let f(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^4 )) with ∣x∣>1 and n integr natural 1)find a explicit form for f(x) 2) determine also g(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^5 )) 3)find the values of integrals ∫_(−∞) ^(+∞) (dt/((t^2 −2t +3)^4 )) and ∫


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Question Number 70870 by Abdo msup. last updated on 09/Oct/19

$l e t f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + x^{2})}^{4}} w i t h ∣ x ∣> 1$ $a n d n i n t e g r n a t u r a l$ $1) f i n d a e x p l i c i t f o r m f o r f (x)$ $2) d e t e r m i n e a l s o g (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + x^{2})}^{5}}$ $3) f i n d t h e v a l u e s o f i n t e g r a l s$ $\int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 3)}^{4}} a n d \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 3)}^{5}}$
Commented bymathmax by abdo last updated on 13/Oct/19
$1) f(x)=∫_(−∞) ^(+∞) (dt/((t^2 −2t +x^2 )^4 )) let ϕ(z)=(1/((z^2 −2z +x^2 )^4 )) z^2 −2z+x^2 =0 →Δ^′ =1−x^2 <0 ⇒Δ^′ =(i(√(x^2 −1)))^2 z_1 =1+i(√(x^2 −1)) and z_2 =1−i(√(x^2 −1)) ∣z_1 ∣=(√(1+x^2 −1))=∣x∣ ⇒z_1 =∣x∣ e^(iarctan((√(x^2 −1)))) z_2 =∣x∣ e^(−iarctan((√(x^2 −1)))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,∣x∣e^(iarctan((√(x^2 −1)))) ) Res(ϕ,∣x∣ e^(iarctan((√(x^2 −1)))) =lim_(z→∣x∣ e^(iarctan((√(x^2 −1)))) ) (1/((4−1)!)){(z−∣x∣e^(iarctan((√(x^2 −1)))) )^4 ϕ(z)}^((3)) =lim_(z→(...)) (1/(3!)){ (1/((z−z_2 )^4 ))}^((3)) ....be continued...$
$1) f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + x^{2})}^{4}} l e t φ (z) = \frac{1}{{(z^{2} - 2 z + x^{2})}^{4}}$ $z^{2} - 2 z + x^{2} = 0 \to Δ^{^{'}} = 1 - x^{2} < 0 \Rightarrow Δ^{^{'}} = {(i \sqrt{x^{2} - 1})}^{2}$ $z_{1} = 1 + i \sqrt{x^{2} - 1} a n d z_{2} = 1 - i \sqrt{x^{2} - 1}$ $∣ z_{1} ∣= \sqrt{1 + x^{2} - 1} =∣ x ∣ \Rightarrow z_{1} =∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})}$ $z_{2} =∣ x ∣ e^{- i a r c t a n (\sqrt{x^{2} - 1})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})})$ $R e s (φ, ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})}$ $= {l i m}_{z \to∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})}} \frac{1}{(4 - 1)!} {{(z - ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1})})}^{4} φ (z)}^{(3)}$ $= {l i m}_{z \to (. . .)} \frac{1}{3!} {\frac{1}{{(z - z_{2})}^{4}}}^{(3)} . . . . b e c o n t i n u e d . . .$
Commented bymathmax by abdo last updated on 13/Oct/19
${(z−z_2 )^(−4) }^((3)) ={−4(z−z_2 )^(−5) }^((2)) ={20(z−z_2 )^(−6) }^((1)) ={−120(z−z_2 )^(−7) } ⇒ Res(ϕ,∣x∣ e^(iarctan((√(x^2 −1))) )=(1/6)(−120)(z_1 −z_2 )^(−7) =−20 (2iIm(z_1 ))^(−7) =−20(2isin(∣x∣arctan((√(x^2 −1))))^(−7) =((−20)/((2i)^7 sin^7 (∣x∣ arctan((√(x^2 −1)))))) =((−20)/(2^7 (−i)sin^7 (∣x∣arctan((√(x^2 −1))))) i^7 =(i^2 )^3 i =−i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((2^2 ×5)/(2^3 ×2^4 (i)sin^7 (∣x∣ arctan((√(x^2 −1)))))) =((5π)/(16 sin^7 (∣x∣ arctan((√(x^2 −1))))) ⇒ f(x)=((5π)/(16 sin^7 (∣x∣arctan((√(x^2 −1)))))$
${{(z - z_{2})}^{- 4}}^{(3)} = {- 4 {(z - z_{2})}^{- 5}}^{(2)} = {20 {(z - z_{2})}^{- 6}}^{(1)}$ $= {- 120 {(z - z_{2})}^{- 7}} \Rightarrow$ $R e s (φ, ∣ x ∣ e^{i a r c t a n (\sqrt{x^{2} - 1}}) = \frac{1}{6} (- 120) {(z_{1} - z_{2})}^{- 7}$ $= - 20 {(2 i I m (z_{1}))}^{- 7} = - 20 (2 i s i n {(∣ x ∣ a r c t a n (\sqrt{x^{2} - 1}))}^{- 7}$ $= \frac{- 20}{{(2 i)}^{7} {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1}))} = \frac{- 20}{2^{7} (- i) {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1})}$ $i^{7} = {(i^{2})}^{3} i = - i$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{2^{2} \times 5}{2^{3} \times 2^{4} (i) {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1}))}$ $= \frac{5 π}{16 {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1})} \Rightarrow$ $f (x) = \frac{5 π}{16 {s i n}^{7} (∣ x ∣ a r c t a n (\sqrt{x^{2} - 1})}$
Commented bymathmax by abdo last updated on 13/Oct/19

$2) w e h a v e f^{^{'}} (x) = - \int_{- \infty}^{+ \infty} \frac{4 (2 x) {(t^{2} - 2 t + x^{2})}^{3}}{{(t^{2} - 2 t + x^{2})}^{8}} d t$ $= - 8 x \int_{- \infty}^{+ \infty} \frac{d x}{{(t^{2} - 2 t + x^{2})}^{5}} = - 8 x g (x) \Rightarrow g (x) = - \frac{1}{8 x} f^{^{'}} (x)$ $r e s t t h e c a l c u l u s o f f^{^{'}} (x) . . . .$
Commented bymathmax by abdo last updated on 13/Oct/19

$\int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 3)}^{4}} = f (\sqrt{3}) = \frac{5 π}{16 {s i n}^{7} (\sqrt{3} a r c t a n (\sqrt{3 - 1})}$ $= \frac{5 π}{16 {s i n}^{7} (\sqrt{3} a r c t a n (2))}$
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