....nice calculus... evaluation: Ω=∫_0 ^( ∞) t^2 e^(−t) ln(t)dt=?? solution: f(s)=∫_0 ^( ∞) t^(2+s) e^(−t) dt Ω=f ′(0)=... f(s)=Γ(3+s) f ′(s)=Γ′(3+s)=ψ(3+s)Γ(3+s) f ′(0)=ψ(3)Γ(3)=2((3/2) −γ) =3−2γ ∴ Ω=∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 129972 by mnjuly1970 last updated on 21/Jan/21

$. . . . n i c e c a l c u l u s . . .$ $e v a l u a t i o n :$ $Ω = \int_{0}^{\infty} t^{2} e^{- t} l n (t) d t = ? ?$ $s o l u t i o n :$ $f (s) = \int_{0}^{\infty} t^{2 + s} e^{- t} d t$ $Ω = f^{'} (0) = . . .$ $f (s) = Γ (3 + s)$ $f^{'} (s) = Γ^{'} (3 + s) = ψ (3 + s) Γ (3 + s)$ $f^{'} (0) = ψ (3) Γ (3) = 2 (\frac{3}{2} - γ)$ $= 3 - 2 γ$ $∴ Ω = \int_{0}^{\infty} t^{2} e^{- t} l n (t) = 3 - 2 γ . . .$
	Answered by Ar Brandon last updated on 21/Jan/21
	$Ω=∫_0 ^∞ t^2 e^(−t) ln(t)dt=∫_0 ^∞ t^2 e^(−t) ∫_0 ^∞ ((e^(−y) −e^(−ty) )/y)dydt =∫_0 ^∞ ∫_0 ^∞ t^2 {((e^(−t−y) −e^(−ty−t) )/y)}dydt =∫_0 ^∞ ∫_0 ^∞ (1/y){t^2 e^(−t) ∙e^(−y) −t^2 e^(−(y+1)t) }dtdy =∫_0 ^∞ (1/y)Γ(3)e^(−y) dy−∫_0 ^∞ ∫_0 ^∞ (1/y)∙((m^2 e^(−m) )/((y+1)^3 ))dmdy =2∫_0 ^∞ (e^(−y) /y)dy−∫_0 ^∞ ((Γ(3))/(y(y+1)^3 ))dy =2{e^(−y) lny+∫e^(−y) lnydy}_0 ^∞ −2∫_0 ^∞ {(1/y)−(1/(y+1))−(1/((y+1)^2 ))−(1/((y+1)^3 ))}dy =−2γ+[2e^(−y) lny−2lny+2ln(y+1)−(2/(y+1))−(1/((y+1)^2 ))]_0 ^∞ =−2γ+3$
	$Ω = \int_{0}^{\infty} t^{2} e^{- t} \ln (t) dt = \int_{0}^{\infty} t^{2} e^{- t} \int_{0}^{\infty} \frac{e^{- y} - e^{- ty}}{y} dydt$ $= \int_{0}^{\infty} \int_{0}^{\infty} t^{2} {\frac{e^{- t - y} - e^{- ty - t}}{y}} dydt$ $= \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{y} {t^{2} e^{- t} \cdot e^{- y} - t^{2} e^{- (y + 1) t}} dtdy$ $= \int_{0}^{\infty} \frac{1}{y} Γ (3) e^{- y} dy - \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{y} \cdot \frac{m^{2} e^{- m}}{{(y + 1)}^{3}} dmdy$ $= 2 \int_{0}^{\infty} \frac{e^{- y}}{y} dy - \int_{0}^{\infty} \frac{Γ (3)}{y {(y + 1)}^{3}} dy$ $= 2 {e^{- y} lny + \int e^{- y} lnydy}_{0}^{\infty}$ $- 2 \int_{0}^{\infty} {\frac{1}{y} - \frac{1}{y + 1} - \frac{1}{{(y + 1)}^{2}} - \frac{1}{{(y + 1)}^{3}}} dy$ $= - 2 γ + {[{2 e}^{- y} lny - 2 lny + 2 \ln (y + 1) - \frac{2}{y + 1} - \frac{1}{{(y + 1)}^{2}}]}_{0}^{\infty}$ $= - 2 γ + 3$
	Commented by mnjuly1970 last updated on 21/Jan/21

	$n i c e v e r y n i c e$ $t h a n k y o u m r b r a n d o n . . .$
	Commented by Ar Brandon last updated on 21/Jan/21
	You're welcome Sir��
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum