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Question Number 66694 by mathmax by abdo last updated on 18/Aug/19

l e t f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{4} + x^{2} + a)} w i t h a \in] \frac{1}{4}, + \infty [

1) c a l c u l a t e f (a)

2) f i n d a l s o g (a) = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{4} + x^{2} + a)}^{2}}

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

\int_{0}^{\infty} \frac{d x}{{(x^{4} + x^{2} + 1)}^{2}}

4) d e v e l o p p f a t i n t e g r s e r i e .

Commented by mathmax by abdo last updated on 19/Aug/19

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

x^{4} + x^{2} + a = 0 \Rightarrow t^{2} + t + a = 0 (t = x^{2})

Δ = 1 - 4 a < 0 \Rightarrow Δ = {(i \sqrt{4 a - 1})}^{2} \Rightarrow t_{1} = \frac{- 1 + i \sqrt{4 a - 1}}{2} a n d

t_{2} = \frac{- 1 - i \sqrt{4 a - 1}}{2} \Rightarrow t^{2} + t + a = (t - t_{1}) (t - t_{2}) = (x^{2} - t_{1}) (x^{2} - t_{2}) \Rightarrow

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

= \frac{1}{i \sqrt{4 a - 1}} {\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}} - \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - {\bar{t}}_{1}}}

= \frac{1}{i \sqrt{4 a - 1}} {2 i I m (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}})} = \frac{2}{\sqrt{4 a - 1}} I m (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}})

l e t c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}} l e t w (z) = \frac{1}{z^{2} - t_{1}} p o l e s o f w ?

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

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

r e s i d u s t h e o r e m g i v e

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

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

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

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

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

{c o s}^{2} (\frac{α}{2}) = \frac{1 + c o s (α)}{2} \Rightarrow c o s (\frac{α}{2}) = \bar{+} \sqrt{\frac{1 + c o s α}{2}}

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

c o s (\frac{α}{2}) = \bar{+} \sqrt{\frac{1 + \frac{1}{2 \sqrt{a}}}{2}} = \bar{+} \sqrt{\frac{1 + 2 \sqrt{a}}{4 \sqrt{a}}} = \bar{+} \frac{1}{2} \sqrt{\frac{1 + 2 \sqrt{a}}{\sqrt{a}}}

b u t f (a) > 0 \Rightarrow c o s (\frac{α}{2}) = - \frac{1}{2} \frac{\sqrt{1 + 2 \sqrt{a}}}{(^{4} \sqrt{a})} \Rightarrow

f (a) = \frac{- 2 π}{\sqrt{4 a - 1} {(^{4} \sqrt{a})}^{2}} \times \frac{- 1}{2} \sqrt{1 + 2 \sqrt{a}} = \frac{π \sqrt{1 + 2 \sqrt{a}}}{\sqrt{a} \sqrt{4 a - 1}} \Rightarrow

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

Commented by mathmax by abdo last updated on 19/Aug/19

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

g (a) = - f^{^{'}} (a) f (a) i s k n o w n r e s t t o c a l c u l a t e f^{^{'}} (a)

Commented by mathmax by abdo last updated on 19/Aug/19

3) \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 3} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 3} = \frac{1}{2} f (3)

= \frac{π}{2 \sqrt{3}} \sqrt{\frac{1 + 2 \sqrt{3}}{11}}