1-x-x-3-dx-1-pi-2-12-

Question Number 143728 by Dwaipayan Shikari last updated on 17/Jun/21

\int_{1}^{\infty} \frac{{x}}{x^{3}} d x = 1 - \frac{π^{2}}{12}

Answered by mathmax by abdo last updated on 17/Jun/21

I = \int_{1}^{\infty} \frac{{x}}{x^{3}} dx = \int_{1}^{\infty} \frac{x - [x]}{x^{3}} dx = \int_{1}^{\infty} \frac{dx}{x^{2}} - \int_{1}^{\infty} \frac{[x]}{x^{3}} dx

= {[- \frac{1}{x}]}_{1}^{\infty} - \int_{1}^{\infty} \frac{[x]}{x^{3}} dx = 1 - \int_{1}^{\infty} \frac{[x]}{x^{3}} dx we have

\int_{1}^{\infty} \frac{[x]}{x^{3}} dx = \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{n}{x^{3}} dx = \sum_{n = 1}^{\infty} n {[\frac{1}{- 3 + 1} x^{- 3 + 1}]}_{n}^{n + 1}

= \sum_{n = 1}^{\infty} n {[- \frac{1}{{2 x}^{2}}]}_{n}^{n + 1} = - \frac{1}{2} \sum_{n = 1}^{\infty} n {\frac{1}{n^{2}} - \frac{1}{{(n + 1)}^{2}}}

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

\frac{2 x + 1}{x {(x + 1)}^{2}} = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{{(x + 1)}^{2}}

a = 1, c = 1 \lim_{x \to + \infty} xf (x) = 0 = a + b \Rightarrow b = - 1 \Rightarrow

\frac{2 n + 1}{n {(n + 1)}^{2}} = \frac{1}{n} - \frac{1}{n + 1} + \frac{1}{{(n + 1)}^{2}} \Rightarrow

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

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

= 1 + \frac{π^{2}}{6} - 1 = \frac{π^{2}}{6} \Rightarrow I = - \frac{1}{2} (\frac{π^{2}}{6}) = - \frac{π^{2}}{12}

Commented by mathmax by abdo last updated on 17/Jun/21

I = 1 - \frac{π^{2}}{12}

Commented by Dwaipayan Shikari last updated on 17/Jun/21

T h a n k s s i r