prove that ∫_0 ^x e^(−t^2 ) dt =((√π)/2) −(e^(−x^2 ) /(√π)) ∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 )) dt with x>0


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Question Number 31105 by abdo imad last updated on 02/Mar/18

$p r o v e t h a t \int_{0}^{x} e^{- t^{2}} d t = \frac{\sqrt{π}}{2} - \frac{e^{- x^{2}}}{\sqrt{π}} \int_{0}^{\infty} \frac{e^{- x^{2} t^{2}}}{1 + t^{2}} d t w i t h x > 0$
Commented byabdo imad last updated on 05/Mar/18

$l e t p u t f (x) = \int_{0}^{\infty} \frac{e^{- x^{2} t^{2}}}{1 + t^{2}} d t w i t h x > 0 a f t e r v e r i f y i n g t h e$ $c o n d i t i o n s o f d e r i v a l i t y o f f o n] 0, + \infty [w e g e t$ $f^{^{'}} (x) = - \int_{0}^{\infty} \frac{2 t^{2} x e^{- x^{2} t^{2}}}{1 + t^{2}} d t = - 2 x \int_{0}^{\infty} \frac{t^{2} e^{- x^{2} t^{2}}}{1 + t^{2}} d t$ $= - 2 x \int_{0}^{\infty} \frac{(1 + t^{2} - 1) e^{- x^{2} t^{2}}}{1 + t^{2}} d t = - 2 x \int_{0}^{\infty} e^{- x^{2} t^{2}} d t + 2 x \int_{0}^{\infty} \frac{e^{- x^{2} t^{2}}}{1 + t^{2}} d t$ $= 2 x f (x) - 2 x \int_{0}^{\infty} e^{- {(x t)}^{2}} d t = 2 x f (x) - 2 x \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{x}$ $= 2 x f (x) - \sqrt{π} \Rightarrow f^{^{'}} (x) - 2 x f (x) = - \sqrt{π}$ $e . h \Rightarrow f^{^{'}} - 2 x f = 0 \Rightarrow \frac{f^{^{'}}}{f} = 2 x \Rightarrow l n ∣ f ∣= x^{2} \Rightarrow f (x) = k e^{x^{2}}$ $m v c m e t h o d \Rightarrow k^{^{'}} e^{x^{2}} + 2 k x e^{x^{2}} - 2 x k e^{x^{2}} = - \sqrt{π} \Rightarrow$ $k^{^{'}} e^{x^{2}} = - \sqrt{π} \Rightarrow k^{^{'}} = - \sqrt{π} e^{- x^{2}} \Rightarrow k (x) = \int_{0}^{x} - \sqrt{π} e^{- t^{2}} d t + λ$ $λ = k (0) = f (0) = \frac{π}{2} \Rightarrow k (x) = \frac{π}{2} - \sqrt{π} \int_{0}^{x} e^{- t^{2}} d t \Rightarrow$ $\int_{0}^{\infty} \frac{e^{- x^{2} t^{2}}}{1 + t^{2}} d t = (\frac{π}{2} - \sqrt{π} \int_{0}^{x} e^{- t^{2}} d t) e^{x^{2}} \Rightarrow$ $e^{- x^{2}} \int_{0}^{\infty} \frac{e^{- x^{2} t^{2}}}{1 + t^{2}} d t = \frac{π}{2} - \sqrt{π} \int_{0}^{x} e^{- t^{2}} d t \Rightarrow$ $\sqrt{π} \int_{0}^{x} e^{- t^{2}} d t = \frac{π}{2} - e^{- x^{2}} \int_{0}^{\infty} \frac{e^{- x^{2} t^{2}}}{1 + t^{2}} d t \Rightarrow$ $\int_{0}^{x} e^{- t^{2}} d t = \frac{\sqrt{π}}{2} - \frac{e^{- x^{2}}}{\sqrt{π}} \int_{0}^{\infty} \frac{e^{- x^{2} t^{2}}}{1 + t^{2}} d t .$
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