calculatef-0-arctan-x-2-x-2-9-dx-with-real-

Question Number 74343 by mathmax by abdo last updated on 22/Nov/19

c a l c u l a t e f (α) = \int_{0}^{\infty} \frac{a r c t a n (α x^{2})}{x^{2} + 9} d x w i t h α r e a l .

Commented by mathmax by abdo last updated on 24/Nov/19

c a s e 1 α > 0 w e h a v e 2 f (α) = \int_{- \infty}^{+ \infty} \frac{a r c t a n (α x^{2})}{x^{2} + 9} d x

=_{x = 3 t} \int_{- \infty}^{+ \infty} \frac{a r c t a n (9 α t^{2})}{9 (t^{2} + 1)} (3) d t = \frac{1}{3} \int_{- \infty}^{+ \infty} \frac{a r c t a n (9 α t^{2})}{t^{2} + 1} d t

x \to \frac{a r c t a n (9 α t^{2})}{t^{2} + 1} i s p o s i t i v e \Rightarrow \int_{- \infty}^{+ \infty} (\dots) d t ⩾ 0 l e t

W (z) = \frac{a r c t a n (9 α z^{2})}{z^{2} + 1} \Rightarrow W (z) = \frac{a r c t a n (9 α z^{2})}{(z - i) (z + i)} a n d

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i) = 2 i π \frac{∣ a r c t a n (9 α (- 1)) ∣}{2 i}

= π a r c t a n (9 α) (b u t t h i s r e s u l t e e d a p r o o f) \Rightarrow

f (α) = \frac{π}{6} a r c t a n (9 α)

c a s e 2 α < 0 l e t α^{^{'}} = - α > 0 \Rightarrow f (α) = \int_{0}^{\infty} \frac{a r c t a n (- α^{^{'}} x^{2})}{x^{2} + 9} d x

= - \int_{0}^{\infty} \frac{a r c t a n (α^{^{'}} x^{2})}{x^{2} + 9} = - \frac{π}{6} a r c t a n (9 α^{^{'}}) = \frac{π}{6} a r c t a n (9 α)