Obtain the value of $\int_0^1f(x) \ dx$, where $f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}

Question

Is the function Riemann integrable? If yes, obtain the value of $\int_0^1f(x) \ dx$

$f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}<x\le\frac{1}{n}\\ 0, & x=0 \end{cases}$

My attempt

$f$ is bounded and monotonically increasing on $[0,1]$. Also, $f$ has infinite discontinuities but only one limit point. Therefore $f$ is Riemann integrable. Now, to calculate the integration

$\int_0^1f(x) \ dx=\int_{1/2}^{1}1 \ dx + \int_{1/3}^{1/2}\frac{1}{2} \ dx + \int_{1/4}^{1/3}\frac{1}{3} \ dx+...$

$=\sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1}$

How do I proceed from here? How do I calculate these summations? I know $\sum \frac{1}{n}$ is $\log 2$, but not the other two summations.

Answer 1

Yes, it is Riemann integrable and $$\int_0^1 f(x) dx=\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\frac{\pi^2}{6}-1.$$ where we used the Basel problem and Mengoli's telescopic series.

Answer 2

Since $$ (0,1]=\bigcup_{n=1}^\infty\left(\dfrac{1}{n+1},\dfrac{1}{n}\right]=\bigcup_{n=1}^\infty I_n, $$ and $$ I_n\cap I_m=\emptyset\quad \forall m,n\in \mathbb{N}, $$ we have \begin{eqnarray} \int_0^1f(x)\,dx&=&\int_{[0,0]}f(x)\,dx+\int_{(0,1]}f(x)\,dx\\ &=&\sum_{n=1}^\infty \int_{I_n}f(x)\,dx\\ &=&\sum_{n=1}^\infty\dfrac{1}{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\\ &=&\sum_{n=1}^\infty\dfrac{1}{n^2}-\sum_{n=1}^\infty\dfrac{1}{n(n+1)}\\ &=&\sum_{n=1}^\infty\dfrac{1}{n^2}-\sum_{n=1}^\infty\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\\ &=&\dfrac{\pi^2}{6}-1 \end{eqnarray}

Answer 3

since $(0,1) = \bigcup_{n=1}^{\infty} ( \frac1{n+1} , \frac1{n })$. (pairwise disjoint union)

$$\int_{0}^1f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}\frac1ndx=\sum_{n=1}^{ \infty} \frac1{n^2 }- \frac1{(n+1)n}= \frac{\pi^2}{6}-1 $$

Given that $ \frac1{(n+1)n}= \frac1{n}- \frac1{n+1}$ then $$\sum_{n=1}^{ \infty} \frac1{(n+1)n}=\sum_{n=1}^{ \infty} (\frac1{n}- \frac1{n+1})=1$$