If-0-e-x-2-dx-pi-2-then-0-e-ax-2-dx-a-gt-0-is-

Question Number 53695 by gunawan last updated on 25/Jan/19

If \underset{0}{\int^{\infty}} e^{- x^{2}} d x = \sqrt{\frac{π}{2}}, then \underset{0}{\int^{\infty}} e^{- {a x}^{2}} d x,

a > 0 is

Commented by Abdo msup. last updated on 25/Jan/19

f i r s t \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} (s o t h e r e i s a e r r o r a t t h e Q .)

a n d \int_{0}^{\infty} e^{- {a x}^{2}} d x =_{\sqrt{a} x = t} \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{a}}

= \frac{1}{\sqrt{a}} \int_{0}^{\infty} e^{- t^{2}} d t = \frac{1}{\sqrt{a}} \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{2 \sqrt{a}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

t = {a x}^{2} x = \frac{\sqrt{t}}{\sqrt{a}} d x = \frac{1}{2 \sqrt{a}} \times \frac{1}{\sqrt{t}} d t

\int_{0}^{\infty} e^{- t} \times \frac{d t}{2 \sqrt{a}} \times \frac{1}{\sqrt{t}}

\frac{1}{2 \sqrt{a}} \int_{0}^{\infty} e^{- t} \times t^{\frac{1}{2} - 1} d t

g a m m a f u n c t i o n \int_{0}^{\infty} e^{- t} t^{n - 1} d t = ⌈ (n)

⌈ (n + 1) = n! [f a c t o r i a l o f n]

⌈ (\frac{1}{2}) = \sqrt{π}

s o

\frac{1}{2 \sqrt{a}} \int_{0}^{\infty} e^{- t} \times t^{\frac{1}{2} - 1} d t

= \frac{1}{2 \sqrt{a}} \times \sqrt{π}

l e t c a l c u l a t e \int_{0}^{\infty} e^{- x^{2}} d x

y = x^{2} x = \sqrt{y} d x = \frac{1}{2 \sqrt{y}} d y

\int_{0}^{\infty} e^{- y} \times \frac{1}{2} \times y^{- \frac{1}{2}} d y

\frac{1}{2} \int_{0}^{\infty} e^{- y} y^{\frac{1}{2} - 1} d y

\frac{1}{2} \times ⌈ (\frac{1}{2}) = \frac{\sqrt{π}}{2}

s o i t h i n k \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} (\boldsymbol n o t \frac{\sqrt{π}}{\sqrt{2}})