1-2-logu-u-1-u-1-1-du-

Question Number 101793 by Dwaipayan Shikari last updated on 04/Jul/20

\int_{1}^{2} \frac{l o g u}{(\sqrt{u - 1}) (\sqrt{u - 1} + 1)} d u

Answered by mathmax by abdo last updated on 05/Jul/20

I = \int_{1}^{2} \frac{\ln (u)}{\sqrt{u - 1} (\sqrt{u - 1} + 1)} du changement \sqrt{u - 1} = x give u - 1 = x^{2} \Rightarrow

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

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

= 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n} with A_{n} = \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx by parts we get

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

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

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

= \frac{1}{n + 1} - \int_{0}^{1} \frac{x^{n}}{x^{2} + 1} dx we have

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

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

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

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

I = 2 \ln 2 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} - 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} - 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} (\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + n + 1})

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

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

= - (- \frac{1}{2}) \times \frac{π^{2}}{6} = \frac{π^{2}}{12}

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

\dots be continued \dots

Commented by 1549442205 last updated on 06/Jul/20

Thank you sir . Please, can you show me

about the function ξ (n) that used in solution ?