Logarithmic Integral Inequality

Question

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: $$ n^ne^{1-n} \leq n! \leq (n+1)^{n+1}e^{-n} $$

My apologies if this is trivial but could someone please help bridge the gap?

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
Try to draw a picture and compare the sum with the area under logarithm curve. Like in the pictures here or here. — Martin Sleziak, Mar 05 '15 at 11:40

score 3 · Answer 1 · answered Feb 19 '15 at 17:01

3

hint: $\displaystyle \sum_{k=1}^n \int_{k-1}^k \log xdx \leq \displaystyle \sum_{k=1}^n \log k \leq \displaystyle \sum_{k=1}^n \int_{k}^{k+1} \log xdx$

answered Feb 19 '15 at 17:01

DeepSea

77,651

Alex · Accepted Answer · 2015-02-19T18:57:23.030

3

One of the ways to apply this hint is notice that $n! = e^{\log n!} = e^{\sum_{k=1}^{n} \log k} <e^{\int_{1}^{n+1} \log x dx} = e^{(n+1) \log (n+1) -n} = e^{\log (n+1)^{n+1}} \cdot e^{-n}$

edited Feb 19 '15 at 18:57

answered Feb 19 '15 at 17:47

Alex

19,262

Logarithmic Integral Inequality

Update 1

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