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Question Number 100575 by bshahid010@gmail.com last updated on 27/Jun/20

Answered by Ar Brandon last updated on 27/Jun/20

Let I=lim_(x→∞) ∫_0 ^x e^x^2  dx=∫_0 ^∞ e^x^2  dx  ⇒I= { ((∫_0 ^∞ e^x^2  dx)),((∫_0 ^∞ e^y^2  dy)) :}  ⇒I^2 =∫_0 ^∞ ∫_0 ^∞ e^(x^2 +y^2 ) dxdy  ⇒I^2 =∫_0 ^(π/2) ∫_0 ^∞ re^r^2  drdθ=[(θ/2)]_0 ^(π/2) [(e^r^2  /1)]_0 ^∞ =lim_(r→∞) (π/4)(e^r^2  −1)  Let J=lim_(x→∞) ∫_0 ^x e^(2x^2 ) dx  ⇒J= { ((∫_0 ^∞ e^(2x^2 ) dx)),((∫_0 ^∞ e^(2y^2 ) dy)) :}  ⇒J^2 =∫_0 ^∞ ∫_0 ^∞ e^(2(x^2 +y^2 )) dxdy  ⇒J^2 =∫_0 ^(π/2) ∫_0 ^∞ re^(2r^2 ) drdθ=[(θ/4)]_0 ^(π/2) [(e^(2r^2 ) /1)]_0 ^∞ =lim_(r→∞) (π/8)(e^(2r^2 ) −1)  ⇒lim_(x→∞) (((∫_0 ^x e^x^2  dx)^2 )/(∫_0 ^x e^(2x^2 ) ))=lim_(r→∞) (((π/4)(e^r^2  −1))/(√((π/8)(e^(2r^2 ) −1))))=lim_(r→∞) ((π(1−(1/e^r^2  )))/(√(2π(1−(1/e^(2r^2 ) )))))                                      =(π/(√(2π)))=(√(π/2))

$$\mathrm{Let}\:\mathcal{I}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\Rightarrow\mathcal{I}=\begin{cases}{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}}\\{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{y}^{\mathrm{2}} } \mathrm{dy}}\end{cases}\:\:\Rightarrow\mathcal{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \mathrm{dxdy} \\ $$$$\Rightarrow\mathcal{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \mathrm{re}^{\mathrm{r}^{\mathrm{2}} } \mathrm{drd}\theta=\left[\frac{\theta}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\frac{\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } }{\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} =\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\pi}{\mathrm{4}}\left(\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } −\mathrm{1}\right) \\ $$$$\mathrm{Let}\:\mathcal{J}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{2x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\Rightarrow\mathcal{J}=\begin{cases}{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{2x}^{\mathrm{2}} } \mathrm{dx}}\\{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{2y}^{\mathrm{2}} } \mathrm{dy}}\end{cases}\:\:\Rightarrow\mathcal{J}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)} \mathrm{dxdy} \\ $$$$\Rightarrow\mathcal{J}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \mathrm{re}^{\mathrm{2r}^{\mathrm{2}} } \mathrm{drd}\theta=\left[\frac{\theta}{\mathrm{4}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\frac{\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } }{\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} =\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\pi}{\mathrm{8}}\left(\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } −\mathrm{1}\right) \\ $$$$\Rightarrow\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\right)^{\mathrm{2}} }{\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{2x}^{\mathrm{2}} } }=\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{\pi}{\mathrm{4}}\left(\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } −\mathrm{1}\right)}{\sqrt{\frac{\pi}{\mathrm{8}}\left(\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } −\mathrm{1}\right)}}=\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\pi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } }\right)}{\sqrt{\mathrm{2}\pi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } }\right)}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\sqrt{\mathrm{2}\pi}}=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$

Commented by Coronavirus last updated on 27/Jun/20

    Clean

$$\:\:\:\:\mathcal{C}{lean} \\ $$

Commented by Ar Brandon last updated on 27/Jun/20

Ouais Corona, sdk ? C'est Einstein. Avec ton nom bizarre là.��

Commented by Coronavirus last updated on 27/Jun/20

ça va je profite du savoir de ce forum

Commented by Ar Brandon last updated on 27/Jun/20

Super ��

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