advanced-calculus-prove-that-R-e-e-x-2x-x-2-dx-1-2-pi-2-6-6-

Question Number 130011 by mnjuly1970 last updated on 21/Jan/21

\dots a d v a n c e d c a l c u l u s \dots

p r o v e t h a t ::

Φ = \int_{R} e^{(- e^{x} + 2 x)} x^{2} d x = {(1 - γ)}^{2} + \frac{π^{2} - 6}{6}

Answered by mindispower last updated on 21/Jan/21

n o t i n t e g r a b l o v e r R

Commented by mnjuly1970 last updated on 21/Jan/21

t h a n k y o u f o r y o u r m e n t i o n

i c o r r e c t e d i t \dots

Answered by mindispower last updated on 22/Jan/21

e^{x} = t

\Leftrightarrow \int_{0}^{\infty} e^{- t} . t . {(l n (t))}^{2} d t = Φ

Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t \Rightarrow Γ^{″} (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} {l n}^{2} (t) d t

Φ = Γ^{″} (2)

Γ^{'} (x) = Γ (x) Ψ (x) \Rightarrow Γ^{″} (x) = Γ (x) Ψ^{'} (x) + Ψ^{2} (x)

Φ = Γ (2) Ψ^{'} (2) + Ψ {(2)}^{2}

= 1 . \sum_{n ⩾ 1} \frac{1}{{(n + 1)}^{2}} + {(Ψ (1) + 1)}^{2}

= ζ (2) - 1 + {(1 - γ)}^{2} = \frac{π^{2} - 6}{6} + {(1 - γ)}^{2}

Commented by mnjuly1970 last updated on 22/Jan/21

t h a n k s a l o t s i r p o w e r

g r a t e f u l \dots

Commented by mindispower last updated on 23/Jan/21

p l e a s u r s i r