Convexity of $\frac x{\sin x},\,\forall x\in[0,\pi)$

Question

How would one prove the following statement?

$f(x)=\frac x{\sin x}$ is convex in $[0,\pi)$.

After having posted the question, I came upon two proofs which I have now posted in the answer below. I would like to know other proofs as well, even the brute force ones. Can we utilize the identity $\frac{\pi x}{\sin(\pi x)}=\Gamma(1+x)\Gamma(1-x)$ where $\Gamma$ is the gamma function? How about the fact that $f$ is the reciprocal of the sinc function with all the nice properties? Or can we represent $f$ by an integral $\int g(t,x)d\mu(t)$ where $g(t,x)$ is convex with respect to $x$ for every $t$ and $\mu(t)$ is a measure of $t$?

Answer 1

Here is a brute force proof via taking the second derivative. \begin{align} f''(x) &= \frac1{\sin^3x}\big(x(1+\cos^2 x)-2\sin x\cos x\big) \\ &= \frac1{\sin^3 x}\int_0^x \sin t(3\sin t-2t\cos t)\,\mathrm dt \\ &>0 \end{align} since $\tan t>\frac23t,\forall x\in\big[0,\frac\pi2\big)$ and $\cos t<0,\,\forall x\in\big(\frac\pi2,\pi\big)$.

Answer 2

Here is an elegant proof given by @deyore.

We show the convexity of $f$ via that of $\ln f(x)$. Take the second derivative of the latter. $$\frac{d^2}{dx^2}\ln f(x)=-\frac1{x^2}+\frac1{(\sin x)^2}>0,\ \forall x\in[0,\pi),$$ since $0<\sin x<x$ there.

Answer 3

Let us say that a function $f(x)$ is horribly convex if it is analytic in a neighbourhood of the origin and the coefficients of its Maclaurin series are non-negative. Quite clearly, the space of horribly convex functions is closed with respect to the multiplication by $x^n$, differentiation and termwise-integration. By the Weierstrass product for the sine function $$ \sin(z) = z\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right) $$ and by applying the logarithmic derivative $\frac{d}{dz}\log(\cdot)$ to both sides we get $$ z\cot(z) = 1+\sum_{n\geq 1}\frac{2z}{n^2 \pi^2 - z^2}=1-2\sum_{n\geq 1}\frac{\zeta(2n)}{\pi^{2n}}z^{2n} $$ already exploited by Euler in one of his proofs of $\zeta(2)=\frac{\pi^2}{6}$ (see my notes).
The fact that $1-z\cot z$ is a horribly convex function is now evident and it can be used for strenghtening the Huygens and Shafer-Fink inequalities, for instance. The trigonometric identity $$ \frac{1}{\sin(x)}=\cot(x/2)-\cot(x) $$ then implies that $\frac{z}{\sin z}$ is a horribly convex function, in particular a convex function over $(0,\pi)$, since the radius of convergence of the Maclaurin series clearly equals $\pi$ ($z=\pm \pi$ are the poles closest to the origin). In equivalent terms, convexity is a straightforward consequence of Herglotz' decomposition or of $\operatorname*{Res}_{z=k\pi}\frac{1}{\sin(z)}=(-1)^k$. Of course $$ \frac{d^2}{dx^2}\log\left(\frac{x}{\sin x}\right) = \frac{1}{\sin^2 x}-\frac{1}{x^2}>0$$ is just as effective.