$\frac{\log(x)}{1-x}$ is increasing

Question

How can I show that the function $h(x):= \frac{\log(x)}{1-x}$ is increasing on $\mathbb{R}_{>0}$? The derivative of $h$ is $h'(x)= \frac{x\log(x)-x+1}{x(1-x)^2}$ but I don't see why $h'(x)$ is always non-negative.

Answer 1

It is enough to show that $x\log(x)-x+1\geq0$ for all $x>0$. Define $g(x)=x\log(x)-x$. Then, $$g^\prime(x)=\log(x)$$ So, for $x\geq1$, we have $g^\prime(x)\geq0$ and so $g$ is increasing on $[1,\infty)$. Hence, for $x\geq 1$, we have $g(x)\geq g(1)=-1$.

On the other hand, for $0< x\leq 1$, note that $g^\prime(x)\leq 0$ so $g$ is decreasing. Hence, $g(x)\geq g(1)=-1$ for all $0<x\leq 1$.

To conclude, we have that for all $x>0$ that $g(x)\geq -1$ or $x\log(x)-x+1\geq0$.

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Note. The function $h(x)$ is not defined at $x=1$ right away, but you can continuously extend it by $h(1)=1$ and likewise $h^\prime(1)=0$. The answer above assumes that.

Hope this helps. :)

Answer 2

Using the well known inequality $\log(x) \ge 1-1/x$ we get that $$ x \log(x) - x + 1 \ge x \left( 1-\frac 1x \right) - x + 1 = 0 $$ for all $x > 0$. This proves that $h'(x) \ge 0$, i.e. that $h$ is increasing.

More precisely, $h'(x) > 0$ except for $x=1$, so that $h$ is strictly increasing.

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Another solution: $$ -h(x) = \frac{\log(x)}{x-1} = \frac{1}{x-1} \int_1^x \frac{dt}{t} = \int_0^1 \frac{ds}{1+s(x-1)} $$ with the substitution $t = 1 + s(x-1)$. The last integral is decreasing in $x$ because the integrand is decreasing in $x$ for every $s$.