Question

Function f(x)=x^3 has odd symmetry.

Fact: if we integrate this function from -k to k for any real number k, we will always get zero, because the negative part (on left) exactly cancels with k=5.

Question: Is the fact still true if we use k= infinity? That is, if we integrate f(x)=x^3 from - infinity to infinity, do we get zero? why or why not?


You may have to sketch a graph but use the example with k=5 integral from -5 to 5 x^3 dx =0

I guess the only hint is try integrating from - infinity to infinity.

Answer 1

you would have to do improper integrals to be satisfied with the results

Answer 2

and you have to take the limits separately, so the answer is a resounding no!

Answer 3

it is not
\[\int_{-\infty}^{\infty}f(x)dx=\lim_{t\to \infty}\int_{-t}^{t}f(x)dx\] you do not take the limits together

Answer 4

you have to take them one at a time, and so since neither converges, you do not get a real number answer

Answer 5

ok but remember although you get zero if you integrate from -t to t, when you take an improper integral you take the limit as t goes to infinity and the limit as t goes to minus infinity separately, not at the same time. in fact you should use two different variables, like
\[\lim_{t\to \infty}\int_0^tf(x)dx+\lim_{s\to -\infty}\int_s^0f(x)dx\]