integrating e^x^2Watch

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#1
not sure how to integrate this

EDIT sorry meant to say
0
11 years ago
#2
Make a double integral in polar coordinates out of it, i.e. start off with then .

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

edit2: Yes, sorry I read your post as e^-x^2, e^x^2 is not 'nice' with these limits.
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11 years ago
#3
Surely it should be ?

Otherwise it will be infinite...
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11 years ago
#4
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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#5
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11 years ago
#6
Well, in that case, go with what TheRandomer said and show your attempt.

Spoiler:
Show
sub u=-x^2
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11 years ago
#7
(Original post by Anon89)
EDIT sorry meant to say
That simplfies things.

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

Spoiler:
Show

Differentiating gives , which is a factor of -2 out from what you want to end up with. You therefore need to divide by -2, giving (you can differentiate this to check that it works)
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11 years ago
#8
(Original post by Daniel Freedman)
That simplfies things.

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

Spoiler:
Show

Differentiating gives , which is a factor of -2 out from what you want to end up with. You therefore need to divide by -2, giving (you can differentiate this to check that it works)
Yep, easiest method to follow. Thats how we did it.
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11 years ago
#9
(Original post by Anon89)
not sure how to integrate this

EDIT sorry meant to say
By the way, the latex code for the infinity sign is \infty.
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11 years ago
#10
heh...
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#11
Thankyou, I was being very stupid and kept thinking i needed to integrate by parts, thats what was confusing me
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