nice-calculus-Evaluate-I-0-1-ln-1-x-x-dx-1-x-2-

Question Number 128977 by mnjuly1970 last updated on 11/Jan/21

\dots n i c e c a l c u l u s \dots

E v a l u a t e :::

I := \int_{0}^{1} l n (\frac{1}{x} - x) \frac{d x}{1 + x^{2}} = ?

* \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots *

Answered by abderrahimelmerrouni last updated on 11/Jan/21

3

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

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

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

\int_{0}^{1} x^{2 n} lnxdx = {[\frac{x^{2 n + 1}}{2 n + 1} lnx]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n}}{2 n + 1} dx

= - \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \int_{0}^{1} \frac{lnx}{1 + x^{2}} dx = - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} = - k (katalan constant)

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

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

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

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

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

= - \frac{1}{2 n + 1} \int_{0}^{1} \sum_{k = 0}^{2 n} x^{k} dx = - \frac{1}{2 n + 1} \sum_{k = 0}^{2 n} {[\frac{x^{k + 1}}{k + 1}]}_{0}^{1}

= - \frac{1}{2 n + 1} \sum_{k = 0}^{2 n} \frac{1}{k + 1} = - \frac{1}{2 n + 1} H_{2 n + 1} \Rightarrow \int_{0}^{1} \frac{\ln (1 - x)}{1 + x^{2}} dx

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

K = \int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} dx =_{x = \tan θ} \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \tan θ)}{1 + \tan^{2} θ} (1 + \tan^{2} θ) d θ

= \int_{0}^{\frac{π}{4}} \ln (1 + \tan θ) d θ the value of this integral is known see the platform \dots

Answered by Lordose last updated on 12/Jan/21

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

Ω = Φ - Ψ

Φ = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} \frac{x^{2 n}}{1 + x^{2}} dx = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} \frac{x^{2 n} - x^{2 n + 2}}{1 - x^{4}}

x = u^{\frac{1}{4}} \Rightarrow dx = \frac{1}{4} u^{- \frac{3}{4}} du

Φ \overset{u = x^{4}}{=} \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} \frac{u^{\frac{2 n - 3}{4}} - u^{\frac{2 n - 1}{4}}}{1 - u} du = \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} (H_{(\frac{2 n - 1}{4})} - H_{(\frac{2 n - 3}{4})})

Φ = \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} (ψ^{0} (\frac{n}{2} + \frac{3}{4}) - ψ^{0} (\frac{n}{2} + \frac{1}{4}))

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

Ω = \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{ψ^{0} (\frac{n}{2} + \frac{3}{4})}{n} - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{ψ^{0} (\frac{n}{2} + \frac{1}{4})}{n} + G = \frac{π \log (2)}{4}

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