0-e-x-2-dx-

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Question Number 202842 by esmaeil last updated on 04/Jan/24

$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Commented by esmaeil last updated on 04/Jan/24

$$\:\cancel{\underline{\mathcal{W}}}{this} \\ $$$$\:\cancel{\underbrace{\mathcal{W}}} \\ $$$$ \\ $$

Answered by MathematicalUser2357 last updated on 04/Jan/24

$$=\frac{\sqrt{\pi}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{886227} \\ $$

Commented by esmaeil last updated on 04/Jan/24

$${tnanks}\:\:\left({solution}?\right) \\ $$

Commented by MathematicalUser2357 last updated on 04/Jan/24

∫e^(−x^2 ) dx=((√π)/2)erf(x)+C ∫_0 ^∞ e^(−x^2 ) dx=[((√π)/2)erf(x)]_0 ^∞ =((√π)/2)

$$\int{e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left({x}\right)+{C} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\left[\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left({x}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$

Answered by aleks041103 last updated on 04/Jan/24

∫_0 ^( ∞) e^(−x^2 ) dx=(1/2)∫_(−∞) ^( ∞) e^(−x^2 ) dx I=∫_(−∞) ^( ∞) e^(−x^2 ) dx=∫_R e^(−x^2 ) dx x↔y I=∫_R e^(−y^2 ) dy ⇒I^2 =(∫_R e^(−x^2 ) dx)(∫_R e^(−y^2 ) dy)= =∫_R ∫_R e^(−x^2 ) e^(−y^2 ) dxdy= =∫∫_R^2 e^(−(x^2 +y^2 )) dxdy Change to polar coord r^2 =x^2 +y^2 dxdy=rdrdθ ⇒I^2 =∫∫_R^2 e^(−(x^2 +y^2 )) dxdy= =∫_0 ^( ∞) (∫_0 ^( 2π) re^(−r^2 ) dθ)dr= =(∫_0 ^( 2π) dθ)(∫_0 ^( ∞) re^(−r^2 ) dr)= =2π (1/2)∫_0 ^( ∞) e^(−r^2 ) d(r^2 )= =π(e^(−0^2 ) −lim_(r→∞) e^(−r^2 ) )=π ⇒I=(√π) ∫_0 ^( ∞) e^(−x^2 ) dx=(I/2) ⇒∫_0 ^( ∞) e^(−x^2 ) dx=((√π)/2)

$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${I}=\int_{−\infty} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\int_{\mathbb{R}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${x}\leftrightarrow{y} \\ $$$${I}=\int_{\mathbb{R}} {e}^{−{y}^{\mathrm{2}} } {dy} \\ $$$$\Rightarrow{I}^{\mathrm{2}} =\left(\int_{\mathbb{R}} {e}^{−{x}^{\mathrm{2}} } {dx}\right)\left(\int_{\mathbb{R}} {e}^{−{y}^{\mathrm{2}} } {dy}\right)= \\ $$$$=\int_{\mathbb{R}} \int_{\mathbb{R}} {e}^{−{x}^{\mathrm{2}} } {e}^{−{y}^{\mathrm{2}} } {dxdy}= \\ $$$$=\int\int_{\mathbb{R}^{\mathrm{2}} } {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy} \\ $$$${Change}\:{to}\:{polar}\:{coord} \\ $$$${r}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${dxdy}={rdrd}\theta \\ $$$$\Rightarrow{I}^{\mathrm{2}} =\int\int_{\mathbb{R}^{\mathrm{2}} } {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy}= \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \left(\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {re}^{−{r}^{\mathrm{2}} } {d}\theta\right){dr}= \\ $$$$=\left(\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {d}\theta\right)\left(\int_{\mathrm{0}} ^{\:\infty} {re}^{−{r}^{\mathrm{2}} } {dr}\right)= \\ $$$$=\mathrm{2}\pi\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} {e}^{−{r}^{\mathrm{2}} } {d}\left({r}^{\mathrm{2}} \right)= \\ $$$$=\pi\left({e}^{−\mathrm{0}^{\mathrm{2}} } −\underset{{r}\rightarrow\infty} {{lim}}\:{e}^{−{r}^{\mathrm{2}} } \right)=\pi \\ $$$$\Rightarrow{I}=\sqrt{\pi} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{{I}}{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$

Commented by esmaeil last updated on 04/Jan/24

$${thanks} \\ $$

Answered by Mathspace last updated on 05/Jan/24

I=∫_0 ^∞ e^(−x^2 ) dx=_(x=(√t)) ∫_0 ^∞ e^(−t) (dt/(2(√t))) =(1/2)∫_0 ^∞ t^(−(1/2)) e^(−t) dt =(1/2)∫_0 ^∞ t^((1/2)−1) e^(−t) dt=(1/2)Γ((1/2)) =(1/2)(√π)=((√π)/2)

$${I}=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}=_{{x}=\sqrt{{t}}} \:\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} \frac{{dt}}{\mathrm{2}\sqrt{{t}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {t}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{t}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {e}^{−{t}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$

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