# Why is $\int\limits_{1}^{n} \log x \,dx \le \sum\limits_{x = 1}^{n}\log x$?

It has been a long time since I studied integrals, so this question may sound stupid. I was going through this wiki page, and came across the following inequality:

$$\int_{1}^{n} \log x \,dx \le \sum_{x = 1}^{n}\log x$$

Why is this inequality true?

Now, my understanding is that the integral on the left side is the area under the curve of $y = \log x$ from $x = 1 \text{ to } n$. Also, this curve is always increasing, and non-negative in $[1 \dots \infty)$. As for the summation, it can be graphically/geometrically represented as rectangles of width 1 on the X-axis, and height $\log x$ for integral values of $x$. So, $\log 3$ can be represented as a rectangle from $x = 3 \text{ to } 4$ of height $\log 3$. Then the summation is the total area of these rectangles. Since the last rectangle goes upto $n+1$, it is beyond the upper limit of the integral. Apart from that, all other rectangles have area lesser than the corresponding sections of the curve (since it is non-negative and increasing in the any section after $x = 1$). So, we get the obvious inequality:

$$\int_{1}^{n} \log x \,dx \ge \sum_{x = 1}^{n-1}\log x$$

How does the addition of the $\log n$ term to the right hand side change the inequality? How can the first inequality be proven?

• Your question made me notice the inequality $\frac{n^n}{e^{n-1}}\leq n!$
– Amr
Jan 24 '13 at 17:17
– A.D
Jan 24 '13 at 17:23

Since $\log x$ is increasing, the integral from $k-1$ to $k$ is $\le \log k$. Add up, noting that $\log 1=0$.
The integral include all the points of summation and some points that are not contained in summation (such as $1/2, 1/3$ etc). Again $\log x > 0$ . So your inequality sign support the fact.