I am being asked to calculate the integral $\int_{a}^{b}\ln x~dx$ using the definition of integral (i.e. expressing it as limit of Riemann sums)
Here's what I did:
Let's divide the interval $[a,b]$ into $n$ subintervals of the form $[x_{i},x_{i+1}]$.
Let $x_i=a+\frac{(b-a)i}{n}$ where $i=0,1,2…,n-1. $ be the start points of each subinterval such that $x_0=a$ and $x_n=b$
Let $\Delta x_i=x_i-x_{i-1}=\frac{b-a}{n}$
I defined the integral like this:
$$\begin{align}\int_{a}^{b}\ln x \, dx&=\lim_{n\rightarrow \infty}\sum_{i=o}^{n-1}f(x_{i})\Delta x_{i}\\
&=\lim_{n\rightarrow \infty}\frac{b-a}{n}\sum_{i=0}^{n-1}\ln\left(a+\frac{(b-a)i}{n}\right)\\
&=\lim_{n\rightarrow \infty}\frac{b-a}{n}\ln\left(\prod_{i=0}^{n-1}\left(a+\frac{(b-a)i}{n}\right)\right)\end{align}$$
And I'm stuck here, I don't know how to find this product, any hint would be appreciated , thank you !
Best Answer
Some hints:
Let ${b\over a}=:\rho>1$, and choose an $N\gg1$. Then use the partition $a=x_0<x_1<\ldots<x_N=b$ given by $$x_k:=a\,\rho^{k/N}\qquad(0\leq k\leq N)\ .$$ Then $$R_N:=\sum_{k=0}^{N-1}\log(x_k)(x_{k+1}-x_k)$$ is an admissible Riemann sum for the given integral, i.e., we can be sure that $$\lim_{N\to\infty} R_N=\int_a^b\log x\>dx\ .$$ Now $R_N$ can be computed in an elementary way; you only need a formula for finite sums of the type $\sum_{k=0}^{N-1} k\, q^k$.