calculate-0-sin-arctan-x-2-2-x-2-1-dx-

Question Number 76352 by mathmax by abdo last updated on 26/Dec/19

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

Commented by mathmax by abdo last updated on 29/Dec/19

l e t A = \int_{0}^{\infty} \frac{s i n (a r c t a n (x^{2} + 2))}{x^{2} + 1} d x \Rightarrow

2 A = \int_{- \infty}^{+ \infty} \frac{s i n (a r c t a n (x^{2} + 2))}{x^{2} + 1} d x = I m (\int_{- \infty}^{+ \infty} \frac{e^{i a r c t a n (x^{2} + 2)}}{x^{2} + 1} d x)

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

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i) = 2 i π \times \frac{e^{i a r c t a n (1)}}{2 i} = π e^{i \frac{π}{4}}

= π {\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} \Rightarrow 2 A = \frac{π}{\sqrt{2}} \Rightarrow A = \frac{π}{2 \sqrt{2}}