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

∫_0 ^1 e^(−x^2 ) dx

Answered by smridha last updated on 24/Jun/20

=((√𝛑)/2)erf(1)

Commented byAr Brandon last updated on 24/Jun/20

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

Commented bysmridha last updated on 24/Jun/20

yeah you can say that  the fact is..  erf(x)=(2/(√𝛑))∫_0 ^x e^(−t^2 ) dt  well I have defferent way wait  I post it...

Answered by smridha last updated on 24/Jun/20

=∫_0 ^1 Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!))dx  =1.Σ_(n=0) ^∞ (((−1)^n )/(n!(2n+1)))  =Σ_(n=0) ^∞ ((1/2)/((n+(1/2)))).(((−1)^n )/(n!))  =Σ_(n=0 ) ^∞ ((((1/2))_n )/(((3/2))_n )).(((−1)^n )/(n!))  =M((1/2),(3/2),−1)or _1 F_1 ((1/2);(3/2);−1)  this is called confluent hypergeometric  f^n .  note:(a)=(((a)_n )/((a+1)_n ))(a+n)[pochhammer symbol]

Commented byAr Brandon last updated on 24/Jun/20

Thanks so very much💞