Question

evaluate the integral from negative infinity to infinity of x^3 dx

Answer 1

hint: x^3 is an odd function

Answer 2

@jim_thompson5910 that doesn't seem to relevant here. The integral doesn't converge (although the principal value integral does...).

Answer 3

the symmetry is very relevant

Answer 4

although, it seems like you'd have infinity - infinity, so I'm not sure

Answer 5

Yeah the symmetry tells us that the PV integral converges to zero, but not the regular integral (which I'm assuming is what's being asked for).

Answer 6

it will be converted to even function after integral @SithsAndGiggles....@jim_thompson5910 is right

Answer 7

\[\int\limits_{-\infty}^{\infty}x^3*dx=[\frac{ x^4 }{ 4 }]\]

Answer 8

with given limit

Answer 9

\[\frac{ (\infty)^4 }{ 4 }-\frac{ (-\infty)^4 }{ 4 }\]

Answer 10

\[(-\infty)^4=(\infty)^4\]

Answer 11

so result will be zero

Answer 12

@gorv Then I suppose
\[\lim_{x\to\infty}(x^3-x^2)~\text{ is also }0~?\]

Answer 13

Because as you say, \(\infty-\infty=0\).

Answer 14

well there is a difference between limit and integration

Answer 15

integration calculate area ..and both side having equal area

Answer 16

and for your kind info

Answer 17

\[\infty^4=\infty^4\]

Answer 18

\[\infty^3>\infty^2\]

Answer 19

well check again ..

Answer 20

\[\lim_{x \rightarrow \infty} (x^2-(-x)^2)\]

Answer 21

will be zero for sure ...

Answer 22

@SithsAndGiggles

Answer 23

\[\large\int_{-\infty}^\infty x^3~dx=\lim_{a\to-\infty}\int_a^c x^3~dx+\lim_{b\to\infty}\int_c^b x^3~dx\]
for some valid \(c\) in the domain of \(\large x^3\) (which would be any real number). This is how we treat improper integrals like this one - we must treat the limits of integration as limits themselves.

Besides, integrals are infinite sums of infinitesimal areas, so exactly how is this different from a limit when the definition of the definite integral IS a limit?

As for my counter-example (which apparently didn't land) - my point was that \(\infty-\infty\) is an indeterminate form. We can't say that one infinity necessarily has the same magnitude as another infinity, to put it in layman's terms.

If my reasoning doesn't hold any merit, then perhaps Wolfram's calculation will convince you otherwise: http://www.wolframalpha.com/input/?i=Integrate%5Bx%5E3%2C%7Bx%2C-Infinity%2CInfinity%7D%5D