Hello everone!
I want to ask something about:
"Since $i \log_e i$ is concave upwards, it is easy to show that $$\sum_{i=2}^{n-1} i \log_e i \leq \int_2^n x \log_e x dx \leq \frac{n^2 \log_e n}{2}-\frac{n^2}{4}$$"
$i \log_e i$ is concave upwards because the second derivative of $i \log_e i$ is positive. Correct?
Can you explain the inquality? I don't understand how we show it?
I was thinking about $\int_0^{\infty} A( f(B(x)) ) + C(x) ) dx = \int_0^{\infty} f(x) dx$
The inspiration came from the following 3 integrals :
Lemma
If $f(x)$ is a bounded non-negative function, then
\begin{equation}\int_0^\infty f\left(x+\frac{1}{x}\right)\,\frac{\ln x}{x}\,dx=0\end...
