0-1-log-e-x-1-x-x-2-1-dx-

Question Number 132191 by rs4089 last updated on 12/Feb/21

\int_{0}^{1} \frac{{l o g}_{e} (x + 1)}{x (x^{2} + 1)} d x

Answered by mathmax by abdo last updated on 12/Feb/21

at form of serie I = \int_{0}^{1} \frac{\ln (x + 1)}{x (x^{2} + 1)} dx \Rightarrow I = \int_{0}^{1} \frac{\ln (1 + x)}{x} \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} dx

also \ln^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow \ln (1 + x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1}

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \Rightarrow \frac{\ln (1 + x)}{x} . \frac{1}{1 + x^{2}} = (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n - 1}) (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n})

= (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n}) . (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n}) = (Σ a_{n}) . (Σ b_{n})

= Σ c_{n} with c_{n} = \sum_{i + j = n} a_{i} b_{j} = \sum_{i + j = n} \frac{{(- 1)}^{i}}{i + 1} x^{i} {(- 1)}^{j} x^{2 j}

= \sum_{i = 0}^{n} \frac{{(- 1)}^{i}}{i + 1} x^{i} {(- 1)}^{n - i} x^{2 (n - i)} \Rightarrow

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

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

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

Answered by mathmax by abdo last updated on 12/Feb/21

let try parametric method Φ = \int_{0}^{1} \frac{\ln (1 + x)}{x (x^{2} + 1)} dx let

f (a) = \int_{0}^{1} \frac{\ln (1 + ax)}{x (x^{2} + 1)} with a > 0 \Rightarrow f^{^{'}} (a) = \int_{0}^{1} \frac{x}{(1 + ax) x (x^{2} + 1)} dx

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

F (x) = \frac{α}{ax + 1} + \frac{mx + n}{x^{2} + 1}

α = \frac{1}{(\frac{1}{a^{2}} + 1)} = \frac{a^{2}}{1 + a^{2}}, \lim_{x \to + \infty} xF (x) = 0 = \frac{α}{a} + m \Rightarrow m = - \frac{α}{a} = - \frac{a}{1 + a^{2}}

F (0) = 1 = α + n \Rightarrow n = 1 - α = 1 - \frac{a^{2}}{1 + a^{2}} = \frac{1}{1 + a^{2}} \Rightarrow

F (x) = \frac{a^{2}}{(1 + a^{2}) (ax + 1)} + \frac{- \frac{a}{1 + a^{2}} x + \frac{1}{1 + a^{2}}}{x^{2} + 1}

\Rightarrow \int_{0}^{1} F (x) dx = \frac{a^{2}}{1 + a^{2}} \int_{0}^{1} \frac{dx}{ax + 1} - \frac{1}{1 + a^{2}} \int_{0}^{1} \frac{ax - 1}{x^{2} + 1} dx

= \frac{a}{1 + a^{2}} {[\ln (ax + 1)]}_{0}^{1} - \frac{a}{2 (1 + a^{2})} {[\ln (1 + x^{2})]}_{0}^{1} + \frac{1}{1 + a^{2}} . \frac{π}{4}

= \frac{a}{1 + a^{2}} \ln (a + 1) - \frac{aln (2)}{2 (1 + a^{2})} + \frac{π}{4 (1 + a^{2})} \Rightarrow

f (a) = \int \frac{aln (a + 1)}{1 + a^{2}} da - \frac{\ln (2)}{2} \int \frac{ada}{1 + a^{2}} + \frac{π}{4} arctana + c

= \int \frac{aln (1 + a)}{1 + a^{2}} da - \frac{\ln (2)}{4} \ln (1 + a^{2}) + \frac{π}{4} \arctan (a) + C

\int \frac{a}{a^{2} + 1} \ln (1 + a) da = \frac{1}{2} \ln (a^{2} + 1) \ln (1 + a) - \int \frac{1}{2} \ln (a^{2} + 1) \frac{da}{1 + a}

= \frac{1}{2} \ln (1 + a) \ln (1 + a^{2}) - \frac{1}{2} \int \frac{\ln (1 + a^{2})}{1 + a} da \Rightarrow

f (a) = \int_{0}^{a} \frac{xln (1 + x)}{1 + x^{2}} dx - \frac{\ln (2)}{4} \ln (1 + a^{2}) + \frac{π}{4} \arctan (a) + C

C = f (0) = 0

f (1) = \int_{0}^{1} \frac{xln (1 + x)}{1 + x^{2}} dx - \frac{\ln^{2} (2)}{4} + \frac{π^{2}}{16} = Φ

we have by parts \int_{0}^{1} \frac{xln (1 + x)}{1 + x^{2}} dx = {[\frac{1}{2} \ln (1 + x) \ln (1 + x^{2})]}_{0}^{1}

- \frac{1}{2} \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x} dx = \frac{1}{2} \ln^{2} (2) - \frac{1}{2} \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x} dx

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

u_{n} = \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx = {[\frac{x^{n + 1}}{n + 1} \ln (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} \times \frac{2 x}{1 + x^{2}} dx

= \frac{\ln (2)}{n + 1} - \frac{2}{n + 1} \int_{0}^{1} \frac{x^{n + 2}}{1 + x^{2}} dx \dots . . be continued \dots