Evaluate integrals to Riemann sum for $\sin x$ and $\ln x$

Question

How can i evaluate to Riemann sum:

i tried this $$\int_0^1 x^3 dx$$

I used $$ \sum_{n=1}^{n}\frac{i^3}{n^3}=1^3+2^3+..+n^3 = \frac{n^2(n+1)^2}{4}$$

$$\int_0^1 x^3 dx = \lim_{x\rightarrow\infty}\frac{1-0}{x^{3+1}}\cdot \frac{x^2(x+1)^2}{4}=\frac{1}{4} $$

$$ \lim_{x\rightarrow\infty}\frac{x^2(x+1)^2}{4x^4}=\lim_{x\rightarrow\infty}\frac{x^4 + 2x^3+x^2}{4x^4}=\frac{1}{4}$$

How can i evaluate those?

$$\int_0^\pi \sin x dx$$

$$\int_1^2 \ln x dx$$

Answer 1

The sine integral has been treated here:

Riemann sum of $\sin(x)$

For the integral $I:=\int_1^2\log x\ dx$ we choose a partition of $[1,2]$ into $n$ subintervals of unequal length by putting $$x_k:=2^{k/n}\qquad(0\leq k\leq n)\ .$$ The longest of these intervals is the last one, and its length is $$2-2^{(n-1)/n}<2^{1/n}-1\to 0\qquad(n\to\infty)\ .$$ A Riemann sum for I can then be set up as follows: $$R_n=\sum_{k=1}^n \log\bigl(2^{k/n}\bigr)\>\bigl(2^{k/n}-2^{(k-1)/n}\bigr)={\log 2\over n}\sum_{k=1}^n k\>\bigl(2^{k/n}-2^{(k-1)/n}\bigr)\ .$$ The sum on the right can be evaluated in closed terms, and one obtains $$R_n=\log 2\>\left(2-{1/n\over 2^{1/n}-1}\right)\ .$$ It follows that $$I=\lim_{n\to\infty} R_n=\log 4 -1\ .$$

Answer 2

The Riemann sum for $\int_0^\pi{\sin{x}}dx$ is written in the form: $$\int_0^\pi{\sin{x}}dx=\lim_{n\to\infty}\sum_{k=1}^{n}{\Delta x\sin{x_k}}=\lim_{n\to\infty}\frac{\pi}{n}\sum_{k=1}^{n}{\sin{\frac{\pi k}{n}}}.$$ The most sound way I know how to approach this problem now is to let $\theta=\pi/n$ and consider Euler's identity: $e^{i\theta}=\cos\theta+i\sin\theta$. That is, $\sin\theta=\Im\{e^{i\theta}\}$, where $\Im\{z\}$ denotes the imaginary part of $z$. Using this, we have $$\lim_{n\to\infty}\frac{\pi}{n}\sum_{k=1}^{n}{\sin{\frac{\pi k}{n}}}=\lim_{n\to\infty}\frac{\pi}{n}\sum_{k=1}^{n}{\Im\{e^{ik\theta}\}}=\lim_{n\to\infty}\frac{\pi}{n}\Im\{\sum_{k=1}^{n}{e^{ik\theta}}\}.$$ The trick now is to evaluate the sum, which is really now a geometric sum, simplify it, and take its imaginary part.

As for the logarithm, we have \begin{align} \int_{1}^{2}\ln x dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}{\ln{\frac{k}{n}}}&=\lim_{n\to\infty}\frac{1}{n}\ln\left(\prod_{k=1}^{n}\frac{k}{n}\right)\\&=\lim_{n\to\infty}\frac{1}{n}\ln\frac{n!}{n^n}\\&=\lim_{n\to\infty}\ln\frac{\sqrt[n]{n!}}{n}\\&=\ln\left(\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}\right) \end{align} And now all that is needed is to find a way to evaluate the final limit (perhaps Stirling's approximation could be used here). Another way to tackle this is to use $x_k=2^{k/n}$ which gives $\Delta x_k=x_{k+1}-x_k=2^{(k+1)/n}-2^{k/n}=2^{k/n}(2^{1/n}-1)$, and then evaluate $$\lim_{n\to\infty}\sum_{k=1}^{n}2^{k/n}(2^{1/n}-1)\ln{2^{k/n}}=\ln2\lim_{n\to\infty}\frac{(2^{1/n}-1)}{n}\sum_{k=1}^{n}k2^{k/n},$$ Which actually turns into a rather nice Arithmeto-Geometric series.

Hope this helps!