Calculate 𝛗=∫_0 ^( 1) (( ln(x ))/(1 + x^( 2) )) dx =^? −G ∼ Solution ∼ 𝛗 = ∫_0 ^( 1) {Σ_(k=0) ^∞ (−1)^( k) x^( 2k) ln(x) }dx = Σ_(k=0) ^∞ (−1 )^( k) ∫_0 ^( 1) x^( 2k) ln(x) dx =^(i.b.p) Σ_(k=0) ^∞ (−1 )^( k) {[(x^( 1+2k) /(1+2k)) ln(x ) ]_0 ^1 −(1/(1+2k))∫_0 ^( 1) x^( 2k) dx } = Σ_(k=0) ^∞ (( (−1)^(1+k) )/(( 1 + 2k )^( 2) )) = − G ( Catalan constant ) ∴ 𝛗 = − G ■ m.n


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Question Number 177604 by mnjuly1970 last updated on 07/Oct/22
$Calculate 𝛗=∫_0 ^( 1) (( ln(x ))/(1 + x^( 2) )) dx =^? −G ∼ Solution ∼ 𝛗 = ∫_0 ^( 1) {Σ_(k=0) ^∞ (−1)^( k) x^( 2k) ln(x) }dx = Σ_(k=0) ^∞ (−1 )^( k) ∫_0 ^( 1) x^( 2k) ln(x) dx =^(i.b.p) Σ_(k=0) ^∞ (−1 )^( k) {[(x^( 1+2k) /(1+2k)) ln(x ) ]_0 ^1 −(1/(1+2k))∫_0 ^( 1) x^( 2k) dx } = Σ_(k=0) ^∞ (( (−1)^(1+k) )/(( 1 + 2k )^( 2) )) = − G ( Catalan constant ) ∴ 𝛗 = − G ■ m.n$
$Calculate$ $ϕ = \int_{0}^{1} \frac{\ln (x)}{1 + x^{2}} d x \overset{?}{=} - G$ $\sim Solution \sim$ $ϕ = \int_{0}^{1} {\underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} x^{2 k} \ln (x)} d x$ $= \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} \int_{0}^{1} x^{2 k} \ln (x) d x$ $\overset{i . b . p}{=} \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} {{[\frac{x^{1 + 2 k}}{1 + 2 k} \ln (x)]}_{0}^{1} - \frac{1}{1 + 2 k} \int_{0}^{1} x^{2 k} d x}$ $= \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{1 + k}}{{(1 + 2 k)}^{2}} = - G (C a t a l a n c o n s t a n t)$ $∴ ϕ = - G ◼ m . n$
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