Prove that $f(t)=\frac{\int_{1}^{b}{x^{1+t}\,dx}}{\int_{1}^{b}{x^{t}\,dx}}$ is an increasing function. Let $b>1$ and define $f:\mathbb{R}\to\mathbb{R}$ by $$f(t)=\frac{\int_{1}^{b}{x^{1+t}\,dx}}{\int_{1}^{b}{x^{t}\,dx}}.$$
I would like to prove that $f$ is an increasing function for all $t$.
Ideas so far: Certainly, it should be true, as $f$ represents the centre of mass of the region beneath the graph of $y=x^t$ over the interval $[1,b]$. So physically, it's obvious. Aside from a vague physical proof, I can show by differentiation that the derivative of $f$ satisfies  $$f'(t)=(t+1)-t\frac{f(t)}{f(t+1)},$$
But I'm not sure where to go with this. 
 A: This answer is an alternative proof of the log-convexity result given above by Jack D'Aurizio. Defining $g(t)=\int_{1}^{b}{x^t}\,dx$, by the Cauchy Schwarz inequality we see that
\begin{align*}
g''(t)\cdot g(t)&=\int_{1}^{b}{ x^t(\log{x})^2}\,dx\cdot\int_{1}^{b}{x^t}\,dx\\
&=\int_{1}^{b}{( x^\frac{t}{2}\log{x})^2}\,dx\cdot\int_{1}^{b}{(x^\frac{t}{2})^2}\,dx\\
&\geq \left(\int_{1}^{b}{x^t\log{x}}\,dt\right)^2\\
&=(g'(t))^2.
\end{align*}
This proves the log-convexity of $g$ and, as D'Aurizio points out, this implies that $f(t)=\frac{g(t+1)}{g(t)}$ is an increasing function. 
A: Key point: your $f$ is a ratio of moments.
$$g(t) = \int_{1}^{b} x^t\,dx $$
is quite trivially a continuous and positive function. $g(t)$ is midpoint-log-convex by the Cauchy-Schwarz inequality:
$$ g(t_1) g(t_2) \geq g\left(\frac{t_1+t_2}{2}\right)^2. $$
By putting together the continuity and the midpoint-log-convexity we get the log-convexity of $g$, which implies that $f(t)=\frac{g(t+1)}{g(t)}$ is an increasing function, as wanted.
