Question

help please

determine whether each integral is convergent or divergent. evalute those that are convergent

Answer 1

\[\int\limits_{-\infty}^{\infty} xe ^{-x^2}\]

Answer 2

@Astrophysics

Answer 3

@IrishBoy123

Answer 4

There are several ways to think about this. You can either do a substitution \(x^2=u\) or you can admire that it's an odd function which means the left half is the negative of the right half, so the whole thing cancels out to zero.

Answer 5

The suggestion above to use the odd function is not correct. Take for example the divergent series
\[
\int_{-\infty}^{\infty} x dx
\]

Answer 6

I mean divergent integral

Answer 7

@eliesaab
\(\large \int_{-\infty}^{\infty} x dx\)


\(\large = \lim\limits_{t \to \infty} \left[ \dfrac{x^2}{2} \right]_{-t}^{t}\)



Whereas @marcelie

\( \large \int\limits_{-\infty}^{\infty} x e ^{-x^2} dx \)

\( \large = -\dfrac{1}{2} \lim\limits_{t \to \infty} \left [ e ^{-x^2} \right ]_{-t}^{ t }\)

@jango_IN_DTOWN posted something uber-similar today


@HolsterEmission ?!?!


i can't believe that just changing the form of the integral [basically, using limit language] just makes it alright; but this seems to me to be the case.

Answer 8

For the definition of improper integral it should
be \[\lim_{t->\infty, c->\infty} \left[\frac {x^2}2\right]^t_{-c}
\]
The above limit does not exist.

Answer 9

You cannot restrict only to the case t=c

Answer 10

@IrishBoy123

Answer 11

i think i figured it out lol

Answer 12

I agree with @eliesaab - in general, the argument that the function is odd is not enough to determine that the integral converges. You also need the fact that \(f(x)\to0\) as \(x\to\pm\infty\) (and \(f\) must be at least piecewise continuous, I would imagine), similar to how \(\sum\limits_na_n\) has the necessary (but not sufficient) condition \(a_n\to0\) in order to converge.

Answer 13

tuvm