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Question Number 4758 by madscientist last updated on 05/Mar/16

\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}

i s t h i s t r u e, i f s o h o w ?

Answered by Yozzii last updated on 05/Mar/16

L e t f (x) = e^{- x^{2}} (x \in R) . S i n c e f (- x) = e^{- {(- x)}^{2}} = e^{- x^{2}} = f (x)

t h e n f (x) i s a n e v e n f u n c t i o n . I n g e n e r a l,

f o r g (u) b e i n g a n e v e n f u n c t i o n, \int_{- a}^{a} g (u) d u = 2 \int_{0}^{a} g (u) d u

i f t h e i n t e g r a l e x i s t s w h o l l y o n u \in [- a, a] .

N o w l e t I (a) = \int_{- a}^{a} f (x) d x . T h e n, I (a) = 2 \int_{0}^{a} f (x) d x

s i n c e f (x) i s e v e n . L e t u s n o w d e t e r m i n e

l = \lim_{a \to \infty} I (a) = \underset{a \to \infty}{\lim 2} \int_{0}^{a} f (x) d x . \Rightarrow l = 2 \int_{0}^{\infty} e^{- x^{2}} d x .

S u b s t i t u t e u = x^{2} w h e r e x ⩾ 0 . \Rightarrow x = \sqrt{u} .

A l s o, d u = 2 x d x = 2 \sqrt{u} d x \Rightarrow \frac{1}{2} u^{- 1 / 2} d u = d x .

A t x = 0, u = 0^{2} = 0 a n d a s x \to \infty, x^{2} \to \infty

\Rightarrow u \to \infty . l h e n c e b e c o m e s

l = 2 \int_{0}^{\infty} 0.5 u^{- 1 / 2} e^{- u} d u = \int_{0}^{\infty} u^{- 1 / 2} e^{- u} d u .

T h e g a m m a f u n c t i o n Γ i s g i v e n b y t h e

f o r m Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t (x > 0) . S o t r u l y

l = Γ (1 / 2) . F r o m t h e e q u a t i o n f o r 0 < x < 1

Γ (x) Γ (1 - x) = \frac{π}{s i n π x}, l e t x = 1 / 2 .

\Rightarrow Γ^{2} (0.5) = \frac{π}{s i n 0.5 π} = π \Rightarrow Γ (0.5) = \pm \sqrt{π} .

S i n c e f (x) > 0 f o r a l l x ⩾ 0, t h e i n t e g r a l

o u g h t t o b e n o n - n e g a t i v e . H e n c e, Γ (0.5) = \sqrt{π}

o n l y . \Rightarrow l = \sqrt{π} . B u t l i s e q u i v a l e n t t o t h e

f o r m \int_{- \infty}^{\infty} e^{- x^{2}} d x . S o, \int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π} .

T h e m a t h e m a t i c a l s t a t e m e n t i s t r u e .

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