Explaining the convergence of $\int_{0}^{1} \sin \left(\frac{1}{x} \right) \, dx$ using the properties of essential singularities

Question

Because the function $\frac{1}{z}$ has a pole at $z=0$, the integral $\int_{0}^{1} \frac{dx}{x} $ doesn't converge.

On the other hand, the integral $\int_{0}^{1} \sin \left(\frac{1}{x} \right) \, dx $ converges even though the function $\sin \left(\frac{1}{z} \right)$ has an essential singularity at $z=0$.

Can an essential singularity be a weaker singularity than an pole?

EDIT:

Originally I had $\int_{0}^{1} \sin \left(\frac{1}{x^{2}} \right) \, dx$ (which converges despite the fact that $\sin \left(\frac{1}{z^{2}} \right)$ has an essential singularity at $z=0$).

robjohn · Accepted Answer · 2013-04-07T18:09:10.593

10

Essential singularities have the property of being both small and large near the singularity (see the Picard Theorems or, as Martin comments, the Casorati-Weierstrass Theorem). In this case, you've chosen a path to the singularity where its size is not big.

If instead, you chose to approach along the line $z=(1+i)t$ ($t\in\mathbb{R}$), then $$ \sin\left(\frac1{z^2}\right)=-i\sinh\left(\frac1{2t^2}\right) $$ which grows very fast near $z=0$, faster than any power of $\left|\frac1z\right|$.

For the modified question

Approach along the line $z=it$ ($t\in\mathbb{R}$), then $$ \sin\left(\frac1z\right)=-i\sinh\left(\frac1t\right) $$ which grows very fast near $z=0$, faster than any power of $\left|\frac1z\right|$.

edited Apr 07 '13 at 18:09

answered Apr 07 '13 at 16:48

robjohn

345,667

Do you mean that the value of $f(z)$ is both small and large near the essential singularity? So it's all about how you approach it? – Random Variable Apr 07 '13 at 16:54
Yes. According to the Big Picard Theorem, all values (except possibly one) are attained infinitely often in any neighborhood of an essential singularity. – robjohn Apr 07 '13 at 17:02
And for a pole, the value of $f(z)$ is large in any punctured neighborhood? – Random Variable Apr 07 '13 at 17:14
Yes. For some $n$, $f(z)(z-z_0)^n$ is analytic in a neighborhood of a pole at $z_0$. The least such $n$ is called the order of the pole. – robjohn Apr 07 '13 at 17:20
Instead of Picard's theorems, Casorati-Weierstrass would be enough for the qualitative statement in the first paragraph. – Martin Apr 07 '13 at 18:03
@Martin: indeed, thanks. Picard is just what came to mind as I was writing. I will include your link. – robjohn Apr 07 '13 at 18:07

Explaining the convergence of $\int_{0}^{1} \sin \left(\frac{1}{x} \right) \, dx$ using the properties of essential singularities

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