Question

How do you evaluate the integral `int x^5dx` from `-oo` to `oo`?

Answer 1

To attempt to evaluate `int_-oo^oo f(x) dx` we choose a number `c` and evaluate, separately, `int_-oo^c f(x) dx` and `int_c^oo f(x) dx`

Explanation:

In order for `int_-oo^oo f(x) dx` to converge, both of the integrals on the half-lines must converge.

We'll use `c=0` (because it's easy to work with).

`int_-oo^0 x^5 dx = lim_(ararr-oo) int_a^0 x^5 dx`

`= lim_(ararr-oo)` `{: x^6/6]_a^0`

`= lim_(ararr-oo) a^6/6`

This limit does not exist, so the integral on the half-line diverges.

Therefore, the integral on the real line diverges.

Note

Since `x^5` is an odd function it is tempting to reason as follows.

For every positive number `a`, we have

`int_-a^a x^5 dx = 0`

So we must also have the integral from `-oo` to `oo` is `0`.

This reasoning FAILS because we define `int_-oo^oo f(x) dx` as above.