$$ \int^1_0 e^{-x^2} \, \mathrm{d} x $$
It seems that needs more than 30 word to make a discription of this problem,but actually that all included in the title. Thank you for your answer.
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1$\int_0^1e^{-x^2}dx=\sqrt\pi{\text erf}(1)/2$ and it appears be that no closed form for ${\text erf}(1)$ is known (http://math.stackexchange.com/questions/355307/non-trivial-values-of-error-function-operatornameerfx). Maybe integrating term by term and using the trick http://en.wikipedia.org/wiki/Proof_that_e_is_irrational will work. – Martín-Blas Pérez Pinilla May 06 '14 at 10:22
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@Nameless，@Martín-Blas Pérez Pinilla,thank you ! – user148210 May 06 '14 at 10:34
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Please edit the question into the body --- it shouldn't just be in the title. – Gerry Myerson May 06 '14 at 11:22
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Excuse me,it's the first time I use this web. – user148210 May 06 '14 at 11:23
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Related on MathOverflow: Is the integral of $e^{-x^2}$ from $0$ to $1$ known to be irrational? – The Amplitwist Dec 26 '23 at 19:23
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EDIT: the proof isn't correct, the factors $(2b-1), (2b-3),\cdots$ spoil it. I will try to fix it. $$ I=\int_0^1e^{-x^2}\,dx= \int_0^1\sum_{n=0}^\infty{(-x^2)^n\over n!}\,dx= \int_0^1\sum_{n=0}^\infty{(-1)^nx^{2n}\over n!}\,dx = \\ = \sum_{n=0}^\infty\int_0^1{(-1)^nx^{2n}\over n!}\,dx= \sum_{n=0}^\infty{(-1)^n\over(2n+1)n!}. $$ If $I={a\over b}\in\Bbb Q$, consider $$ X=(2b+1)b!\left(I-\sum_{n=0}^b{(-1)^n\over(2n+1)n!}\right)= \\ = a(2b+1)(b-1)!-\sum_{n=0}^b{(-1)^n(2b+1)b!\over(2n+1)n!}= \sum_{n=b+1}^\infty{(-1)^n(2b+1)b!\over(2n+1)n!}, $$ and the $|\ |$ of the last series will be $<1$ (is alternating).
Martín-Blas Pérez Pinilla
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@user, are you familiar with the standard proof that $e$ is irrational? This is a simple-but-clever variation on that. – Gerry Myerson May 06 '14 at 11:23