# Riemann integrable?

Consider the function: $f:[0,\frac{1}{2\pi}]\to\mathbb{R}:f(x):=x\cos(1/x)$

In general, every continuous function on a compact interval is Riemann integrable.
However, for the tagged partitions: $$x_0:=0,x_1:=\frac{1}{2\pi N},\ldots,x_N:=\frac{1}{2\pi},t_0:=0,t_1:=\frac{1}{2\pi N},\ldots,t_{N-1}:=\frac{1}{4\pi}$$ the Riemann sums diverge: $$\sum_i f(t_i)(x_{i+1}-x_i)=\sum_{k=1}^N \frac{1}{k}\to\infty$$

How can this happen?

• What is your question? – Pascal Engeler Jun 9 '14 at 19:57
• How is $\;\pi\;$ a partition of $\;[0,1]\;$ ?? – DonAntonio Jun 9 '14 at 20:05
• How can this happen while the theorem states that any Riemann sums converge? – C-Star-W-Star Jun 9 '14 at 20:06
• The partition is meant as $[0,\frac{1}{2\pi N}),\ldots,[\frac{1}{2\pi},1]$. – C-Star-W-Star Jun 9 '14 at 20:09
• Are you "jumping" between $\;\frac1{2\pi}\cong 0.159\;$ and all the way to $\;1\;$ ? That's not a valid partition as in the limit (Riemann Integralwise) partitions must both tend to infinite partition points and at the same time the mesh of the partition (i.e., the maximum of all the subintervals' lengths) must go to zero ! – DonAntonio Jun 9 '14 at 20:15

The function $\;x\cos\frac1x\;$ is continuous and thus bounded at the unit integral (even at zero if we define its value as zero there) and thus Riemann integrable there.

The actual limit of Riemann Sums is

$$\lim_{\begin{cases}n\to\infty\\||\Delta_x||\to 0\end{cases}}\sum_{i=1}^nf(c_i)\left(x_i-x_{i-1}\right)\;,\;\;\text{with}\;\;||\Delta_x||:=\max_i||x_i-x_{i-1}||$$

and thus your proposed partitions are illegal.

• What do you mean by illegal? – C-Star-W-Star Jun 9 '14 at 20:20
• Read my answer about what have to fulfill partitions when doing Riemann Sums: if you always end the partitions in the subinterval $\;\left[\frac1{2\pi}\;,\;\;1\right]\;$ then the lengths of the subintervals of the partitions in the limit do not go to zero, as required. – DonAntonio Jun 9 '14 at 20:22
• Now it does but the problem is still there... – C-Star-W-Star Jun 9 '14 at 20:24
• No, it doesn't: you still stop the last subinterval at $\;\frac1{2\pi}\;$ ...! To put weird numbers $\;1,2,..\;$ to the right of $\;\pi\;$ there doesn't change that...unless you mean something I can't understand. – DonAntonio Jun 9 '14 at 20:26
• I guess I don't understand your point about gaps... – C-Star-W-Star Jun 9 '14 at 20:29