prove-that-I-0-ln-x-1-e-x-dx-1-2-ln-2-2-

Question Number 151447 by mnjuly1970 last updated on 21/Aug/21

p r o v e t h a t \dots

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

Answered by Lordose last updated on 21/Aug/21

I = \int_{0}^{\infty} \frac{\ln (x)}{1 + e^{x}} dx \overset{u = e^{- x}}{=} \int_{0}^{1} \frac{uln (\ln (\frac{1}{u}))}{1 + u} \cdot \frac{du}{u}

I = \int_{0}^{1} \frac{\ln (\ln (\frac{1}{u}))}{1 + u} du = ∣ \frac{\partial}{\partial a} \int_{0}^{1} \frac{\ln^{a} (\frac{1}{u})}{1 + u} du ∣_{a = 1}

I (a) = \int_{0}^{1} \frac{{(\ln (\frac{1}{u}))}^{a}}{1 + u} du \overset{x = - \ln (u)}{=} \int_{0}^{\infty} \frac{x^{a} e^{- x}}{1 + e^{- x}} dx

I (a) = \int_{0}^{\infty} x^{a} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{k + 1} e^{- kx} dx

I (a) \overset{y = kx}{=} \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k + 1}}{k^{a + 1}} \int_{0}^{\infty} y^{a} e^{- y} dy = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k + 1}}{k^{a + 1}} Γ (a + 1)

I (a) = η (a + 1) Γ (a + 1)

I^{'} (a) = η (a + 1) Γ (a + 1) ψ (a + 1) + η^{'} (a + 1) Γ (a + 1)

I = \lim_{a \to 0} (I^{'} (a))

Ψ (1) = - γ, η (1) = \ln (2), η^{'} (1) = γ \ln (2) - \frac{\ln^{2} (2)}{2}

Ω = - γ \ln (2) + γ \ln (2) - \frac{\ln^{2} (2)}{2} = - \frac{\ln^{2} (2)}{2} ▴ ▴ ▴

\emptyset sE

Commented by mnjuly1970 last updated on 21/Aug/21

v e r y n i c e m a s t e r l o r d o s e . .

Commented by Tawa11 last updated on 22/Aug/21

And Q 151641

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