let f(a) =∫_(−∞) ^(+∞) (dx/((x^4 +x^2 +a))) with a∈](1/4),+∞[ 1) calculate f(a) 2)find also g(a) =∫_(−∞) ^(+∞) (dx/((x^4 +x^2 +a)^2 )) 3) find the value of integrals ∫_0 ^∞ (dx/((x^4 +x^2 +3))) and ∫_0 ^∞ (dx/((x^4 +x^2 +1)^2 )) 4) developp f at integrserie.


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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
$1) f(a) =∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +a)) x^4 +x^2 +a =0 ⇒t^2 +t +a=0 (t=x^2 ) Δ=1−4a<0 ⇒Δ =(i(√(4a−1)))^2 ⇒t_1 =((−1+i(√(4a−1)))/2) and t_2 =((−1−i(√(4a−1)))/2) ⇒t^2 +t+a =(t−t_1 )(t−t_2 )=(x^2 −t_1 )(x^2 −t_2 ) ⇒ f(a) =∫_(−∞) ^(+∞) (dx/((x^2 −t_1 )(x^2 −t_2 ))) = ∫_(−∞) ^(+∞) {(1/(x^2 −t_1 ))−(1/(x^2 −t_2 ))}dx =(1/(i(√(4a−1)))){ ∫_(−∞) ^(+∞) (dx/(x^2 −t_1 )) −∫_(−∞) ^(+∞) (dx/(x^2 −t_1 ^− ))} =(1/(i(√(4a−1)))){2i Im( ∫_(−∞) ^(+∞) (dx/(x^2 −t_1 )))} =(2/(√(4a−1))) Im(∫_(−∞) ^(+∞) (dx/(x^2 −t_1 ))) let calculate ∫_(−∞) ^(+∞) (dx/(x^2 −t_1 )) let w(z) =(1/(z^2 −t_1 )) poles of w? ∣t_1 ∣ =(1/2)(√(1+4a−1))=(√a) ⇒t_1 =(√a)e^(−iarctan((√(4a−1)))) ⇒(√t_1 )=^4 (√a)e^(−(i/2)arctan((√(4a−1)))) ⇒ w(z) =(1/((z−^4 (√a)e^(−(i/2)arctan((√(4a−1)))) )(z+^4 (√a)e^(−(i/2)arctan((√(4a−1)))) ))) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,−^4 (√a)e^(−(i/2)arctan((√(4a−1)))) ) =2iπ×(1/(−2(^4 (√a))e^(−(i/2)arctan((√(4a−1)))) )) =−((iπ)/((^4 (√a)))) e^((i/2)rctan((√(4a−1)))) =((−iπ)/((^4 (√a)))){ cos(((arctan((√(4a−1))))/2))+i sin(((arctan((√(4a−1)))/2))} ⇒ f(a) =(2/(√(4a−1)))(((−π)/((^4 (√a)))) cos(((arctan((√(4a−1))))/2))) =((−2π)/((√(4a−1))(^4 (√a)))) cos(((arctan((√(4a−1))))/2)) cos^2 ((α/2)) =((1+cos(α))/2) ⇒cos((α/2))=+^− (√((1+cosα)/2)) here α=arctan(√(4a−1)) ⇒ cosα =(1/(√(1+((√(4a−1)))^2 ))) =(1/(2(√a))) ⇒ cos((α/2)) =+^− (√((1+(1/(2(√a))))/2))=+^− (√((1+2(√a))/(4(√a))))=+^− (1/2)(√((1+2(√a))/(√a))) but f(a)>0 ⇒cos((α/2))=−(1/2)((√(1+2(√a)))/((^4 (√a)))) ⇒ f(a) =((−2π)/((√(4a−1))(^4 (√a))^2 ))×((−1)/2)(√(1+2(√a)))=((π(√(1+2(√a))))/((√a)(√(4a−1)))) ⇒ f(a) =(π/(√a))(√((1+2(√a))/(4a−1))) with a>(1/4)$
$1) f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + a}$ $x^{4} + x^{2} + a = 0 \Rightarrow t^{2} + t + a = 0 (t = x^{2})$ $Δ = 1 - 4 a < 0 \Rightarrow Δ = {(i \sqrt{4 a - 1})}^{2} \Rightarrow t_{1} = \frac{- 1 + i \sqrt{4 a - 1}}{2} a n d$ $t_{2} = \frac{- 1 - i \sqrt{4 a - 1}}{2} \Rightarrow t^{2} + t + a = (t - t_{1}) (t - t_{2}) = (x^{2} - t_{1}) (x^{2} - t_{2}) \Rightarrow$ $f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - t_{1}) (x^{2} - t_{2})} = \int_{- \infty}^{+ \infty} {\frac{1}{x^{2} - t_{1}} - \frac{1}{x^{2} - t_{2}}} d x$ $= \frac{1}{i \sqrt{4 a - 1}} {\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}} - \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - {\bar{t}}_{1}}}$ $= \frac{1}{i \sqrt{4 a - 1}} {2 i I m (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}})} = \frac{2}{\sqrt{4 a - 1}} I m (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}})$ $l e t c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - t_{1}} l e t w (z) = \frac{1}{z^{2} - t_{1}} p o l e s o f w ?$ $∣ t_{1} ∣ = \frac{1}{2} \sqrt{1 + 4 a - 1} = \sqrt{a} \Rightarrow t_{1} = \sqrt{a} e^{- i a r c t a n (\sqrt{4 a - 1})}$ $\Rightarrow \sqrt{t_{1}} =^{4} \sqrt{a} e^{- \frac{i}{2} a r c t a n (\sqrt{4 a - 1})} \Rightarrow w (z) = \frac{1}{(z -^{4} \sqrt{a} e^{- \frac{i}{2} a r c t a n (\sqrt{4 a - 1})}) (z +^{4} \sqrt{a} e^{- \frac{i}{2} a r c t a n (\sqrt{4 a - 1})})}$ $r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (w, -^{4} \sqrt{a} e^{- \frac{i}{2} a r c t a n (\sqrt{4 a - 1})})$ $= 2 i π \times \frac{1}{- 2 (^{4} \sqrt{a}) e^{- \frac{i}{2} a r c t a n (\sqrt{4 a - 1})}} = - \frac{i π}{(^{4} \sqrt{a})} e^{\frac{i}{2} r c t a n (\sqrt{4 a - 1})}$ $= \frac{- i π}{(^{4} \sqrt{a})} {c o s (\frac{a r c t a n (\sqrt{4 a - 1})}{2}) + i s i n (\frac{a r c t a n (\sqrt{4 a - 1}}{2})} \Rightarrow$ $f (a) = \frac{2}{\sqrt{4 a - 1}} (\frac{- π}{(^{4} \sqrt{a})} c o s (\frac{a r c t a n (\sqrt{4 a - 1})}{2}))$ $= \frac{- 2 π}{\sqrt{4 a - 1} (^{4} \sqrt{a})} c o s (\frac{a r c t a n (\sqrt{4 a - 1})}{2})$ ${c o s}^{2} (\frac{α}{2}) = \frac{1 + c o s (α)}{2} \Rightarrow c o s (\frac{α}{2}) = \bar{+} \sqrt{\frac{1 + c o s α}{2}}$ $h e r e α = a r c t a n \sqrt{4 a - 1} \Rightarrow c o s α = \frac{1}{\sqrt{1 + {(\sqrt{4 a - 1})}^{2}}} = \frac{1}{2 \sqrt{a}} \Rightarrow$ $c o s (\frac{α}{2}) = \bar{+} \sqrt{\frac{1 + \frac{1}{2 \sqrt{a}}}{2}} = \bar{+} \sqrt{\frac{1 + 2 \sqrt{a}}{4 \sqrt{a}}} = \bar{+} \frac{1}{2} \sqrt{\frac{1 + 2 \sqrt{a}}{\sqrt{a}}}$ $b u t f (a) > 0 \Rightarrow c o s (\frac{α}{2}) = - \frac{1}{2} \frac{\sqrt{1 + 2 \sqrt{a}}}{(^{4} \sqrt{a})} \Rightarrow$ $f (a) = \frac{- 2 π}{\sqrt{4 a - 1} {(^{4} \sqrt{a})}^{2}} \times \frac{- 1}{2} \sqrt{1 + 2 \sqrt{a}} = \frac{π \sqrt{1 + 2 \sqrt{a}}}{\sqrt{a} \sqrt{4 a - 1}} \Rightarrow$ $f (a) = \frac{π}{\sqrt{a}} \sqrt{\frac{1 + 2 \sqrt{a}}{4 a - 1}} w i t h a > \frac{1}{4}$
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

$3) \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 3} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 3} = \frac{1}{2} f (3)$ $= \frac{π}{2 \sqrt{3}} \sqrt{\frac{1 + 2 \sqrt{3}}{11}}$
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