4
$\begingroup$

I want to prove that function \begin{equation} f(x)=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt \end{equation}

is quasi-concave. One approach is to obtain the closed form of the integral (provided below) and then prove that the result is quasi-concave. I tried this but it seems to be difficult. Do you have any idea how to prove the quasi-concavity?

\begin{align} f(x)&=\frac{1}{x+1} \int\limits_0^x \log \left(1+\frac{1}{x+1+t} \right)~dt\\ &= \frac{1}{x+1} \left[ \log(2x + 2) + (2x+1)\log \left(1+\frac{1}{2x+1} \right) -\log(x+2) - (x+1)\log \left( 1+\frac{1}{x+1} \right) \right] \end{align}

$\endgroup$
1
$\begingroup$

For $x>-1/2$, we have $$ f(x)=\ln 4-\frac{x+2}{x+1}\, \ln \frac{x+2}{x+1} -\frac{2 x+1}{x+1} \ln\frac{2 x+1}{x+1} $$ and hence $$ f'(x)=\frac1{(x+1)^2}\ln \left(1+\frac{1-x}{2 x+1}\right), $$ which is $>0$ for $x\in(-1/2,1)$ and $<0$ for $x>1$. So, $f$ increases on $(-1/2,1]$ and decreases on $[1,\infty)$. So, $f$ is quasi-concave.

$\endgroup$
2
$\begingroup$

f(x) is increasing from 0 1o 1, where it reaches its global maximum, and decreasing from 1 to $\infty$. It is concave from 0 to 2.00841388..., and convex beyond that (note that the 2nd derivative of f(x) reaches its maximum at 3.01890145886..., after which it declines to 0 in the limit as x goes to $\infty$). Hence quasi-concave.

I leave it to you to prove these statements to your satisfaction.

$\endgroup$
2
$\begingroup$

The derivative is: ${d f(x) \over d x} = \frac{-\log (x+1)+\log (x+2)+\log \left(\frac{x+1}{2 x+1}\right)}{(x+1)^2}$, which is:

df(x)/dx

and hence is both positive and negative, as Mark L. Stone wrote.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.