Can you solve this Veeky Forums?

Even if the domain was specified been curves have finite area.

...

This......

>e is a variable

[eqn]\int e^{-x^2}de={{e^{1-x^2}}\over{1-x^2}}+C[/eqn]

Well, I can solve it if it's a definite integral over all the reals. In that case it's [math] \sqrt{\pi} [/math] .

>definite integral over all the reals
Don't you mean indefinite integral?

Is there any way to 'prove' that a function has no elementary antiderivative? And how can you find a 'special' function (erf) to write it down?

About 1.77263...
Since there is no dx I just summed e^-x^2 from negative infinity to infinity (?).

Yea or sqrt of pi...

[math]\int_\mathcal{M}\;d\alpha=\oint_{\delta \mathcal{M}} \;\alpha[/math]

no you fuckstick

do you even know the difference between definite and indefinite integrals?

>And how can you find a 'special' function (erf) to write it down?
That's easy. You define the function derp(x) as the integral from 0 to x of f(a) da, for your tricky function f. Boom, you now have an antiderivative.

Yes, of course that's cheating, but that's pretty much how functions like erf come to be.

...

>It's neither riemann nor lesbegue integrable function.
IT IS A CONTINUOUS FUNCTION AND THE INTEGRAL OVER R IS FINITE.

HOW FUCKING MORE REIMANN INTEGRABLE COULD IT GET????????

It just doesn't have an analytical anti-derivative, thats all.

>And how can you find a 'special' function (erf) to write it down?
Just define it as the anti derivative, which is known to exist LOL.

now, can you solve this for x Veeky Forums?

-2/3 you massive brainlet

Come on Veeky Forums, it's three time easier than triple integrals ;^)

You didn't define your notation properly, it's not for me to make assumptions

I found the autist

What if 2 was the variable?

No, I meant the integral [math] \int_{-\infty}^{\infty} e^{-x^2} dx [/math]

you mean
[math]\lim_{a \to -\infty} \lim_{b \to \infty} \int _a^b{e^{-x^2}dx} [/math]
right?

No need to write that, as it converges

Why not just say -a instead of defining b with a separate limit?

[math] \int e^{-x^2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1) n!} [/math]

Now, get off my board

Jesus I'm not even a STEM major and I can do this shit.

You square the whole value thus making a double integral. Then change everything into polar coordinates via the Jacobian. Integration leads to a derivation of an altered version of the standard normal distribution.

It's a joy to do that problem. All these bullshit answers make me upset.

Also the derivation can be found in Probability Theory by Hoel Port and Stone

>call other people brainlets
>forget about other solution
Bitch please

>I'm not even a STEM major
Yeah, we can tell. Its the indefinite integral you absolute brainlet

This is the only proper answer in this thread if you dont want to use the error function like
did

Whats that snake sign in the start?

That "S" stands for "Sum," but it's a really big, really small sum.

It's called a snaek doggo
It makes the equation that follows "cute"

(sqrt(pi)/2 ) erf(x) + constant

d2=0

wtf is r

You know you integrate with respect to x^2 and what not, right?

>square
>combine
>polar
>???
>win

en.wikipedia.org/wiki/Error_function

en.m.wikipedia.org/wiki/Liouville's_theorem_(differential_algebra)

not without [math]d\hat\zeta[/math] at the end

It's e. Check mate.

It's √(π)

Try solving it with the wonders of complex analysis. You can solve wack ass integrals in R with Cauchy's residue theorem

Could U-Substitution work here? Why or why not?

define x = constant
then its (e^(-(x^2)))(variable) + constant

is this a joke