maple says:

$$\int_{-\infty}^\infty \sin(x)/x^2$$, based on CauchyPrincipalValue);

is $$0$$ as we expect. (And the non-principal-value integral is "undefined".)

This one has a pole of order $$2$$, and Maple says

$$\int_{-\infty}^\infty \sin(x)/x^3$$,,CauchyPrincipalValue=true);

is $$\infty$$.

And for higher integer powers, Maple has alternately $$0$$ and $$\infty$$ for the Cauchy principal values.