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))

Commented byCoronavirus last updated on 27/Jun/20

    Clean

Commented byAr Brandon last updated on 27/Jun/20

Ouais Corona, sdk ? C'est Einstein. Avec ton nom bizarre lร .๐Ÿ˜€

Commented byCoronavirus last updated on 27/Jun/20

รงa va je profite du savoir de ce forum

Commented byAr Brandon last updated on 27/Jun/20

Super ๐Ÿ˜‰