let f(x) = ∫_(−∞) ^(+∞) ((arctan (xt^2 ))/(1+2t^2 ))dt 1) find a explicite form of f(x) 2) calculate ∫_0 ^∞ ((arctan(t^2 ))/(1+2t^2 ))dt and ∫


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Question Number 42605 by maxmathsup by imad last updated on 28/Aug/18

$l e t f (x) = \int_{- \infty}^{+ \infty} \frac{a r c t a n ({x t}^{2})}{1 + 2 t^{2}} d t$ $1) f i n d a e x p l i c i t e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (t^{2})}{1 + 2 t^{2}} d t a n d \int_{0}^{\infty} \frac{a r c t a n (2 t^{2})}{1 + 2 t^{2}} d t$
Commented by maxmathsup by imad last updated on 30/Aug/18
$1) we have f^′ (x) = ∫_(−∞) ^(+∞) (t^2 /((1+x^2 t^4 )(1+2t^2 )))dt let ϕ(z) = (z^2 /((1 +x^2 z^4 )(1+2z^2 ))) we have ϕ(z) = (z^2 /((xz^2 −i)(xz^2 +i)((√2)z−i)((√2)z+i))) =(z^2 /(2x^2 (z^2 −(i/x))(z^2 +(i/x))(z−(i/(√2)))(z+(i/(√2))))) (x>0) = (z^2 /(2x^2 (z−(1/(√x))e^((iπ)/4) )(z+(1/(√x))e^((iπ)/4) )(z−(1/(√x))e^(−((iπ)/4)) )(z+(1/(√x))e^(−((iπ)/4)) )(z−(i/(√2)))(z+(i/(√2))))) so the poles of ϕ are +^− (1/(√x)) e^((iπ)/4) , +^− (1/(√x))e^(−((iπ)/4)) and +^− (i/(√2)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,(1/(√x))e^((iπ)/4) ) +Res(ϕ,−(1/(√x)) e^(−((iπ)/4)) ) +Res(ϕ,(i/(√2)))} Res(ϕ,(e^((iπ)/4) /(√x))) = (i/(2x^3 ((2/(√x))e^((iπ)/4) )(((2i)/x))(1+((2i)/x)))) = ((x^2 (√x))/(8 x^3 (x+2i))) e^(−((iπ)/4)) = ((√x)/(8x(x+2i))) e^(−((iπ)/4)) Res(ϕ,−(e^(−((iπ)/4)) /(√x))) = ((−i)/(2x^3 (((−2)/(√x)) e^(−((iπ)/4)) )(((−2i)/(√x)))(1−((2i)/x)))) =−((√x)/(8x(x−2i))) e^((iπ)/4) Res(ϕ,(i/(√2))) = ((−1)/(4x^2 (((2i)/(√2)))(−(1/2)−(i/x))(−(1/2) +(i/x)))) = ((−1)/(4i(√2)x^2 ((1/4)+(1/x^2 )))) =((−1)/(i(√2)(4x^2 )(((x^2 +4)/(4x^2 ))))) = (i/((√2)(x^2 +4))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { ((x^2 (√x))/(8x(x+2i))) e^(−((iπ)/4)) −((√x)/(8x(x−2i))) e^((iπ)/4) +(i/((√2)(x^2 +4)))} ....be continued...$
$1) w e h a v e f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{t^{2}}{(1 + x^{2} t^{4}) (1 + 2 t^{2})} d t l e t$ $φ (z) = \frac{z^{2}}{(1 + x^{2} z^{4}) (1 + 2 z^{2})} w e h a v e$ $φ (z) = \frac{z^{2}}{({x z}^{2} - i) ({x z}^{2} + i) (\sqrt{2} z - i) (\sqrt{2} z + i)} = \frac{z^{2}}{2 x^{2} (z^{2} - \frac{i}{x}) (z^{2} + \frac{i}{x}) (z - \frac{i}{\sqrt{2}}) (z + \frac{i}{\sqrt{2}})} (x > 0)$ $= \frac{z^{2}}{2 x^{2} (z - \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) (z - \frac{i}{\sqrt{2}}) (z + \frac{i}{\sqrt{2}})} s o t h e$ $p o l e s o f φ a r e \bar{+} \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}, \bar{+} \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}} a n d \bar{+} \frac{i}{\sqrt{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) + R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) + R e s (φ, \frac{i}{\sqrt{2}})}$ $R e s (φ, \frac{e^{\frac{i π}{4}}}{\sqrt{x}}) = \frac{i}{2 x^{3} (\frac{2}{\sqrt{x}} e^{\frac{i π}{4}}) (\frac{2 i}{x}) (1 + \frac{2 i}{x})} = \frac{x^{2} \sqrt{x}}{8 x^{3} (x + 2 i)} e^{- \frac{i π}{4}}$ $= \frac{\sqrt{x}}{8 x (x + 2 i)} e^{- \frac{i π}{4}}$ $R e s (φ, - \frac{e^{- \frac{i π}{4}}}{\sqrt{x}}) = \frac{- i}{2 x^{3} (\frac{- 2}{\sqrt{x}} e^{- \frac{i π}{4}}) (\frac{- 2 i}{\sqrt{x}}) (1 - \frac{2 i}{x})} = - \frac{\sqrt{x}}{8 x (x - 2 i)} e^{\frac{i π}{4}}$ $R e s (φ, \frac{i}{\sqrt{2}}) = \frac{- 1}{4 x^{2} (\frac{2 i}{\sqrt{2}}) (- \frac{1}{2} - \frac{i}{x}) (- \frac{1}{2} + \frac{i}{x})} = \frac{- 1}{4 i \sqrt{2} x^{2} (\frac{1}{4} + \frac{1}{x^{2}})}$ $= \frac{- 1}{i \sqrt{2} (4 x^{2}) (\frac{x^{2} + 4}{4 x^{2}})} = \frac{i}{\sqrt{2} (x^{2} + 4)} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{x^{2} \sqrt{x}}{8 x (x + 2 i)} e^{- \frac{i π}{4}} - \frac{\sqrt{x}}{8 x (x - 2 i)} e^{\frac{i π}{4}} + \frac{i}{\sqrt{2} (x^{2} + 4)}}$ $. . . . b e c o n t i n u e d . . .$
Commented by maxmathsup by imad last updated on 30/Aug/18

$\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (- 2 i I m (\frac{\sqrt{x}}{8 x (x - 2 i)} e^{\frac{i π}{4}})) - \frac{2 π}{\sqrt{2} (x^{2} + 4)}$ $= 4 π I m (\frac{\sqrt{x}}{8 x (x - 2 i)} e^{\frac{i π}{4}}) - \frac{2 π}{\sqrt{2} (x^{2} + 4)} = \frac{π \sqrt{x}}{2 x} I m (\frac{e^{\frac{i π}{4}}}{x - 2 i}) - \frac{2 π}{\sqrt{2} (x^{2} + 4)} b u t$ $\frac{e^{\frac{i π}{4}}}{x - 2 i} = \frac{(x + 2 i) e^{\frac{i π}{4}}}{x^{2} + 4} = \frac{(x + 2 i) (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}{x^{2} + 4} = \frac{1}{\sqrt{2}} \frac{(x + 2 i) (1 + i)}{(x^{2} + 4)}$ $= \frac{1}{\sqrt{2}} \frac{x + i x + 2 i - 2}{x^{2} + 4} = \frac{1}{\sqrt{2}} \frac{x - 2 + i (x + 2)}{x^{2} + 4} \Rightarrow I m (. . .) = \frac{x + 2}{\sqrt{2} (x^{2} + 4)} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{π \sqrt{x}}{2 x} \frac{x + 2}{\sqrt{2} (x^{2} + 4)} - \frac{2 π}{\sqrt{2} (x^{2} + 4)} = f^{^{'}} (x) \Rightarrow$ $f^{^{'}} (x) = \frac{π}{2 \sqrt{2} \sqrt{x}} \frac{x + 2}{(x^{2} + 4)} - \frac{2 π}{\sqrt{2} (x^{2} + 4)} \Rightarrow f (x) = \frac{π}{2 \sqrt{2}} \int \frac{x + 2}{\sqrt{x} (x^{2} + 4)} - π \sqrt{2} \int \frac{d x}{x^{2} + 4} + c$
Commented by maxmathsup by imad last updated on 30/Aug/18

$c h a n g e m e n t \sqrt{x} = t g i v e \int \frac{x + 2}{\sqrt{x} (x^{2} + 4)} d x = \int \frac{t^{2} + 2}{t (t^{4} + 4)} (2 t) d t$ $= \int \frac{2 t^{2} + 4}{t^{4} + 4} d t \Rightarrow f (x) = \int \frac{2 t^{2} + 4}{t^{4} + 4} d t - π a r c t a n x + c . . . b e c o n t i n u e d . . .$
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