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Question-200802




Question Number 200802 by mnjuly1970 last updated on 23/Nov/23
Answered by witcher3 last updated on 23/Nov/23
∫_0 ^∞ e^(−x^2 ) dx=(1/2)∫_0 ^∞ t^((1/2)−1) e^(−t) dt,x^2 =t  =((Γ((1/2)))/2)  ∅=(((Γ((1/2)))/2))^3 =((π(√π))/8)
$$\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt},\mathrm{x}^{\mathrm{2}} =\mathrm{t} \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$\emptyset=\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\pi\sqrt{\pi}}{\mathrm{8}} \\ $$

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