calculate-0-1-Li-2-1-x-4-dx-

Question Number 162002 by mnjuly1970 last updated on 25/Dec/21

c a l c u l a t e

Ω = \int_{0}^{1} {Li}_{2} (1 - x^{4}) d x = ?

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Answered by Lordose last updated on 27/Dec/21

Ω = \int_{0}^{1} {Li}_{2} (1 - x^{4}) dx

Ω \overset{x = x^{\frac{1}{4}}}{=} \frac{1}{4} \int_{0}^{1} x^{\frac{1}{4} - 1} {Li}_{2} (1 - x) dx

Ω \overset{IBP}{=} \frac{1}{4} ({4 x}^{\frac{1}{4}} {Li}_{2} (1 - x) ∣_{0}^{1} - 4 \int_{0}^{1} \frac{x^{\frac{1}{4}} \log (x)}{1 - x} dx)

Ω = - \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{n + \frac{5}{4} - 1} \log (x) \overset{IBP \times 2}{=} \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{5}{4})}^{2}}

N . B :: ψ^{(m)} (z) = {(- 1)}^{m + 1} m! \underset{k = 0}{\sum^{\infty}} \frac{1}{{(z + k)}^{m + 1}}

Ω = ψ^{(1)} (\frac{5}{4}) = 8 G + π^{2} - 16

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