Anon89
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#1
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#1
not sure how to integrate this
\frac{2}{\sqrt{2\pi}}\int_{0}^{\  propto}e^{x^2}dx
please help with detailed steps as finding it hard to follow the solution i have

EDIT sorry meant to say
\frac{2}{\sqrt{2\pi}}\int_{0}^{\  propto}xe^{-x^2}dx
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nota bene
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Make a double integral in polar coordinates out of it, i.e. start off with I=\int_0^{\infty} e^{-x^2}dx then I^2=e^{-x^2}e^{-y^2}dxdy.

Does this make sense to you?

edit: The Randomer: Good luck with that; this (indefinite) integral is not possible to express in terms of elementary functions. You can try as much as you want, but you won't succeed:p:

edit2: Yes, sorry I read your post as e^-x^2, e^x^2 is not 'nice' with these limits.
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Dystopia
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Surely it should be e^{-x^{2}}?

Otherwise it will be infinite...
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nota bene
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Is there any chance you can link to the solution you have - because I don't really know what you want to do with this integral as I believe it diverges...
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Anon89
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please see edit, mistyped question
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nota bene
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Well, in that case, go with what TheRandomer said and show your attempt.

Spoiler:
Show
sub u=-x^2
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Daniel Freedman
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(Original post by Anon89)
EDIT sorry meant to say
\frac{2}{\sqrt{2\pi}}\int_{0}^{\  propto}xe^{-x^2}dx
That simplfies things.

Remember that  e^x integrates to  e^x . Your answer (for the indefinite integral) is going to have  e^{-x^2} in it. Differentiate  e^{-x^2} to see by what factor you are out, and you can then divide to adjust the constant.

Spoiler:
Show


Differentiating  e^{-x^2} gives  -2x e^{-x^2} , which is a factor of -2 out from what you want to end up with. You therefore need to divide by -2, giving  -\frac{1}{2} e^{-x^2} (you can differentiate this to check that it works)
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The Sherminator
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(Original post by Daniel Freedman)
That simplfies things.

Remember that  e^x integrates to  e^x . Your answer (for the indefinite integral) is going to have  e^{-x^2} in it. Differentiate  e^{-x^2} to see by what factor you are out, and you can then divide to adjust the constant.

Spoiler:
Show


Differentiating  e^{-x^2} gives  -2x e^{-x^2} , which is a factor of -2 out from what you want to end up with. You therefore need to divide by -2, giving  -\frac{1}{2} e^{-x^2} (you can differentiate this to check that it works)
Yep, easiest method to follow. Thats how we did it.
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Daniel Freedman
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(Original post by Anon89)
not sure how to integrate this
\frac{2}{\sqrt{2\pi}}\int_{0}^{\  propto}e^{x^2}dx
please help with detailed steps as finding it hard to follow the solution i have

EDIT sorry meant to say
\frac{2}{\sqrt{2\pi}}\int_{0}^{\  propto}xe^{-x^2}dx
By the way, the latex code for the infinity sign is \infty.
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TheRandomer
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heh...
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Anon89
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#11
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Thankyou, I was being very stupid and kept thinking i needed to integrate by parts, thats what was confusing me
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