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How can I determine the anti-derivative of e^(x^2)dx ?
I've hopelessly tried everything I've learnt in my calculus class , and nothing seems to work.
So I've been reading a bit about Gaussian integrals, and the idea just crossed my mind. What if I can apply the same method to this integral ? Yet I must say , the reason behind changing a single integral into a double one never did make much sense to me. Afterall , we're working with one variable here!
I never really thought about solving this integral before, so I wrote out a solution for it to see if i could do anything. I use integration by parts and get it as an infinite sum:
where c is our constant of integration (don't forget that ) and (2n - 1)!! is a double factorial, which means the product (2n - 1)(2n - 3)(2n - 5)...5 × 3 × 1. Note that (-1)!! = 0!! = 1.
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I was trying to solve an easy differential equation in general chemistry trying multiple ways when I got stuck on this integral. I cant say I understood what u did above ( sorry , but is that advanced calculus or something? ) , but I did try expanding 2^(x^2).
I used a maclaurin series for 2^(x^2) and then integrated it. I got an infinite series : x/0! + (x^3)/(1!3) + (x^5)/(2!5) + ... + ∑x^(2n+1)/n!(2n+1) + c , n[0, ∞]
This is very different from your result. It makes no sense to me , as how am I supposed to know to where this series converges anyway ?
*correction : I used the M.S. for e^(x^2) not 2^(x^2).
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