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Question Number 69564 by mathmax by abdo last updated on 25/Sep/19

l e t f (a) = \int_{0}^{\infty} \frac{d x}{x^{4} - 2 x^{2} + a} w i t h a r e a l a n d a > 1

1) d e t e r m i n e a e x p l i c i t f o r m f o r f (a)

2) c a l c u l a t e g (a) = \int_{0}^{\infty} \frac{d x}{{(x^{4} - 2 x^{2} + a)}^{2}}

3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d x}{x^{4} - 2 x^{2} + 3}

a n d \int_{0}^{\infty} \frac{d x}{{(x^{4} - 2 x^{2} + 3)}^{2}}

Commented by mathmax by abdo last updated on 27/Sep/19

1) f (a) = \int_{0}^{\infty} \frac{d x}{x^{4} - 2 x^{2} + a} \Rightarrow 2 f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} - 2 x^{2} + a}

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

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

Δ^{^{'}} = 1 - a < 0 \Rightarrow Δ^{^{'}} = {(i \sqrt{a - 1})}^{2} \Rightarrow t_{1} = 1 + i \sqrt{a - 1}

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

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

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

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

t h e p o l e s o f W a r e \bar{+} a^{\frac{1}{4}} e^{\frac{i}{2} a r c t a n (\sqrt{a - 1})} a n d \bar{+} a^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (\sqrt{a - 1})}

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

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

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

= \frac{e^{- \frac{i}{2} a r c t a n (\sqrt{a - 1})}}{2 a^{\frac{3}{4}} (2 i s i n (a r c t a n \sqrt{a - 1})} = \frac{e^{- \frac{i}{2} a r c t a n (\sqrt{a - 1})}}{4 i a^{\frac{3}{4}} s i n {a r c t a n (\sqrt{a - 1})}}

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

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

= \frac{e^{\frac{i}{2} a r c t a n (\sqrt{a - 1})}}{2 a^{\frac{1}{4}} a^{\frac{1}{2}} (2 i s i n {a r c t a n \sqrt{a - 1})}} = \frac{e^{\frac{i}{2} a r c t a n (\sqrt{a - 1})}}{4 {i a}^{\frac{3}{4}} s i n {a r c t a n \sqrt{a - 1}}}

b u t s i n {a r c t a n \sqrt{a - 1}) = \frac{\sqrt{a - 1}}{\sqrt{1 + a - 1}} = \frac{\sqrt{a - 1}}{\sqrt{a}} = \sqrt{\frac{a - 1}{a}}

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {\frac{e^{- \frac{i}{2} a r c t a n \sqrt{a - 1}}}{4 i a^{\frac{3}{4}} \times \sqrt{\frac{a - 1}{a}}} + \frac{e^{\frac{i}{2} a r c t a n \sqrt{a - 1}}}{4 {i a}^{\frac{3}{4}} \times \sqrt{\frac{a - 1}{a}}}}

= \frac{π}{2 a^{\frac{3}{4}} \frac{\sqrt{a - 1}}{a^{\frac{1}{2}}}} {2 c o s {a r c t a n \sqrt{a - 1}}

= \frac{π}{a^{\frac{1}{4}} \sqrt{a - 1}} \times c o s {a r c t a n \sqrt{a - 1}} b u t

c o s {a r c t a n \sqrt{a - 1}} = \frac{1}{\sqrt{1 + a - 1}} = \frac{1}{\sqrt{a}} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = \frac{π}{(^{4} \sqrt{a}) \sqrt{a^{2} - a}} = 2 f (a) \Rightarrow

f (a) = \frac{π}{2 (^{4} \sqrt{a}) \sqrt{a^{2} - a}} w i t h a > 1

Commented by mathmax by abdo last updated on 27/Sep/19

2) w e h a v e f^{^{'}} (a) = - \int_{0}^{\infty} \frac{d x}{{(x^{4} - 2 x^{2} + a)}^{2}} = - g (a) \Rightarrow

g (a) = - f^{^{'}} (a) r e s t t o c a l c u l a t e f^{^{'}} (a) \dots

Commented by mathmax by abdo last updated on 27/Sep/19

3) \int_{0}^{\infty} \frac{d x}{x^{4} - 2 x^{2} + 3} = f (3) = \frac{π}{2 (^{4} \sqrt{3}) \sqrt{9 - 3}} = \frac{π}{2 (^{4} \sqrt{3}) \sqrt{6}}