caoculate-0-arctan-x-2-1-2x-2-1-dx-

Question Number 73179 by mathmax by abdo last updated on 07/Nov/19

c a o c u l a t e \int_{0}^{\infty} \frac{a r c t a n (x^{2} - 1)}{2 x^{2} + 1} d x

Commented by mathmax by abdo last updated on 07/Nov/19

r e s i d u s m e t h o d b u t n e e d m o r e p r o o f l e t A = \int_{0}^{\infty} \frac{a r c t a n (x^{2} - 1)}{2 x^{2} + 1} d x

\Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{a r c t a n (x^{2} - 1)}{2 x^{2} + 1} d x l e t W (z) = \frac{a r c t a n (z^{2} - 1)}{2 z^{2} + 1} \Rightarrow

W (z) = \frac{a r c t a n (z^{2} - 1)}{2 (z^{2} + \frac{1}{2})} = \frac{a r c t a n (z^{2} - 1)}{2 (z - \frac{i}{\sqrt{2}}) (z + \frac{i}{\sqrt{2}})} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, \frac{i}{\sqrt{2}}) a n d

R e s (W, \frac{i}{\sqrt{2}}) = {l i m}_{z \to \frac{i}{\sqrt{2}}} (z - \frac{i}{\sqrt{2}}) W (z) = \frac{a r c t a n ({(\frac{i}{\sqrt{2}})}^{2} - 1)}{2 (\frac{2 i}{\sqrt{2}})}

= \sqrt{2} \frac{a r c t a n (- \frac{1}{2} - 1)}{4 i} = - \frac{\sqrt{2}}{4 i} a r c t a n (\frac{3}{2}) \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times (- \frac{\sqrt{2}}{4 i}) a r c t a n (\frac{3}{2}) = - \frac{π \sqrt{2}}{2} a r c t a n (\frac{3}{2}) = 2 A

\Rightarrow A = - \frac{π \sqrt{2}}{4} a r c t a n (\frac{3}{2}) a n d i f t h e g r a p h o f t h e f u n c t i o n

x \to \frac{a r c t a n (x^{2} - 1)}{2 x^{2} + 1} i s o n (x^{^{'}} o x) w e t a k e t h e a b s o l u t e v a l u e \dots

Answered by mind is power last updated on 07/Nov/19

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

\int \frac{dx}{{2 x}^{2} + 1} = \frac{1}{\sqrt{2}} \arctan (x \sqrt{2})

a = {[\frac{a r c t a n (x \sqrt{2})}{\sqrt{2}} . a r c t a n (x^{2} - 1)]}_{0}^{+ \infty} - \frac{1}{\sqrt{2}} \int_{0}^{+ \infty} \frac{\arctan (x \sqrt{2}) . 2 xdx}{{(x^{2} - 1)}^{2} + 1}

a = \frac{π^{2}}{4 \sqrt{2}} - \sqrt{2} \int_{0}^{+ \infty} \frac{\arctan (x \sqrt{2}) xdx}{{(x^{2} - 1)}^{2} + 1} = \frac{π^{2}}{4 \sqrt{2}} - \frac{1}{\sqrt{2}} \int_{- \infty}^{+ \infty} \frac{\arctan (x \sqrt{2}) xdx}{{(x^{2} - 1)}^{2} + 1}

on pose f (t) = \int_{- \infty}^{+ \infty} \frac{xarctan (tx)}{{(x^{2} - 1)}^{2} + 1} dx

t ⩾ 0

f^{'} (t) = \int_{- \infty}^{+ \infty} \frac{x^{2} dx}{(1 + x^{2} t^{2}) (x^{2} - 1 - i) (x^{2} - 1 + i)}

pols are x_{0} = + \frac{i}{t}, x^{2} = 1 + i \Rightarrow x_{2} = (2^{\frac{1}{4}}) e^{\frac{i π}{8}}

x = - 1 + i \Rightarrow x_{1} = (2^{\frac{1}{4}}) e^{\frac{i 3 π}{8}}

f^{'} (t) = 2 i π res (f, x_{0}, x_{1} x_{2})

2 i π res (f, \frac{i}{t}) = 2 i π . \frac{\frac{- 1}{t^{2}}}{(2 i) t ({(\frac{1 + t^{2}}{t^{2}})}^{2} + 1)} = \frac{- π t}{({2 t}^{4} + {2 t}^{2} + 1)}

too be continued