calculate-0-cos-2x-dx-x-2-1-2x-2-3-

Question Number 33735 by prof Abdo imad last updated on 22/Apr/18

c a l c u l a t e \int_{0}^{\infty} \frac{c o s (2 x) d x}{(x^{2} + 1) (2 x^{2} + 3)} .

Commented by prof Abdo imad last updated on 25/Apr/18

l e t p u t I = \int_{0}^{\infty} \frac{c o s (2 x)}{(x^{2} + 1) (2 x^{2} + 3)} d x

2 I = \int_{- \infty}^{+ \infty} \frac{c o s (2 x)}{(x^{2} + 1) (2 x^{2} + 3)} d x

= R e (\int_{- \infty}^{+ \infty} \frac{e^{i 2 x}}{(x^{2} + 1) (2 x^{2} + 3)} d x) l e t c o n s i d e r t h e

c o m p l e x f u n c t i o n φ (z) = \frac{e^{i 2 z}}{(z^{2} + 1) (2 z^{2} + 3)}

φ (z) = \frac{e^{i 2 z}}{2 (z^{2} + 1) (z^{2} + \frac{3}{2})} = \frac{e^{i 2 z}}{2 (z - i) (z + i) (z - i \sqrt{\frac{3}{2}) (z + i \frac{\sqrt{3}}{2})}}

s o t h e p o l e s o f φ a r e i, - i, i \sqrt{\frac{3}{2}}, - i \sqrt{\frac{3}{2}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (R e s (φ, i) + R e s (φ, i \sqrt{\frac{3}{2}}))

R e s (φ, i) = \frac{e^{- 2}}{2 (2 i) (- 1 + \frac{3}{2})} = \frac{e^{- 2}}{4 i . \frac{1}{2}} = \frac{e^{- 2}}{2 i}

R e s (φ, i \sqrt{\frac{3}{2}}) = \frac{e^{- 2 \sqrt{\frac{3}{2}}}}{2 (- \frac{3}{2} + 1) (2 i \sqrt{\frac{3}{2}})}

= \frac{e^{- \sqrt{6}}}{- 2 i \frac{\sqrt{3}}{\sqrt{2}}} = - \frac{e^{- \sqrt{6}}}{i \sqrt{6}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{e^{- 2}}{2 i} - \frac{e^{- \sqrt{6}}}{i \sqrt{6}})

= π e^{- 2} - \frac{2 π}{\sqrt{6}} e^{- \sqrt{6}} \Rightarrow

I = \frac{π}{2} e^{- 2} - \frac{π}{\sqrt{6}} e^{- \sqrt{6}} .