1-0-x-1-x-1-lnx-dx-

Question Number 179844 by mathlove last updated on 03/Nov/22

{\int_{0}}^{1} \frac{x - 1}{(x + 1) l n x} d x = ?

Answered by Peace last updated on 03/Nov/22

\int_{0}^{1} \frac{x - x^{s}}{(x + 1) l n (x)} = f (s)

f (0) = Δ = \int_{0}^{1} \frac{x - 1}{l n (x) (x + 1)}

f (1) = 0

f^{'} (s) = \int_{0}^{1} \frac{x^{s}}{1 + x} d x = \int_{0}^{1} \frac{x^{s} - x^{s + 1}}{1 - x^{2}} d x

= \int_{0}^{1} \frac{t^{\frac{s - 1}{2}} - t^{\frac{s}{2}}}{2 (1 - t)} d t = \frac{1}{2} (Ψ (\frac{s}{2} + 1) - Ψ (\frac{s + 1}{2}))

Ψ (s) = - γ + \int_{0}^{1} \frac{1 - x^{s - 1}}{1 - x} d x

Ψ (s) = \frac{Γ^{'} (s)}{Γ (s)}

f (s) = l n (\frac{Γ (\frac{s + 2}{2})}{Γ (\frac{s + 1}{2})}) + c

f (1) = l n (\frac{Γ (\frac{3}{2})}{Γ (1)}) + c = 0 \Rightarrow c = l n (\frac{1}{Γ (\frac{3}{2})}) = l n (\frac{2}{\sqrt{π}})

Δ = f (0) = l n (\frac{Γ (1)}{Γ (\frac{1}{2})}) + l n (\frac{2}{\sqrt{π}}) = l n (\frac{1}{\sqrt{π}}) + l n (\frac{2}{\sqrt{π}}) = l n (\frac{2}{π})

Facebook Tweet Pin