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Question Number 124979 by mathmax by abdo last updated on 07/Dec/20

let f (x) = \arctan (x^{n}) with n natural

1) find f^{(n)} (0) and f^{(n)} (1)

2) developp f at integr serie

3) calculte \int_{0}^{\infty} \frac{f (x)}{x^{n}} dx with n ⩾ 2

Answered by Dwaipayan Shikari last updated on 07/Dec/20

f (x) = {t a n}^{- 1} x^{n}

\int_{0}^{\infty} \frac{{t a n}^{- 1} x^{n}}{x^{n}} d x = - {[n \frac{{t a n}^{- 1} x^{n}}{x^{n - 1}}]}_{0}^{\infty} + n^{2} \int_{0}^{\infty} x^{1 - n} \frac{x^{n - 1}}{1 + x^{2 n}} d x

= n^{2} \int_{0}^{\infty} \frac{d x}{1 + x^{2 n}} x^{2 n} = t \Rightarrow 2 n x^{2 n - 1} = \frac{d t}{d x}

= \frac{n}{2} \int_{0}^{\infty} x^{1 - 2 n} \frac{1}{1 + t} d t = \frac{n}{2} \int_{0}^{\infty} t^{\frac{1}{2 n} - 1} \frac{1}{1 + t} d t \frac{t}{1 + t} = u \Rightarrow \frac{1}{{(1 + t)}^{2}} = \frac{d u}{d t}

= \frac{n}{2} \int_{0}^{1} \frac{u^{\frac{1}{2 n} - 1}}{{(1 - u)}^{\frac{1}{2 n} - 1}} {(1 - u)}^{- 1} d u \frac{u}{1 - u} = t

= \frac{n}{2} \int_{0}^{1} u^{\frac{1}{2 n} - 1} {(1 - u)}^{- \frac{1}{2 n}} d u = \frac{n}{2} Γ (\frac{1}{2 n}) Γ (1 - \frac{1}{2 n}) = \frac{n}{2} . \frac{π}{s i n (\frac{π}{2 n})}

Answered by mathmax by abdo last updated on 07/Dec/20

1) f (x) = \arctan (x^{n}) \Rightarrow f^{(1)} (x) = \frac{{nx}^{n - 1}}{1 + x^{2 n}} \Rightarrow

f^{(p)} (x) = n {(\frac{x^{n - 1}}{x^{2 n} + 1})}^{(p - 1)} with p ⩾ 1

let decompose F (x) = \frac{x^{n - 1}}{x^{2 n} + 1}

z^{2 n} + 1 = 0 \Rightarrow z^{2 n} = e^{i (2 k + 1) π} \Rightarrow z_{k} = e^{i \frac{(2 k + 1) π}{2 n}} and k \in [[0, 2 n - 1]] \Rightarrow

F (x) = \frac{x^{n - 1}}{\prod_{k = 0}^{2 n - 1} (x - z_{k})} = \sum_{k = 0}^{2 n - 1} \frac{a_{k}}{x - z_{k}} we have

a_{k} = \frac{z_{k}^{n - 1}}{2 n z_{k}^{2 n - 1}} = \frac{1}{2 n} \times \frac{z_{k}^{n}}{(- 1)} = - \frac{1}{2 n} z_{k}^{n} = - \frac{1}{2 n} \times e^{i \frac{(2 k + 1) π}{2}}

= - \frac{1}{2 n} e^{ik π} e^{\frac{i π}{2}} = - \frac{1}{2 n} {(- 1)}^{k} i = \frac{i {(- 1)}^{k}}{2 n} \Rightarrow F (x) = \frac{i}{2 n} \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{x - z_{k}} \Rightarrow

f^{(p)} (x) = \frac{i}{2} \sum_{k = 0}^{2 n - 1} {(- 1)}^{k} {\frac{1}{x - z_{k}}}^{(p - 1)}

= \frac{i}{2} \sum_{k = 0}^{2 n - 1} {(- 1)}^{k} \frac{{(- 1)}^{p - 1} (p - 1)!}{{(x - z_{k})}^{p}} \Rightarrow

f^{(n)} (x) = \frac{i}{2} \sum_{k = 0}^{2 n - 1} {(- 1)}^{k} \times \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - z_{k})}^{n}} \Rightarrow

f^{(n)} (0) = \frac{i {(- 1)}^{n - 1}}{2} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{{(- z_{k})}^{n}}

= - \frac{i}{2} (n - 1)! \sum_{k = 0}^{2 n - 1} \frac{{(- 1)}^{k}}{z_{k}^{n}}

f^{(n)} (1) = \frac{i}{2} \sum_{k = 0}^{2 n - 1} {(- 1)}^{k} \frac{{(- 1)}^{n - 1} (n - 1)!}{{(1 - z_{k})}^{n}}

Answered by mathmax by abdo last updated on 07/Dec/20

2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} and f^{) n)} (0) is known

Answered by mathmax by abdo last updated on 07/Dec/20

3) A_{n} = \int_{0}^{\infty} \frac{f (x)}{x^{n}} dx \Rightarrow A_{n} = \int_{0}^{\infty} \frac{\arctan (x^{n})}{x^{n}} dx = \int_{0}^{\infty} x^{- n} \arctan (x^{n}) dx

by psrts A_{n} = {[\frac{1}{1 - n} x^{1 - n} \arctan (x^{n})]}_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{1 - n} x^{1 - n} \times \frac{{nx}^{n - 1}}{1 + x^{2 n}} dx

= \frac{n}{n - 1} \int_{0}^{\infty} \frac{dx}{1 + x^{2 n}} but \int_{0}^{\infty} \frac{dx}{1 + x^{2 n}} =_{x^{2 n} = t \to x = t^{\frac{1}{2 n}}} \int_{0}^{\infty} \frac{1}{2 n} t^{\frac{1}{2 n} - 1} \frac{dt}{1 + t}

= \frac{1}{2 n} \int_{0}^{\infty} \frac{t^{\frac{1}{2 n} - 1}}{1 + t} dt = \frac{1}{2 n} \times \frac{π}{\sin (\frac{π}{2 n})} \Rightarrow A_{n} = \frac{n}{n - 1} \times \frac{π}{2 nsin (\frac{π}{2 n})}

\Rightarrow A_{n} = \frac{π}{2 (n - 1) \sin (\frac{π}{2 n})} (n ⩾ 2)