I-n-0-1-1-x-2-n-dx-prove-that-n-1-I-n-n-pi-

Question Number 208194 by universe last updated on 07/Jun/24

I_{n} = \int_{0}^{\infty} \frac{1}{{(1 + x^{2})}^{n}} d x

p r o v e t h a t \underset{n = 1}{\sum^{\infty}} \frac{I_{n}}{n} = π

Answered by Berbere last updated on 07/Jun/24

I_{n} = \int_{0}^{\infty} \frac{y^{- \frac{1}{2}} d y}{2 {(1 + y)}^{n}} = \frac{1}{2} \int_{0}^{\infty} \frac{y^{\frac{1}{2} - 1}}{{(1 + y)}^{\frac{1}{2} + 1 - \frac{1}{2}}} = \frac{1}{2} β (\frac{1}{2}, n - \frac{1}{2})

\sum_{n ⩾ 1} \frac{I}{n} = \sum_{n ⩾ 1} \frac{1}{2 n} . \frac{Γ (\frac{1}{2}) Γ (n - \frac{1}{2})}{Γ (n)} = Σ \frac{\frac{1}{2} Γ (\frac{1}{2}) Γ (n - \frac{1}{2})}{Γ (n + 1)} = Σ \frac{Γ (\frac{3}{2}) Γ (n - \frac{1}{2})}{Γ (n - \frac{1}{2} + \frac{3}{2})}

= \sum_{n ⩾ 1} β (\frac{3}{2}, n - \frac{1}{2}) \overset{n = 1 + n}{=} \sum_{n ⩾ 0} \int_{0}^{1} t^{n + \frac{1}{2}} {(1 - t)}^{\frac{3}{2}} d t

= \int_{0}^{1} \sqrt{t} {(1 - t)}^{\frac{3}{2}} . \frac{1}{1 - t} d t = \int_{0}^{1} \sqrt{t (1 - t)} d t = β (\frac{1}{2}, \frac{1}{2}) = \frac{π}{s i n (\frac{π}{2})} = π

\sum_{n ⩾ 1} \frac{I_{n}}{n} = π

Answered by namphamduc last updated on 07/Jun/24

No need to use Beta function

I_{n} = \int_{0}^{\infty} \frac{1}{{(1 + x^{2})}^{n}} d x

\Rightarrow S = \underset{n = 1}{\sum^{\infty}} \frac{I_{n}}{n} = \int_{0}^{\infty} (\underset{n = 1}{\sum^{\infty}} \frac{1}{n {(1 + x^{2})}^{n}}) d x = - \int_{0}^{\infty} \ln (1 - \frac{1}{1 + x^{2}}) d x

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

{\begin{cases} u = \ln (\frac{x^{2}}{1 + x^{2}}) \\ d v = d x \end{cases} \Rightarrow {\begin{cases} d u = \frac{2}{x (1 + x^{2})} d x \\ v = x \end{cases}

\Rightarrow S = (- x \ln (\frac{x^{2}}{1 + x^{2}}) + {2 \tan}^{- 1} (x)) ∣_{0}^{\infty} = 2 . \frac{π}{2} = π