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 )) ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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 bymathmax 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 bymathmax 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 bymathmax 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 bymathmax 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 byturbo msup by abdo last updated on 09/Jul/19

	$t h a n k y o u s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum