advanced-calculus-evaluation-0-ln-1-x-x-1-x-2-dx-solution-0-1-ln-1-x-x-1-x-2-dx-1-1-ln-1-x-x-1-x

Question Number 132324 by mnjuly1970 last updated on 13/Feb/21

\dots . a d v a n c e d c a l c u l u s \dots

e v a l u a t i o n :

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

s o l u t i o n :

ϕ = [\int_{0}^{1} \frac{l n (1 + x)}{x (1 + x^{2})} d x = ϕ_{1}] + [\int_{1}^{\infty} \frac{l n (1 + x)}{x (1 + x^{2})} d x = ϕ_{2}]

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

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

∴ ϕ = ϕ_{1} + ϕ_{2} = \int_{0}^{1} (\frac{1}{x} + x) \frac{l n (1 + x)}{1 + x^{2}} d x - Φ

ϕ = \int_{0}^{1} \frac{l n (1 + x)}{x} d x - Φ = - {l i}_{2} (- 1) - Φ

Φ = \int_{0}^{1} \frac{x l n (x)}{1 + x^{2}} d x = \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} {(- 1)}^{n} x^{2 n + 1} l n (x) d x

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {{[\frac{x^{2 n + 2}}{2 n + 2} l n (x)]}_{0}^{1} - \frac{1}{2 n + 2} \int_{0}^{1} x^{2 n + 1} d x}

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{4 {(n + 1)}^{2}} = - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{2}}

= \frac{- 1}{4} η (2) = \frac{- π^{2}}{48} \dots .

∴ ϕ = - {l i}_{2} (- 1) - Φ = \frac{π^{2}}{12} + \frac{π^{2}}{48}

ϕ = \frac{5 π^{2}}{48}

Answered by Dwaipayan Shikari last updated on 13/Feb/21

I (a) = \int_{0}^{1} \frac{t^{a}}{1 + t^{2}} d t = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \int_{0}^{1} t^{2 n + a} d t

= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{(2 n + a + 1)} \Rightarrow I^{'} (a) = \int_{0}^{1} \frac{t^{a} l o g (t)}{1 + t^{2}} d t = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + a + 1)}^{2}}

a = 1 I^{'} (a) = - \frac{1}{4} . \frac{π^{2}}{12} = - \frac{π^{2}}{48}

Commented by Dwaipayan Shikari last updated on 13/Feb/21

\int_{0}^{1} \frac{t^{7} l o g (t)}{1 + t^{2}} d t = - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{{(2 n + 8)}^{2}}

= - (\frac{π^{2}}{12} - (1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16})) = \frac{115}{144} - \frac{π^{2}}{12} \dots

Commented by mnjuly1970 last updated on 13/Feb/21

g r a t e f u l \dots