An integral identity

Prove that for $n \geq 2$ it holds that

$\int_{1}^{n} \frac{n - \left \lfloor x \right \rfloor}{x} \, \mathrm{d}x = \ln n!$

Solution

We have successively:

$\begin{align*} \ln n! &= \sum_{k=2}^{n} \ln k \\ &=\sum_{k=2}^{n} \int_{1}^{k} \frac{\mathrm{d}x}{x} \\ &=\sum_{k=2}^{n} \sum_{j=1}^{k-1} \int_{j}^{j+1} \frac{\mathrm{d}x}{x} \\ &=\sum_{j=1}^{n-1} \sum_{k=j+1}^{n} \int_{j}^{j+1} \frac{\mathrm{d}x}{x} \\ &= \sum_{j=1}^{n-1} \left ( n - \left ( j+1 \right ) +1 \right ) \int_{j}^{j+1} \frac{\mathrm{d}x}{x} \\ &= \sum_{j=1}^{n-1} \left ( n - j \right ) \int_{j}^{j+1} \frac{\mathrm{d}x}{x} \\ &= \sum_{j=1}^{n-1} \int_{j}^{j+1} \frac{n - j}{x} \, \mathrm{d}x \\ &= \int_{1}^{n} \frac{n - \left \lfloor x \right \rfloor}{x} \, \mathrm{d}x \end{align*}$

On the centralizer

A trigonometric series

An integral identity

They might interest you:

On the centralizer

A trigonometric series

On a transcedental number

Sum of reciprocals

Trigonometric integral

The line is the shortest path between two points

On the convergence of a series

Square logarithmic integral

Sinc series

Leave a Reply Cancel reply