# Logarithmic Integral Inequality

Give a convincing argument that the following inequalities are true: $$\int_1^n \log x\mathrm dx \leq \log1 + \log2 + ... \log n \leq \int_1^{n+1}\log x \mathrm dx$$ for any $n \geq 1$ . We are given the hint to observe that: $$\int_{k-1}^k \log x\mathrm dx \leq \log k \leq \int_k^{k+1}\log x\mathrm dx$$

## Update 1

BRIC-Fan's argument makes sense but I'm supposed to use the result of the above inequality to show that: $$n^ne^{1-n} \leq n! \leq (n+1)^{n+1}e^{-n}$$

• If you accept the hint, can't you just add up integrals, e.g., $\int_1^2 + \int_2^3 = \int_1^3$ – Simon S Feb 19 '15 at 17:00
• Since $\log$ is an increasing function, $\log k \cdot ((k+1) - k) \leq \int_{k}^{k+1} \log(x) dx \leq \log(k+1) \cdot ((k+1) - k)$. – Pedro M. Feb 19 '15 at 17:04
One of the ways to apply this hint is notice that $n! = e^{\log n!} = e^{\sum_{k=1}^{n} \log k} <e^{\int_{1}^{n+1} \log x dx} = e^{(n+1) \log (n+1) -n} = e^{\log (n+1)^{n+1}} \cdot e^{-n}$
hint: $\displaystyle \sum_{k=1}^n \int_{k-1}^k \log xdx \leq \displaystyle \sum_{k=1}^n \log k \leq \displaystyle \sum_{k=1}^n \int_{k}^{k+1} \log xdx$