0-1-ln-x-2-1-x-1-dx-

Question Number 96925 by M±th+et+s last updated on 05/Jun/20

\int_{0}^{1} \frac{l n (x^{2} + 1)}{x + 1} d x

Answered by mathmax by abdo last updated on 06/Jun/20

let f (α) = \int_{0}^{1} \frac{\ln (x^{2} + α)}{x + 1} dx we have f (1) = \int_{0}^{1} \frac{\ln (x^{2} + 1)}{x + 1} dx (α > 0)

f^{^{'}} (α) = \int_{0}^{1} \frac{dx}{(x^{2} + α) (x + 1)} let decompose F (x) = \frac{1}{(x + 1) (x^{2} + α)}

F (x) = \frac{a}{x + 1} + \frac{bx + c}{x^{2} + α}

we have a = \frac{1}{1 + α}

\lim_{x \to + \infty} xF (x) = 0 = a + b \Rightarrow b = - \frac{1}{1 + α}

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

F (x) = \frac{1}{(1 + α) (x + 1)} + \frac{- \frac{1}{1 + α} x + \frac{1}{1 + α}}{x^{2} + α} = \frac{1}{1 + α} {\frac{1}{x + 1} - \frac{x}{x^{2} + α} + \frac{1}{x^{2} + α}} \Rightarrow

\int F (x) = \frac{1}{1 + α} {\ln ∣ x + 1 ∣ - \frac{1}{2} \ln (x^{2} + α) + \int \frac{dx}{x^{2} + α}}

\int \frac{dx}{x^{2} + α} =_{x = \sqrt{α} u} \int \frac{\sqrt{α} du}{α (1 + u^{2})} = \frac{1}{\sqrt{α}} \arctan (\frac{x}{\sqrt{α}}) \Rightarrow

\int_{0}^{1} F (x) dx = \frac{1}{1 + α} {{[\ln ∣ \frac{x + 1}{\sqrt{x^{2} + α}} ∣]}_{0}^{1} + {[\frac{1}{\sqrt{α}} \arctan (\frac{x}{\sqrt{α}})]}_{0}^{1}}

= \frac{1}{1 + α} {\ln (\frac{2}{\sqrt{1 + α}}) - \ln (\frac{1}{\sqrt{α}}) + \frac{1}{\sqrt{α}} \arctan (\frac{1}{\sqrt{α}})} \Rightarrow

f (α) = \int \frac{\ln (\frac{2}{\sqrt{1 + α}})}{1 + α} d α + \frac{1}{2} \int \frac{\ln α}{1 + α} d α + \int \frac{\arctan (\frac{1}{\sqrt{α}})}{(1 + α) \sqrt{α}} d α + c

\dots be continued \dots .

Answered by mathmax by abdo last updated on 07/Jun/20

let take a try with series we have

I = \int_{0}^{1} \ln (1 + x^{2}) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} dx = \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx

A_{n} = \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx we have \ln^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{p = 0}^{\infty} {(- 1)}^{p} u^{p} \Rightarrow

\ln (1 + u) = \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p} u^{p + 1}}{p + 1} + c (c = 0)

= \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1} u^{p}}{p} \Rightarrow \ln (1 + x^{2}) = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p} x^{2 p} \Rightarrow

A_{n} = \int_{0}^{1} x^{n} \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p} x^{2 p} dx = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p} \int_{0}^{1} x^{n + 2 p} dx

= \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p (n + 2 p + 1)} \Rightarrow I = \sum_{n = 0}^{\infty} {(- 1)}^{n} (\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p (n + 2 p + 1)})

I = \sum_{n ⩾ 0} \sum_{p ⩾ 1} \frac{{(- 1)}^{n + p - 1}}{p (n + 2 p + 1)}

Commented by M±th+et+s last updated on 07/Jun/20

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