Question

How do the limits of an integral change for an odd function?

Answer 1

they don't change unless you use u-sub

Answer 2

if you are integrating an odd function from \(-a\) to \(a\) then you always get zero
otherwise the limits do not change

Answer 3

Here is my total question: We have an even function f(y). We have:

an integral negative infinity to zero of y*f(y)dy + an integral from 0 to infinity of y*f(y)dy.

I need to show that this sum is zero. A hint was to sub w for -y in first integrand.

I'm trying to understand how my limits go from (negative infinity to zero) TO (zero to infinity). I have seen this on a few websites but none of them clearly explain what is happening.

Thanks!

Answer 4

Well if you have an odd function then you have:
\[\int\limits_{-\infty}^{\infty}f(x)dx=- \int\limits_{- \infty}^0 f(-x)dx+\int\limits_0^{\infty}f(x)dx\]
The the first integral you make a substitution:
\[\phi = -x \implies d \phi = - dx; \implies (-\infty,0] \rightarrow (\infty,0]\]
Giving:
\[- \int\limits_{\infty}^0 f(\phi) (- d \phi)=- \int\limits_0^{\infty}f(\phi)d \phi\]
But since x is just as good as phi they sum to zero.

Answer 5

Just so you know, if a function \(g(x)\) is odd it is not necessarily true that

\[\int\limits_{-\infty}^{\infty}g(x)dx=0\]

take for example

\[\int\limits_{-\infty}^{\infty}\frac{x}{1+x^2}dx\]

Answer 6

Well of course it can only be zero if the integral EXISTS lol. If you take that integral over any symmetric finite bound it will be zero :P
http://www.wolframalpha.com/input/?i=integral+of+x%2F%281%2Bx^2%29%2Cx%2C-10000000%2C10000000