1) calculate by residus method ∫_0 ^∞ (dx/((1+x^2 )^3 )) 2) find the value of ∫


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Question Number 66459 by mathmax by abdo last updated on 15/Aug/19

$1) c a l c u l a t e b y r e s i d u s m e t h o d \int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{3}}$ $2) f i n d t h e v a l u e o f \int_{0}^{1} \frac{1 + x^{4}}{{(1 + x^{2})}^{3}} d x$
Commented by mathmax by abdo last updated on 17/Aug/19
$1)let A =∫_0 ^∞ (dx/((1+x^2 )^3 )) ⇒2A =∫_(−∞) ^(+∞) (dx/((x^2 +1)^3 )) let W(z)=(1/((z^2 +1)^3 )) ⇒W(z)=(1/((z−i)^3 (z+i)^3 )) so the poles of W are +^− i (triples) residus theoreme give ∫_(−∞) ^(+∞) W(z)dz=2iπRes(W,i) Res(W,i)=lim_(z→i) (1/((3−1)!)){(z−i)^3 W(z)}^((2)) =lim_(z→i) (1/2){(z+i)^(−3) }^((2)) =lim_(z→i) (1/2){−3(z+i)^(−4) }^((1)) =lim_(z→i) (1/2){12(z+i)^(−5) } =6(2i)^(−5) =(6/((2i)^5 )) =(6/(2^5 i)) =(3/(16i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(3/(16i)) =((3π)/8) ⇒A =((3π)/(16)) 2)we have ∫_0 ^∞ (dx/((1+x^2 )^3 )) =∫_0 ^1 (dx/((1+x^2 )^3 )) +∫_1 ^(+∞) (dx/((1+x^2 )^3 )) and ∫_1 ^(+∞) (dx/((1+x^2 )^3 )) =_(x=(1/t)) −∫_0 ^1 (−/((1+(1/t^2 ))^3 ))(dt/t^2 ) =∫_0 ^∞ (t^6 /(t^2 (1+t^2 )^3 ))dt =∫_0 ^∞ (t^4 /((1+t^2 )^3 ))dt ⇒∫_0 ^∞ (dx/((1+x^2 )^3 )) =∫_0 ^1 (dx/((1+x^2 )^3 )) +∫_0 ^∞ (x^4 /((1+x^2 )^3 ))dx =∫_0 ^∞ ((1+x^4 )/((1+x^2 )^3 ))dx ⇒∫_0 ^∞ ((1+x^4 )/((1+x^2 )^3 ))=((3π)/(16)) .$
$1) l e t A = \int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{3}} \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{3}}$ $l e t W (z) = \frac{1}{{(z^{2} + 1)}^{3}} \Rightarrow W (z) = \frac{1}{{(z - i)}^{3} {(z + i)}^{3}} s o t h e p o l e s o f W a r e$ $\bar{+} i (t r i p l e s) r e s i d u s t h e o r e m e g i v e \int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)$ $R e s (W, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} W (z)}^{(2)}$ $= {l i m}_{z \to i} \frac{1}{2} {{(z + i)}^{- 3}}^{(2)} = {l i m}_{z \to i} \frac{1}{2} {- 3 {(z + i)}^{- 4}}^{(1)}$ $= {l i m}_{z \to i} \frac{1}{2} {12 {(z + i)}^{- 5}} = 6 {(2 i)}^{- 5} = \frac{6}{{(2 i)}^{5}} = \frac{6}{2^{5} i} = \frac{3}{16 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times \frac{3}{16 i} = \frac{3 π}{8} \Rightarrow A = \frac{3 π}{16}$ $2) w e h a v e \int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{3}} = \int_{0}^{1} \frac{d x}{{(1 + x^{2})}^{3}} + \int_{1}^{+ \infty} \frac{d x}{{(1 + x^{2})}^{3}} a n d$ $\int_{1}^{+ \infty} \frac{d x}{{(1 + x^{2})}^{3}} =_{x = \frac{1}{t}} - \int_{0}^{1} \frac{-}{{(1 + \frac{1}{t^{2}})}^{3}} \frac{d t}{t^{2}} = \int_{0}^{\infty} \frac{t^{6}}{t^{2} {(1 + t^{2})}^{3}} d t$ $= \int_{0}^{\infty} \frac{t^{4}}{{(1 + t^{2})}^{3}} d t \Rightarrow \int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{3}} = \int_{0}^{1} \frac{d x}{{(1 + x^{2})}^{3}} + \int_{0}^{\infty} \frac{x^{4}}{{(1 + x^{2})}^{3}} d x$ $= \int_{0}^{\infty} \frac{1 + x^{4}}{{(1 + x^{2})}^{3}} d x \Rightarrow \int_{0}^{\infty} \frac{1 + x^{4}}{{(1 + x^{2})}^{3}} = \frac{3 π}{16} .$
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