calculate-0-1-x-2-2-x-2-x-4-dx-

Question Number 72396 by mathmax by abdo last updated on 28/Oct/19

c a l c u l a t e \int_{0}^{\infty} \frac{1 + x^{2}}{2 + x^{2} + x^{4}} d x

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

l e t A = \int_{0}^{\infty} \frac{1 + x^{2}}{x^{4} + x^{2} + 2} d x \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{x^{2} + 1}{x^{4} + x^{2} + 2} d x

l e t W (z) = \frac{z^{2} + 1}{z^{4} + z^{2} + 2} p o l e s o f W ?

z^{4} + z^{2} + 2 = 0 \Rightarrow t^{2} + t + 2 = 0 w i t h t = z^{2}

Δ = 1 - 8 = - 7 \Rightarrow t_{1} = \frac{- 1 + i \sqrt{7}}{2} a n d t_{2} = \frac{- 1 - i \sqrt{7}}{2}

∣ t_{1} ∣ = \frac{1}{2} \sqrt{1 + 7} = \frac{1}{2} (2 \sqrt{2}) = \sqrt{2} \Rightarrow t_{1} = \sqrt{2} e^{- i a r c t a n (\sqrt{7})}

t_{2} = c o n j (t_{1}) = \sqrt{2} e^{i a r c t a n (\sqrt{7})} \Rightarrow W (z) = \frac{z^{2} + 1}{(z^{2} - \sqrt{2} e^{i a r c t a n (\sqrt{7})}) (z^{2} - \sqrt{2} e^{- i a r c t a n (\sqrt{7})})}

= \frac{z^{2} + 1}{(z - (^{4} \sqrt{2}) e^{\frac{i}{2} a r c t a n (\sqrt{7})}) (z + (^{4} \sqrt{2}) e^{\frac{i}{2} a r c t a n (\sqrt{7})}) (z - (^{4} \sqrt{2}) e^{- \frac{i}{2} a r c t a n (\sqrt{7})}) (z + (^{4} \sqrt{2}) e^{- \frac{i}{2} a r c t a n (\sqrt{7})})}

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W,^{4} \sqrt{2} e^{\frac{i}{2} a r c t a n (\sqrt{7})}) + R e s (W, -^{4} \sqrt{2} e^{- \frac{i}{2} a r c t a n (\sqrt{7})}}

\dots b e c o n t i n u e d \dots .

Commented by mathmax by abdo last updated on 30/Oct/19

R e s (W,^{4} \sqrt{2} e^{\frac{i}{2} a r c t a n (\sqrt{7})}) = \frac{1 + \sqrt{2} e^{i a r c t a n (\sqrt{7})}}{2^{4} \sqrt{2} e^{\frac{i}{2} a r c t a n (\sqrt{7})} (\sqrt{2}) (2 i) s i n (a r c t a n (\sqrt{7}))}

= \frac{1 + \sqrt{2} e^{i a r c t a n (\sqrt{7})}}{(4 i) \sqrt{2} (^{4} \sqrt{2}) e^{\frac{i}{2} a r c t a n (\sqrt{7})} s i n (a r c t a n \sqrt{7})}

R e s (W, -^{4} \sqrt{2} e^{- \frac{i}{2} a r c t a n (\sqrt{7})}) = \frac{1 + \sqrt{2} e^{- i a r c t a n (\sqrt{7})}}{- \sqrt{2} (2 i) s i n (a r c t a n (\sqrt{7})) (- 2^{4} \sqrt{2} e^{- \frac{i}{2} a r c t a n (\sqrt{7})})}

= \frac{1 + \sqrt{2} e^{- i a r c t a n (\sqrt{7})}}{(4 i) \sqrt{2} (^{4} \sqrt{2}) e^{- \frac{i}{2} a r c t a n (\sqrt{7})} s i n (a r c t a n \sqrt{7})} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {\frac{e^{- \frac{i}{2} a r c t a n (\sqrt{7})} + \sqrt{2} e^{\frac{i}{2} a r c t a n (\sqrt{7})}}{(4 i) \sqrt{2} (^{4} \sqrt{2}) s i n (a r c t a n (\sqrt{7})}

+ \frac{e^{- \frac{i}{2} a r c t a n (\sqrt{7})} + \sqrt{2} e^{- \frac{i}{2} a r c t a n (\sqrt{7})}}{(4 i) \sqrt{2} (^{4} \sqrt{2}) s i n (a r c t a n \sqrt{7})}} r e s t t o c o m p l e t e t h e c a l c u l u s .