advsnced-calculus-calculate-0-e-x-ln-1-1-x-dx-

Question Number 129418 by mnjuly1970 last updated on 15/Jan/21

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

c a l c u l a t e : Ω = \int_{0}^{\infty} e^{- \sqrt{x}} l n (1 + \frac{1}{\sqrt{x}}) d x

Answered by Dwaipayan Shikari last updated on 15/Jan/21

\int_{0}^{\infty} e^{- \sqrt{x}} l o g (1 + \sqrt{x}) - \int_{0}^{\infty} e^{- \sqrt{x}} l o g (\sqrt{x}) d x

= 2 \int_{0}^{\infty} {t e}^{- t} l o g (1 + t) - 2 \int_{0}^{\infty} {t e}^{- t} l o g (t)

= 2 - 2 (Γ^{'} (2)) = 2 - 2 (- γ + 1) = 2 γ

Commented by mnjuly1970 last updated on 15/Jan/21

e x c e l l e n t \dots g r a t e f u l . .

Answered by mindispower last updated on 15/Jan/21

\sqrt{x} = t

\Rightarrow \int_{0}^{\infty} e^{- t} l n (1 + \frac{1}{t}) . 2 t d t

\int l n (1 + t) {t e}^{- t} d t b y p a r t

\int {t e}^{- t} d t = - (t + 1) e^{- t} d t

\int_{0}^{\infty} l n (1 + t) {t e}^{- t} d t = {[- (t + 1) e^{- t} l n (1 + t)]}_{0}^{\infty} + \int \frac{(t + 1) e^{- t}}{t + 1} d t

= \int_{0}^{\infty} e^{- t} d t = Γ (1) = 1

\int_{0}^{\infty} {t e}^{- t} l n (t) d t = \partial_{x} \int_{0}^{\infty} t^{x - 1} e^{- t} d t ∣_{x = 2} = Γ^{'} (2) = Γ (2) Ψ (2)

= Ψ (2) = 1 + Ψ (1) = 1 - γ

\int_{0}^{\infty} e^{- t} l n (1 + \frac{1}{t}) . 2 t d t = 2 \int_{0}^{\infty} e^{- t} l n (1 + t) t d t - 2 \int_{0}^{\infty} e^{- t} t l n (t) d t

= 2.1 - 2 (1 - γ) = 2 γ

Commented by mnjuly1970 last updated on 16/Jan/21

v e r y n i c e a s a l w a y s \dots

Answered by Lordose last updated on 16/Jan/21

Ω = \int_{0}^{\infty} e^{- \sqrt{x}} \ln (1 + \frac{1}{\sqrt{x}}) dx

u = \sqrt{x} \Rightarrow dx = 2 udu

Ω = 2 \int_{0}^{\infty} {ue}^{- u} \ln (1 + \frac{1}{u}) du = 2 \int_{0}^{\infty} {ue}^{- u} \ln (\frac{1 + u}{u}) du

Ω = 2 (\int_{0}^{\infty} {ue}^{- u} \ln (1 + u) du - \int_{0}^{\infty} {ue}^{- u} \ln (u) du)

Φ = \int_{0}^{\infty} {ue}^{- u} \ln (1 + u) du \overset{IBP}{=} ∣ - e^{- u} (1 + u) \ln (1 + u) ∣_{0}^{\infty} + \int_{0}^{\infty} e^{- u} du

Φ = 1

Ω = 2 (Φ - \int_{0}^{\infty} {ue}^{- u} \ln (u) du)

Ω = 2 (1 - \int_{0}^{\infty} {ue}^{- u} \ln (u) du) = 2 - \frac{\partial}{\partial a} ∣_{a = 0} 2 \int_{0}^{\infty} u^{1 + a - 1} e^{- u} du

Ω = 2 - \frac{\partial}{\partial a} ∣_{a = 0} 2 Γ (1 + a) = 2 - 2 Γ (1 + a) ψ^{0} (1 + a) ∣_{a = 0} = 2 - 2 (1 - γ)

Ω = 2 γ

Where γ = Euler mascheroni constant

★ L ϕ rD \emptyset sE

Commented by mnjuly1970 last updated on 16/Jan/21

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