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))](https://www.tinkutara.com/question/Q100592.png)
$$\mathrm{Let}\:\mathcal{I}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\Rightarrow\mathcal{I}=\begin{cases}{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}}\\{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{y}^{\mathrm{2}} } \mathrm{dy}}\end{cases}\:\:\Rightarrow\mathcal{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \mathrm{dxdy} \\ $$$$\Rightarrow\mathcal{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \mathrm{re}^{\mathrm{r}^{\mathrm{2}} } \mathrm{drd}\theta=\left[\frac{\theta}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\frac{\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } }{\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} =\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\pi}{\mathrm{4}}\left(\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } −\mathrm{1}\right) \\ $$$$\mathrm{Let}\:\mathcal{J}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{2x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\Rightarrow\mathcal{J}=\begin{cases}{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{2x}^{\mathrm{2}} } \mathrm{dx}}\\{\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{2y}^{\mathrm{2}} } \mathrm{dy}}\end{cases}\:\:\Rightarrow\mathcal{J}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)} \mathrm{dxdy} \\ $$$$\Rightarrow\mathcal{J}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \mathrm{re}^{\mathrm{2r}^{\mathrm{2}} } \mathrm{drd}\theta=\left[\frac{\theta}{\mathrm{4}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\frac{\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } }{\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} =\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\pi}{\mathrm{8}}\left(\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } −\mathrm{1}\right) \\ $$$$\Rightarrow\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\right)^{\mathrm{2}} }{\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{\mathrm{2x}^{\mathrm{2}} } }=\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{\pi}{\mathrm{4}}\left(\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } −\mathrm{1}\right)}{\:\sqrt{\frac{\pi}{\mathrm{8}}\left(\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } −\mathrm{1}\right)}}=\underset{\mathrm{r}\rightarrow\infty} {\mathrm{lim}}\frac{\pi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } }\right)}{\:\sqrt{\mathrm{2}\pi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2r}^{\mathrm{2}} } }\right)}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\:\sqrt{\mathrm{2}\pi}}=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$
Commented by Coronavirus last updated on 27/Jun/20
$$\:\:\:\:\mathcal{C}{lean} \\ $$
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 ��