calculate-f-0-1-ln-x-4-4-dx-with-gt-0-then-find-the-value-of-0-1-ln-1-x-4-dx-

Question Number 128954 by mathmax by abdo last updated on 11/Jan/21

calculate f (λ) = \int_{0}^{1} \ln (x^{4} + λ^{4}) dx with λ > 0 then find the value of

\int_{0}^{1} \ln (1 + x^{4}) dx

Answered by Lordose last updated on 16/Jan/21

f (λ) = \int_{0}^{1} \ln (x^{4} + λ^{4}) dx = \int_{0}^{1} \ln (λ^{4} ({(\frac{x}{λ})}^{4} + 1)) dx

f (λ) = 4 \int_{0}^{1} \ln (λ) dx + \int_{0}^{1} \ln (1 + \frac{x^{4}}{λ^{4}}) dx

f (λ) = 4 \ln (λ) + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} {(\frac{x}{λ})}^{4 n} dx

f (λ) = 4 \ln (λ) + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n (4 n + 1) λ^{4 n}}

Ω = \int_{0}^{1} \ln (1 + x^{4}) dx = 4 \ln (1) + \underset{n = 1}{\sum^{\infty}} \frac{(- 1)}{n (4 n + 1)}

Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n (4 n + 1)} = \frac{π}{\sqrt{2}} - 4 + \ln (2) + \sqrt{2} \coth^{- 1} (\sqrt{2})

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