Prove that f is not integrable on [0,1]

[issacnewton]

Homework Statement

Let ##f(0) = 0## and ##f(x) = 1/x ## if ##0 < x \leqslant 1##. Show that ##f## is not integrable on ##[0,1]##.
Hint: Show that the first term in the Riemann sum, ##f(x_1^*) ~\Delta x##, can be made arbitrarily large

Homework Equations

Definition of integral using Riemann sum

The Attempt at a Solution

Using the definition, we have $$\int_0^1 f(x) dx = \lim_{n \to\infty} \sum_{i=1}^n f(x_i^*) \Delta x $$ Now I am not sure how the hint could be used here. Should I try to go for a proof by contradiction ?

 

[haruspex]

how the hint could be used here

You need to consider specific choices for x₁ and pick corresponding values for x₁^*.

 

[issacnewton]

If we divide ##[0,1]## into ##n## intervals, then ##\Delta x = \frac{1}{n}## and the first interval would be ##[x_0, x_1] = [0, \frac{1}{n}]##. If we take ##x_1^*## to be right endpoint of the interval, then ##f(x_1^*) = 1/(1/n) = n## and hence ##f(x_1^*)\Delta x = 1## How can this be made arbitrarily large ?

 

[nuuskur]

Nvm, didn't notice your last post :/
The definition of the Riemann integral eventually hinges on a limit. Specifically when all of your segments' width goes to zero. In this case, though, the tighter you pick your partition, the larger the sum will become, unbounded. Resulting series diverges.

 

[PeroK]

If we divide ##[0,1]## into ##n## intervals, then ##\Delta x = \frac{1}{n}## and the first interval would be ##[x_0, x_1] = [0, \frac{1}{n}]##. If we take ##x_1^*## to be right endpoint of the interval, then ##f(x_1^*) = 1/(1/n) = n## and hence ##f(x_1^*)\Delta x = 1## How can this be made arbitrarily large ?

The hint looks like a hindrance to me. Try looking at the Riemann sum for ##n##, not just one term.

 

[haruspex]

The hint looks like a hindrance to me. Try looking at the Riemann sum for ##n##, not just one term.

True, but...

How can this be made arbitrarily large ?

So pick a different x₁*.