let f(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^3 )) with a>0 1) calculate f(a) 2)calculste also g(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^4 )) 3) find the values of integrals ∫_0 ^∞ (dx/((x^2 +1)^3 )) ∫


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Question Number 63782 by mathmax by abdo last updated on 09/Jul/19

$l e t f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{{(a^{2} + x^{2})}^{3}} w i t h a > 0$ $1) c a l c u l a t e f (a)$ $2) c a l c u l s t e a l s o g (a) = \int_{- \infty}^{+ \infty} \frac{d x}{{(a^{2} + x^{2})}^{4}}$ $3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}}$ $\int_{0}^{\infty} \frac{d x}{{(x^{2} + 2)}^{4}}$
Commented by mathmax by abdo last updated on 09/Jul/19
$1) let h(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^2 )) we have h^′ (a) =−∫_(−∞) ^(+∞) ((2(2a)(a^2 +x^2 ))/((a^2 +x^2 )^4 )) =−∫_(−∞) ^(+∞) ((4a)/((a^2 +x^2 )^3 ))dx =−4a f(a) ⇒f(a)=−(1/(4a))h^′ (a) let find h(a) we have h(a) =_(x=at) ∫_(−∞) ^(+∞) ((adt)/(a^4 (1+t^2 )^2 )) =(1/a^3 ) ∫_(−∞) ^(+∞) (dt/((t^2 +1)^2 )) let w(z) =(1/((z^2 +1)^2 )) ⇒f(z) =(1/((z−i)^2 (z+i)^2 )) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,i) Res(w,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 w(z)}^((1)) =lim_(z→i) {(z+i)^(−2) }^((1)) =lim_(z→i) −2(z+i)^(−3) =((−2)/((2i)^3 )) =((−2)/(−8i)) =(1/(4i)) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ(1/(4i)) =(π/2) ⇒h(a) =(π/(2a^3 )) ⇒ h^′ (a)=(π/2) ((−3a^2 )/a^6 ) =((−3π)/(2a^4 )) ⇒f(a) =−(1/(4a))(((−3π)/(2a^4 ))) =((3π)/(8a^5 )) ★f(a) =((3π)/(8a^5 )) ★$
$1) l e t h (a) = \int_{- \infty}^{+ \infty} \frac{d x}{{(a^{2} + x^{2})}^{2}} w e h a v e h^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{2 (2 a) (a^{2} + x^{2})}{{(a^{2} + x^{2})}^{4}}$ $= - \int_{- \infty}^{+ \infty} \frac{4 a}{{(a^{2} + x^{2})}^{3}} d x = - 4 a f (a) \Rightarrow f (a) = - \frac{1}{4 a} h^{^{'}} (a) l e t f i n d h (a)$ $w e h a v e h (a) =_{x = a t} \int_{- \infty}^{+ \infty} \frac{a d t}{a^{4} {(1 + t^{2})}^{2}} = \frac{1}{a^{3}} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{2}}$ $l e t w (z) = \frac{1}{{(z^{2} + 1)}^{2}} \Rightarrow f (z) = \frac{1}{{(z - i)}^{2} {(z + i)}^{2}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (w, i)$ $R e s (w, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} w (z)}^{(1)}$ $= {l i m}_{z \to i} {{(z + i)}^{- 2}}^{(1)} = {l i m}_{z \to i} - 2 {(z + i)}^{- 3} = \frac{- 2}{{(2 i)}^{3}} = \frac{- 2}{- 8 i} = \frac{1}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π \frac{1}{4 i} = \frac{π}{2} \Rightarrow h (a) = \frac{π}{2 a^{3}} \Rightarrow$ $h^{^{'}} (a) = \frac{π}{2} \frac{- 3 a^{2}}{a^{6}} = \frac{- 3 π}{2 a^{4}} \Rightarrow f (a) = - \frac{1}{4 a} (\frac{- 3 π}{2 a^{4}}) = \frac{3 π}{8 a^{5}}$ $★ f (a) = \frac{3 π}{8 a^{5}} ★$
Commented by mathmax by abdo last updated on 09/Jul/19

$2) w e h a v e b y d e r i v a t i o n f^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{3 (2 a) {(a^{2} + x^{2})}^{2}}{{(a^{2} + x^{2})}^{6}} d x$ $= - 6 a \int_{- \infty}^{+ \infty} \frac{d x}{{(a^{2} + x^{2})}^{4}} = - 6 a g (a) \Rightarrow g (a) = - \frac{1}{6 a} f^{^{'}} (a)$ $b u t f (a) = \frac{3 π}{8 a^{5}} \Rightarrow f^{^{'}} (a) = \frac{3 π}{8} (- \frac{5 a^{4}}{a^{10}}) = - \frac{15 π}{8 a^{6}} \Rightarrow$ $g (a) = - \frac{1}{6 a} (- \frac{15 π}{8 a^{6}}) = \frac{15 π}{48 a^{7}}$
Commented by mathmax by abdo last updated on 09/Jul/19

$\Rightarrow g (a) = \frac{3.5 π}{3.16 a^{7}} \Rightarrow g (a) = \frac{5 π}{16 a^{7}} .$
Commented by mathmax by abdo last updated on 09/Jul/19

$3) w e h a v e p r o v e d t h a t \int_{- \infty}^{+ \infty} \frac{d x}{{(a^{2} + x^{2})}^{3}} = \frac{3 π}{8 a^{5}}$ $a = 1 \Rightarrow \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{3}} = \frac{3 π}{8} \Rightarrow 2 \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}} = \frac{3 π}{8} \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}} = \frac{3 π}{16} . a l s o w e h a v e p r o v e d t h a t$ $\int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + a^{2})}^{4}} = \frac{5 π}{16 a^{7}}$ $a = \sqrt{2} \Rightarrow \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2)}^{4}} = \frac{5 π}{16 {(\sqrt{2})}^{7}} \Rightarrow \int_{0}^{\infty} \frac{d x}{{(x^{2} + 2)}^{4}} = \frac{5 π}{32 {(\sqrt{2})}^{7}} .$
	Answered by MJS last updated on 09/Jul/19

	$reduction formula$ $\int \frac{d x}{{({p x}^{2} + q)}^{n}} = \frac{x}{2 q (n - 1) {({p x}^{2} + q)}^{n - 1}} + \frac{2 n - 3}{2 q (n - 1)} \int \frac{d x}{{({p x}^{2} + q)}^{n - 1}}$ $\int \frac{d x}{{(x^{2} + a^{2})}^{3}} =$ $p = 1; q = a^{2}; n = 3$ $= \frac{x}{4 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3}{4 a^{2}} \int \frac{d x}{{(x^{2} + a^{2})}^{2}} =$ $p = 1; q = a^{2}; n = 2$ $= \frac{x}{4 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3}{4 a^{2}} \times \frac{x}{2 a^{2} (x^{2} + a^{2})} + \frac{3}{4 a^{2}} \times \frac{1}{2 a^{2}} \int \frac{d x}{x^{2} + a^{2}} =$ $= \frac{x}{4 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3 x}{8 a^{4} (x^{2} + a^{2})} + \frac{3}{8 a^{4}} \times \frac{1}{a} \arctan \frac{x}{a} =$ $= \frac{x (3 x^{2} + 5 a^{2})}{8 a^{4} {(x^{2} + a^{2})}^{2}} + \frac{3}{8 a^{5}} \arctan \frac{x}{a} + C$ $f (a) = \underset{- \infty}{\int^{+ \infty}} \frac{d x}{{(x^{2} + a^{2})}^{3}} = \frac{3 π}{8 a^{5}}$ $for g (a) we go the same path$ $\int \frac{d x}{{(x^{2} + a^{2})}^{4}} = \frac{x (15 x^{4} + 40 a^{2} x^{2} + 33 a^{4})}{48 a^{6} {(x^{2} + a^{2})}^{3}} + \frac{5}{16 a^{7}} \arctan \frac{x}{2} + C$ $g (a) = \underset{- \infty}{\int^{+ \infty}} \frac{d x}{{(x^{2} + a^{2})}^{4}} = \frac{5 π}{16 a^{7}}$
	Commented by turbo msup by abdo last updated on 09/Jul/19

	$t h a n k y o u s i r .$
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