Question

How do you evaluate the integral `int 1/x` from [0,1]?

Answer 1

It's undefined.

Explanation:

The integral `int_0^1 1/xdx` is equivalent to `int_0^1x^-1dx`.
All integrals of the form `int_0^t x^ndx`, where `t` is a real parameter, are equal to `t^(n+1)/(n+1)`, so if `t = 1` and `n=-1`,
`int_0^1 1/xdx` = `1^0/0`, which is undefined.

Answer 2

Diverging...

Explanation:

`int_0 ^1 1/x dx = [ln(x) ]_0 ^1`

`=> ln1 - ln0 = ln 0`

We know `ln0` is undefined, but `lim_(n to 0^+ ) ln n = -oo`

So hence the integral is diverging...