nice-calculus-evaluation-0-t-2-e-t-ln-t-dt-solution-f-s-0-t-2-s-e-t-dt-f-0-f-s-3-s-f-s-3-s-3-s

Question Number 129972 by mnjuly1970 last updated on 21/Jan/21

\dots . n i c e c a l c u l u s \dots

e v a l u a t i o n :

Ω = \int_{0}^{\infty} t^{2} e^{- t} l n (t) d t = ? ?

s o l u t i o n :

f (s) = \int_{0}^{\infty} t^{2 + s} e^{- t} d t

Ω = f^{'} (0) = \dots

f (s) = Γ (3 + s)

f^{'} (s) = Γ^{'} (3 + s) = ψ (3 + s) Γ (3 + s)

f^{'} (0) = ψ (3) Γ (3) = 2 (\frac{3}{2} - γ)

= 3 - 2 γ

∴ Ω = \int_{0}^{\infty} t^{2} e^{- t} l n (t) = 3 - 2 γ \dots

Answered by Ar Brandon last updated on 21/Jan/21

Ω = \int_{0}^{\infty} t^{2} e^{- t} \ln (t) dt = \int_{0}^{\infty} t^{2} e^{- t} \int_{0}^{\infty} \frac{e^{- y} - e^{- ty}}{y} dydt

= \int_{0}^{\infty} \int_{0}^{\infty} t^{2} {\frac{e^{- t - y} - e^{- ty - t}}{y}} dydt

= \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{y} {t^{2} e^{- t} \cdot e^{- y} - t^{2} e^{- (y + 1) t}} dtdy

= \int_{0}^{\infty} \frac{1}{y} Γ (3) e^{- y} dy - \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{y} \cdot \frac{m^{2} e^{- m}}{{(y + 1)}^{3}} dmdy

= 2 \int_{0}^{\infty} \frac{e^{- y}}{y} dy - \int_{0}^{\infty} \frac{Γ (3)}{y {(y + 1)}^{3}} dy

= 2 {e^{- y} lny + \int e^{- y} lnydy}_{0}^{\infty}

- 2 \int_{0}^{\infty} {\frac{1}{y} - \frac{1}{y + 1} - \frac{1}{{(y + 1)}^{2}} - \frac{1}{{(y + 1)}^{3}}} dy

= - 2 γ + {[{2 e}^{- y} lny - 2 lny + 2 \ln (y + 1) - \frac{2}{y + 1} - \frac{1}{{(y + 1)}^{2}}]}_{0}^{\infty}

= - 2 γ + 3

Commented by mnjuly1970 last updated on 21/Jan/21

n i c e v e r y n i c e

t h a n k y o u m r b r a n d o n \dots

Commented by Ar Brandon last updated on 21/Jan/21

You're welcome Sir��