0-1-e-x-2-dx-

Question Number 100047 by Ar Brandon last updated on 24/Jun/20

\int_{0}^{1} e^{- x^{2}} dx

Answered by smridha last updated on 24/Jun/20

= \frac{\sqrt{π}}{2} e r f (1)

Commented by Ar Brandon last updated on 24/Jun/20

Thanks, but how do we arrive there ? Or is it just a theory ?

Commented by smridha last updated on 24/Jun/20

y e a h y o u c a n s a y t h a t

t h e f a c t i s . .

e r f (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t

w e l l I h a v e d e f f e r e n t w a y w a i t

I p o s t i t \dots

Answered by smridha last updated on 24/Jun/20

= \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n}}{n!} d x

= 1 . \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n! (2 n + 1)}

= \underset{n = 0}{\sum^{\infty}} \frac{\frac{1}{2}}{(n + \frac{1}{2})} . \frac{{(- 1)}^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{{(\frac{3}{2})}_{n}} . \frac{{(- 1)}^{n}}{n!}

= M (\frac{1}{2}, \frac{3}{2}, - 1) o r_{1} F_{1} (\frac{1}{2}; \frac{3}{2}; - 1)

t h i s i s c a l l e d c o n f l u e n t h y p e r g e o m e t r i c

f^{n} .

n o t e : (a) = \frac{{(a)}_{n}}{{(a + 1)}_{n}} (a + n) [p o c h h a m m e r s y m b o l]

Commented by Ar Brandon last updated on 24/Jun/20

Thanks so very much��