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-0-arctan-2t-2-1-2t-2-dt-

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) 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)}}

\dots . b e c o n t i n u e d \dots

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 (\dots) = \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 \dots b e c o n t i n u e d \dots