....advanced calculus... evaluation : 𝛗=∫_(0^( ) ) ^( ∞) ((ln(1+x))/(x(1+x^2 ))) dx solution: 𝛗=[∫_0 ^( 1) ((ln(1+x))/(x(1+x^2 )))dx=𝛗_1 ]+[∫_1 ^( ∞) ((ln(1+x))/(x(1+x^2 ))) dx=𝛗_2 ] 𝛗_2 =^(x=(1/t)) ∫_0 ^( 1) ((ln(1+(1/t)))/((1/t)(1+(1/t^2 ))))(dt/t^2 )=∫_0 ^( 1) ((tln(1+(1/t)))/(1+t^2 ))dt =∫_0 ^( 1) ((tln(1+t))/(1+t^2 ))dt−∫_0 ^( 1) ((tln(t))/(1+t^2 ))dt ∴ 𝛗=𝛗_1 +𝛗_2 =∫_0 ^( 1) ((1/x)+x)((ln(1+x))/(1+x^2 ))dx−Φ 𝛗=∫_0 ^( 1) ((ln(1+x))/x)dx−Φ=−li_2 (−1)−Φ Φ=∫_0 ^( 1) ((xln(x))/(1+x^2 ))dx=Σ_(n=0) ^∞ ∫_0 ^( 1) (−1)^n x^(2n+1) ln(x)dx =Σ_(n=0) ^∞ (−1)^n {[(x^(2n+2) /(2n+2)) ln(x)]_0 ^1 −(1/(2n+2))∫_0 ^( 1) x^(2n+1) dx} =Σ_(n=0) ^∞ (−1)^(n+1) (1/(4(n+1)^2 ))=−(1/4) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =((−1)/4) η(2)=((−π^2 )/(48)) .... ∴ 𝛗=−li_2 (−1)−Φ=(π^2 /(12))+(π^2 /(48)) 𝛗=((5π^2 )/(48))


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Question Number 132324 by mnjuly1970 last updated on 13/Feb/21
$....advanced calculus... evaluation : 𝛗=∫_(0^( ) ) ^( ∞) ((ln(1+x))/(x(1+x^2 ))) dx solution: 𝛗=[∫_0 ^( 1) ((ln(1+x))/(x(1+x^2 )))dx=𝛗_1 ]+[∫_1 ^( ∞) ((ln(1+x))/(x(1+x^2 ))) dx=𝛗_2 ] 𝛗_2 =^(x=(1/t)) ∫_0 ^( 1) ((ln(1+(1/t)))/((1/t)(1+(1/t^2 ))))(dt/t^2 )=∫_0 ^( 1) ((tln(1+(1/t)))/(1+t^2 ))dt =∫_0 ^( 1) ((tln(1+t))/(1+t^2 ))dt−∫_0 ^( 1) ((tln(t))/(1+t^2 ))dt ∴ 𝛗=𝛗_1 +𝛗_2 =∫_0 ^( 1) ((1/x)+x)((ln(1+x))/(1+x^2 ))dx−Φ 𝛗=∫_0 ^( 1) ((ln(1+x))/x)dx−Φ=−li_2 (−1)−Φ Φ=∫_0 ^( 1) ((xln(x))/(1+x^2 ))dx=Σ_(n=0) ^∞ ∫_0 ^( 1) (−1)^n x^(2n+1) ln(x)dx =Σ_(n=0) ^∞ (−1)^n {[(x^(2n+2) /(2n+2)) ln(x)]_0 ^1 −(1/(2n+2))∫_0 ^( 1) x^(2n+1) dx} =Σ_(n=0) ^∞ (−1)^(n+1) (1/(4(n+1)^2 ))=−(1/4) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =((−1)/4) η(2)=((−π^2 )/(48)) .... ∴ 𝛗=−li_2 (−1)−Φ=(π^2 /(12))+(π^2 /(48)) 𝛗=((5π^2 )/(48))$
$. . . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t i o n :$ $ϕ = \int_{0^{}}^{\infty} \frac{l n (1 + x)}{x (1 + x^{2})} d x$ $s o l u t i o n :$ $ϕ = [\int_{0}^{1} \frac{l n (1 + x)}{x (1 + x^{2})} d x = ϕ_{1}] + [\int_{1}^{\infty} \frac{l n (1 + x)}{x (1 + x^{2})} d x = ϕ_{2}]$ $ϕ_{2} \overset{x = \frac{1}{t}}{=} \int_{0}^{1} \frac{l n (1 + \frac{1}{t})}{\frac{1}{t} (1 + \frac{1}{t^{2}})} \frac{d t}{t^{2}} = \int_{0}^{1} \frac{t l n (1 + \frac{1}{t})}{1 + t^{2}} d t$ $= \int_{0}^{1} \frac{t l n (1 + t)}{1 + t^{2}} d t - \int_{0}^{1} \frac{t l n (t)}{1 + t^{2}} d t$ $∴ ϕ = ϕ_{1} + ϕ_{2} = \int_{0}^{1} (\frac{1}{x} + x) \frac{l n (1 + x)}{1 + x^{2}} d x - Φ$ $ϕ = \int_{0}^{1} \frac{l n (1 + x)}{x} d x - Φ = - {l i}_{2} (- 1) - Φ$ $Φ = \int_{0}^{1} \frac{x l n (x)}{1 + x^{2}} d x = \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} {(- 1)}^{n} x^{2 n + 1} l n (x) d x$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {{[\frac{x^{2 n + 2}}{2 n + 2} l n (x)]}_{0}^{1} - \frac{1}{2 n + 2} \int_{0}^{1} x^{2 n + 1} d x}$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{4 {(n + 1)}^{2}} = - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{2}}$ $= \frac{- 1}{4} η (2) = \frac{- π^{2}}{48} . . . .$ $∴ ϕ = - {l i}_{2} (- 1) - Φ = \frac{π^{2}}{12} + \frac{π^{2}}{48}$ $ϕ = \frac{5 π^{2}}{48}$
	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

	$\int_{0}^{1} \frac{t^{7} l o g (t)}{1 + t^{2}} d t = - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{{(2 n + 8)}^{2}}$ $= - (\frac{π^{2}}{12} - (1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16})) = \frac{115}{144} - \frac{π^{2}}{12} . . .$
	Commented by mnjuly1970 last updated on 13/Feb/21

	$g r a t e f u l . . .$
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