calculate-arctan-2x-2-1-x-2-dx-

Question Number 84759 by mathmax by abdo last updated on 15/Mar/20

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

Answered by mind is power last updated on 17/Mar/20

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

f (0) = 0

f^{'} (t) = \int_{- \infty}^{+ \infty} \frac{x^{2} d x}{(1 + x^{2}) (1 + t^{2} x^{4})} =

f^{'} (t) = 2 i π R e s (i, \frac{e^{i \frac{π}{4}}}{\sqrt{t}}, \frac{e^{i \frac{3 π}{4}}}{\sqrt{t}})

= 2 i π . \frac{- 1}{2 i (1 + t^{2})} + 2 i π . \frac{\frac{i}{t}}{(1 + \frac{i}{t}) (4 \sqrt{t} e^{\frac{3 i π}{4}})} + 2 i π . \frac{\frac{- i}{t}}{(1 - \frac{i}{t}) 4 \sqrt{t} e^{i \frac{π}{4}}}

= - \frac{π}{1 + t^{2}} + \frac{2 π}{(t + i) (4 \sqrt{t} e^{- \frac{i π}{4}})} + \frac{2 π}{(t - i) (4 \sqrt{t} e^{i \frac{π}{4}})}

- \frac{π}{1 + t^{2}} + 4 π . R e (\frac{1}{(t + i) (4 \sqrt{t} e^{- \frac{i π}{4}})})

- \frac{π}{1 + t^{2}} + 4 π . R e (\frac{(t - i) (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})}{(t^{2} + 1) (4 \sqrt{t})})

- \frac{π}{1 + t^{2}} + π . \frac{\sqrt{2}}{2} . \frac{\sqrt{t}}{t^{2} + 1} + \frac{π \sqrt{2}}{2 \sqrt{t} (1 + t^{2})}

f (t) = - π a r c t a n (t) + \frac{π}{\sqrt{2}} {\int \frac{\sqrt{t} d t}{t^{2} + 1} + \int \frac{d t}{\sqrt{t} (1 + t^{2})}

\int (\sqrt{t} + \frac{1}{\sqrt{t}}) \frac{d t}{(1 + t^{2})}

= \int \frac{t + 1}{\sqrt{t} (1 + t^{2})} d t u = \sqrt{t} \Rightarrow

= 2 \int \frac{u^{2} + 1}{(1 + u^{4})} d u = \int \frac{2 (u^{2} + 1)}{(u^{2} - \sqrt{2} u + 1) (u^{2} + \sqrt{2} u + 1)} d u

= \int \frac{d u}{u^{2} - \sqrt{2} u + 1} + \int \frac{d u}{u^{2} + \sqrt{2} u + 1}

= \int \frac{d u}{{(u - \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} + \int \frac{d u}{{(u + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}}

= \int \frac{2 d u}{{(\sqrt{2} u - 1)}^{2} + 1} + \int \frac{2 d u}{{(\sqrt{2} u + 1)}^{2} + 1}

= \sqrt{2} {{t a n}^{-} (\sqrt{2} u - 1) + {t a n}^{-} (\sqrt{2} u + 1)}

f (t) = - π {t a n}^{-} (t) + π {{t a n}^{-} (\sqrt{2 t} - 1) + {t a n}^{-} (\sqrt{2 t} + 1)} + c

f (0) = 0 \Rightarrow

c = 0

f (t) = - π {t a n}^{-} (t) + π {{t a n}^{-} (\sqrt{2 t} - 1) + {t a n}^{-} (\sqrt{2 t} + 1)

f (2) = \int \frac{{t a n}^{-} (2 x^{2})}{1 + x^{2}} d x = - π {t a n}^{-} (2) + π {{t a n}^{-} (3) + {t a n}^{-} (5)}

= π {- {t a n}^{-} (2) + {t a n}^{-} (3) + {t a n}^{-} (5)}

Commented by abdomathmax last updated on 18/Mar/20

t h a n k y o u s i r .