Bounding $\sin(\pi z)$ away from $0$ on a circular contour in $\Bbb C$

Question

I'm currently trying to show the following:

Let $C_R$ be the circle of radius $R=N+\frac{1}{2}$ for some $N\in\Bbb N$. Then $$\lim_{R\to\infty}\ \int_{C_R}\frac{1}{z^2\sin(\pi z)}\,dz=0.$$

This will follow immediately if we can prove $\sin(\pi z)$ is bounded away from zero on the contour. To that end, I've shown $|\sin(\pi z)|\geq1$ for $z\in C_R$ is equivalent to proving the inequality

$$\cosh(x)\geq 2+\cos\left(\sqrt{(2\pi R)^2-x^2}\right)$$

for $-2\pi R\leq x\leq2\pi R$.

This brings me to my question:

How can we prove this inequality?

I've thought of trying to use Taylor series here, but $x$ isn't necessarily close to $0$. I also thought of this as a Calculus I style problem (after all, we're just minimizing a single-variable function), but the derivative was too convoluted to reasonably work with unless I missed something.

The plots that I know how to make on Mathematica agree with my computations, so I have reason to believe this is the correct idea, I'm just not sure how to formally prove it.

Answer 1

Yes, $|\sin(z)| \ge 1$ does hold on circles with radius $N+1/2$ centered at the origin ($N$ a positive integer), see the second part of this answer.

In order to get some positive lower bound on those circles one can proceed as follows. The idea is to distinguish whether $z$ is “close to the real axis” or not.

For $z = x+iy$ with $x, y \in \Bbb R $ is $$ |\sin(\pi z)|^2 = \sin^2(\pi x) + \sinh^2(\pi y) \, . $$

If $|y| \ge 1/4$ then $$ |\sin(\pi z)| \ge |\sinh(\pi y)| \ge \sinh( \frac \pi 4) > 0 \, . $$

And if $|z| = N + 1/2$ with $|y| \le 1/4$ then $$ N + \frac 14 \le |z| - |y| \le |x| \le N + \frac 12 $$ so that $$ |\sin(\pi z)| \ge |\sin(\pi x) | \ge \sin( \frac \pi 4) > 0 \, . $$

This proves $|\sin(\pi z)| \ge \sin( \frac \pi 4) = 1/\sqrt 2$ for all $z$ on the circle.

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The (sharp) bound $|\sin(z)| \ge 1$ on the circles with radius $N+1/2$ requires a bit more work. We have

$$ \begin{align} |\sin(\pi z)|^2 &= \sin^2(\pi x) + \sinh^2(\pi y) \\ &= \cosh^2(\pi y) - \cos^2(\pi x) \\ &= \frac 12 \bigl( \cosh(2\pi y) - \cos(2\pi x)\bigr) \end{align} $$ and the goal is to prove that $$ \tag{*} \cosh(2\pi y) - \cos(2\pi x) \ge 2 $$ for $x^2+y^2 = (N+1/2)^2$ and all positive integers $N$. (Which is the same inequality that you got.)

It suffices to consider non-negative $x$. The idea is indeed to use Taylor series, but not at $x=0$ but at $x=N+1/2$.

With $d = N+1/2 - x$ is $$ 0 \le d \le N + \frac 12 \, ,\\ x = N + \frac 12 - d \, ,\\ y^2 = 2 d \left( N + \frac 12 \right) - d^2 \, . $$

Then $$ \cosh(2\pi y) \ge 1 + \frac{(2 \pi y)^2}{2!} = 1 + 2 \pi^2 \bigl(2d \left( N + \frac 12 \right) - d^2 \bigr) $$ and $$ - \cos(2\pi x) = \cos(2 \pi d) \ge 1 - \frac{(2 \pi d)^2}{2!} = 1 - 2 \pi^2 d^2 \, . $$ Adding the last two inequalities gives $$ \cosh(2\pi y)- \cos(2\pi x) \ge 2 + 2 \pi^2 \bigl(2d \left( N + \frac 12 \right) - 2d^2 \bigr) \\ = 2 + 4 \pi^2 d \bigl( N + \frac 12 - d\bigr) \ge 2 \, . $$ This proves $(*)$ and we are done.

Answer 2

Note that $|\sin z|^2= \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y$ is a convex combination of $\sinh^2 y$ and $\cosh ^2 y$ (whose convex weights are $\cos^2 x$ and $\sin^2 x$). Thus it is straightforward to bound this expression on horizontal and vertical line segments where either $y>0$ is constant or $x$ is constant.

The sneaky trick is to deform $C$ onto such a grid, then estimate the integral. Of course it is best to use $x= (N+(1/2)) \pi$ as your first vertical segment, because there we know that the convex combination places all its weight purely on the larger term $\cosh^2 y \geq 1$.