My objective is evaluate the integral $\int_1^x t^{-1} dt$ using the Riemman sums definitions: $\int_{a}^{b}f(x)dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*) \Delta x$, where a partition of $[a,b]$ is chosen.
$[1,x] = \displaystyle \bigg[1,1+\frac{x-1}{n}\bigg]\cup\bigg[1+\frac{x-1}{n},1+\frac{2(x-1)}{n}\bigg]\cup … \cup\bigg[1+\frac{(n-1)(x-1)}{n},x\bigg]$,
$\displaystyle\Delta x = \frac{x-1}{n}$, $\displaystyle x_k^*=1+k\frac{x-1}{n}=\frac{n+kx-k}{n}$
Then:
$\displaystyle\int_{1}^{x} \frac{1}{t} dt = \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{x^*_k}\Delta x=\lim_{n \to \infty}\sum_{k=1}^{n}\bigg( \frac{n}{n+kx-k}\bigg)\frac{x-1}{n}=\lim_{n \to \infty}\sum_{k=1}^n\frac{x-1}{n+kx-k}$
I already verified in Wolfram that the last limit indeed equals to $\ln|x|$, but I do not know how to evaluate it. Solving the summation first also seems complicated, as Wolfram shows digamma functions… is there a better way to avoid this obstacle, maybe changing the partition? But if there is a way of solving this limit I would also appreciate.
Thanks.
Best Answer
Hint:
Use the partition $1, x^{1/n}, x^{2/n}, \ldots, x$ and note that $n(x^{1/n}-1) \to \log x$.