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*.