How can I integrate an odd function with limits from -A to A?

[stunner5000pt]

$$\int_{-\infty}^{infty} s e^{-\frac{2s^2}{N}} ds$$

how do i integrate here?? I don't think the 'trick' of differentiating wrt N would work here since the limits of integration are all space...

any ideas??

 

[StatusX]

substitution

 

[stunner5000pt]

ok this is substitution i did

let x^2 = u
then 2xdx = du

$$I = \frac{1}{2\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} e^{\frac{-u}{2 \sigma^2}} du$$

what happens in $$\left[ e^{-u} \right]_{-\infty}^{\infty} \rightarrow \infty$$

something isn't right ...?

its supposed to be zero, no?

 

[dextercioby]

The Cauchy principal value of that integral is zero and that can be seen since you're integrating an odd function on an interrval symmetric wrt zero on the real axis.

Daniel.

 

[HallsofIvy]

When you change variables, change the limits of integration too.
Setting u= x² in
$$\int_{-\infty}^{\infty} s e^{-\frac{2s^2}{N}} ds$$
(I would have been inclined to let u be the whole $\frac{2s^2}{N}$.)
does NOT give
$$I = \frac{1}{2\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} e^{\frac{-u}{2 \sigma^2}} du$$
you have the wrong limits of integration.

Actually, you don't need to use substitution at all. The integral of any odd function from -A to A is what?