calculate-0-xarctan-2x-x-2-1-2-dx-

Question Number 135368 by Bird last updated on 12/Mar/21

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

Answered by Dwaipayan Shikari last updated on 12/Mar/21

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

I^{'} (a) = \int_{0}^{\infty} \frac{x^{2}}{{(1 + a^{2} x^{2})}^{2} {(x^{2} + 1)}^{2}} d x = \int_{0}^{\infty} \frac{1}{(1 + a^{2} x^{2}) (1 + x^{2})} - \frac{1}{(1 + a^{2} x^{2}) {(1 + x^{2})}^{2}}

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

= \frac{2}{a^{2} - 1} (\frac{π}{2} - \frac{π}{2 a}) - \frac{1}{2} \int_{0}^{\infty} \frac{t^{\frac{1}{2} - 1}}{{(1 + t)}^{\frac{3}{2} + \frac{1}{2}}} d t

= \frac{π}{a (a + 1)} - \frac{1}{2} . \frac{Γ (\frac{3}{2}) Γ (\frac{1}{2})}{Γ (2)} = \frac{π}{a + 1} - \frac{π}{4}

I (a) = π l o g (\frac{a}{a + 1}) - \frac{π}{4} + C \frac{π}{2} \int_{0}^{\infty} \frac{x}{{(x^{2} + 1)}^{2}} d x = \frac{π}{4} \int_{1}^{\infty} \frac{d t}{t^{2}} = \frac{π}{4}

\lim_{z \to \infty} I (z) = - \frac{π}{4} + C = \frac{π}{4} \Rightarrow C = \frac{π}{2}

I (a) = π l o g (a + 1) + \frac{π}{4} \Rightarrow I (2) = π l o g (3) + \frac{π}{4}