calculate-0-e-2x-ln-1-e-3x-dx-

Question Number 141931 by mathmax by abdo last updated on 24/May/21

calculate \int_{0}^{\infty} e^{- 2 x} \ln (1 + e^{3 x}) dx

Answered by mindispower last updated on 24/May/21

l n (1 + e^{3 x}) = 3 x + l n (1 + e^{- 3 x})

= \int_{0}^{\infty} 3 {x e}^{- 2 \boldsymbol x} \boldsymbol d x + \int_{0}^{\infty} e^{- 2 x} . Σ \frac{{(- 1)}^{k}}{k + 1} e^{- 3 (1 + k) x} d x

= \int_{0}^{\infty} \frac{3}{4} {y e}^{- y} d y + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{k + 1} \int_{0}^{\infty} e^{- (5 + 3 k) x} d x

= \frac{3}{4} Γ (2) + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{k + 1} . \frac{1}{5 + 3 k}

= \frac{3}{4} - \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{2 (5 + 3 k)} + \frac{3}{2} \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{k + 1}

= \frac{3}{4} + \frac{3}{2} l n (2) + \frac{1}{2} (- \sum_{k ⩾ 0} \frac{1}{6 k + 5} + \frac{1}{6 k + 8})

= \frac{3}{2} l n (2 e^{\frac{1}{2}}) + \frac{1}{12} (\sum_{k ⩾ 0} \frac{1}{(k + \frac{5}{8}) (k + \frac{8}{6})})

= \frac{3}{2} l n (2 e^{\frac{1}{2}}) + \frac{1}{12} \frac{Ψ (\frac{8}{6}) - Ψ (\frac{5}{6})}{\frac{1}{6}}

= \frac{3}{2} l n (2 e^{\frac{1}{2}}) + \frac{1}{2} Ψ (\frac{4}{3}) - \frac{1}{2} Ψ (\frac{5}{6})

Ψ (\frac{4}{3}) = Ψ (\frac{1}{3}) + 3

Ψ (\frac{1}{3}) = - γ - l n (6) - \frac{π}{2} c o t (\frac{π}{3}) + 2 c o s (\frac{2 π}{3}) l n (s i n (\frac{π}{3}))

= - γ - l n (6) - \frac{π}{2 \sqrt{3}} - l n (\frac{\sqrt{3}}{2})

Ψ (\frac{5}{6}) = - γ - l n (12) - \frac{π}{2} c o t (\frac{5 π}{6}) + 2 c o s (\frac{2 π}{6}) l n (s i n (\frac{π}{6}))

+ 2 c o s (\frac{4 π}{6}) l n (s i n (\frac{2 π}{6}))

= - γ - l n (12) + \frac{π}{2} \sqrt{3} + l n (\frac{1}{2}) - l n (\frac{\sqrt{3}}{2})

Commented by Mathspace last updated on 24/May/21

t h s n k s s i r

Answered by mathmax by abdo last updated on 26/May/21

Φ = \int_{0}^{\infty} e^{- 2 x} \log (1 + e^{3 x}) dx by parts

Φ = {[- \frac{1}{2} e^{- 2 x} \log (1 + e^{3 x})]}_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} e^{- 2 x} \times \frac{{3 e}^{3 x}}{1 + e^{3 x}} dx

= \frac{\log 2}{2} + \frac{3}{2} \int_{0}^{\infty} \frac{e^{x}}{1 + e^{3 x}} dx but

\int_{0}^{\infty} \frac{e^{x}}{1 + e^{3 x}} dx =_{e^{x} = t} \int_{0}^{\infty} \frac{t}{1 + t^{3}} \times \frac{dt}{t} = \int_{1}^{\infty} \frac{dt}{t^{3} + 1}

F (t) = \frac{1}{t^{3} + 1} = \frac{1}{(t + 1) (t^{2} - t + 1)} = \frac{a}{t + 1} + \frac{bt + c}{t^{2} - t + 1}

a = \frac{1}{3}, \lim_{t \to + \infty} tF (t) = 0 = a + b \Rightarrow b = - \frac{1}{3}

F (0) = 1 = a + c \Rightarrow c = 1 - \frac{1}{3} = \frac{2}{3} \Rightarrow

F (t) = \frac{1}{3 (t + 1)} + \frac{- \frac{1}{3} t + \frac{2}{3}}{t^{2} - t + 1} = \frac{1}{3 (t + 1)} - \frac{1}{3} \frac{t - 2}{t^{2} - t + 1} \Rightarrow

\int_{1}^{\infty} F (t) dt = {[\frac{1}{3} \log ∣ \frac{t + 1}{\sqrt{t^{2} - t + 1}} ∣]}_{1}^{\infty} (\to 0) + \frac{2}{3} \int_{1}^{\infty} \frac{dt}{t^{2} - t + 1}

\int_{1}^{\infty} \frac{dt}{t^{2} - t + 1} = \int_{1}^{\infty} \frac{dt}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}} =_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} y \to y = \frac{2 t - 1}{\sqrt{3}}}

= \frac{4}{3} \int_{\frac{1}{\sqrt{3}}}^{\infty} \frac{1}{y^{2} + 1} \frac{\sqrt{3}}{2} dy = \frac{2}{\sqrt{3}} {[arctany]}_{\frac{1}{\sqrt{3}}}^{\infty} = \frac{2}{\sqrt{3}} (\frac{π}{2} - \frac{π}{6}) = \frac{2}{\sqrt{3}} . \frac{π}{3}

= \frac{2 π}{3 \sqrt{3}} \Rightarrow Φ = \frac{\log 2}{2} + \frac{3}{2} . \frac{2 π}{3 \sqrt{3}} \Rightarrow Φ = \frac{\log 2}{2} + \frac{π}{\sqrt{3}}