0-1-log-1-x-x-2-x-3-x-4-dx-x-pi-2-15-

Question Number 104644 by Rohit@Thakur last updated on 22/Jul/20

\int_{0}^{1} \frac{l o g (1 - x + x^{2} - x^{3} + x^{4}) d x}{x} = - \frac{π^{2}}{15}

Answered by mathmax by abdo last updated on 23/Jul/20

I = \int_{0}^{1} \frac{\ln (1 - x + x^{2} - x^{3} + x^{4})}{x} dx \Rightarrow I = \int_{0}^{1} \frac{\ln (\frac{1 - {(- x)}^{5}}{1 - (- x)})}{x} dx

= \int_{0}^{1} \frac{\ln (1 + x^{5}) - \ln (1 + x)}{x} dx = \int_{0}^{1} \frac{\ln (1 + x^{5})}{x} dx - \int_{0}^{1} \frac{\ln (1 + x)}{x} dx

we have \ln^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} for ∣ u ∣< 1 \Rightarrow

\ln (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n + 1}}{n + 1} + c (c = 0) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} u^{n}}{n} \Rightarrow

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

= - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = - δ (2) = - (2^{1 - 2} - 1) ξ (2) = - (- \frac{1}{2}) \times \frac{π^{2}}{6} = \frac{π^{2}}{12}

\ln (1 + x^{5}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{5 n} \Rightarrow \frac{\ln (1 + x^{5})}{x} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{5 n - 1} \Rightarrow

\int_{0}^{1} \frac{\ln (1 + x^{5})}{x} dx = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (5 n)} = - \frac{1}{5} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{5} \times \frac{π^{2}}{12} = \frac{π^{2}}{60}

\Rightarrow I = \frac{π^{2}}{60} - \frac{π^{2}}{12} = (\frac{1}{60} - \frac{1}{12}) π^{2} = \frac{1 - 5}{60} \times π^{2} = - \frac{4 π^{2}}{60} = - \frac{π^{2}}{15}

the result is proved .