Convexity of a function (volume-to-area ratio) I am working on some estimates on the sphere and I come up with the function $h(t):=\frac{\int_0^t \sin^{m}r \,dr}{\sin ^m t}$ on $(0, \pi)$, where $m\ge 1$ is an integer. 
(This is the volume-to-area ratio of the geodesic ball of radius $t$ in the $(m+1)$-dimensional sphere, if you want to know.)
Graphing the function suggests that $h(t)$ is convex for any $m$. But I am unable to prove this. Differentiating it two times is somewhat messy, and you have to compare the integral of $\sin ^m r$ with the function itself anyway, which I have no clue (though I can prove that $h(t)$ is increasing if that helps).
Any idea?
 A: Right, the first thing to note is that this function is of the form $f/f'$, where of course $f(t) = \int_0^t \sin^m{r} \, dr$. Since this is obviously smooth away from the endpoints, it is sufficient to show that the second derivative is positive.
It is easy to check that $f$ satisfies the differential equation
$$ (\sin{t})f'' = (m\cos{t}) f'. $$
Differentiating and using this,
$$ \left( \frac{f}{f'} \right)' = 1-\frac{ff''}{f'^2} = 1-m\cot{t} \frac{f}{f'}, $$
and differentiating again and using the relation again,
$$ \left( \frac{f}{f'} \right)'' = \dotsb = -m\cot{t} + m((m+1)\cot^2{t}+1)\frac{f}{f'}. $$
Multiplying by the positive $(\sin^2{t})f'/[m(1+m\cos^2{t})] = (\sin^{m+2}{t})/[m(1+m\cos^2{t})]$ gives
$$ g(t) = f(t)-\frac{\cos{t}\sin^{m+1}{t}}{1+m\cos^2{t}}, $$
and so we want to prove that this is positive. Since $g(0)=0$, $g(t) = \int_0^t g'(r) \, dr$ so it is enough to show that the $g'$ is nonnegative. A bit of work shows that
$$ g'(t) = \frac{2\sin^{m+2}{t}}{(1+m\cos^2{t})^2} \geq 0, $$
as required.
