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e-x-2-dx-can-we-get-a-close-form-of-this-integral-or-analytic-solution-




Question Number 64579 by Tawa1 last updated on 19/Jul/19
∫ e^x^2   dx    can we get a close form of this integral or analytic solution
$$\int\:\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{close}\:\mathrm{form}\:\mathrm{of}\:\mathrm{this}\:\mathrm{integral}\:\mathrm{or}\:\mathrm{analytic}\:\mathrm{solution} \\ $$
Commented by mathmax by abdo last updated on 19/Jul/19
at form of serie  ∫ e^x^2  dx =∫ Σ_(n=0) ^∞  (x^(2n) /(n!))dx =Σ_(n=0) ^∞  (1/(n!))∫ x^(2n) dx   =Σ_(n=0) ^∞  (x^(2n+1) /((2n+1)n!)) +c    with radius of convergence R=+∞
$${at}\:{form}\:{of}\:{serie} \\ $$$$\int\:{e}^{{x}^{\mathrm{2}} } {dx}\:=\int\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{{n}!}{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\int\:{x}^{\mathrm{2}{n}} {dx}\: \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\:+{c}\:\:\:\:{with}\:{radius}\:{of}\:{convergence}\:{R}=+\infty \\ $$

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