let f(x) =∫_0 ^(+∞) (dt/((t^2 +x^2 )^3 )) with x>0 1) find a explicit form off (x) 1) calculate ∫_0 ^∞ (dx/((t^2 +3)^3 )) and ∫_0 ^∞ (dt/((t^2 +4)^3 )) 2) find the value of A(θ) =∫_0 ^∞ (dt/((t^2 +sin^2 θ)^3 )) with 0<θ<π.


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Question Number 57899 by maxmathsup by imad last updated on 13/Apr/19

$l e t f (x) = \int_{0}^{+ \infty} \frac{d t}{{(t^{2} + x^{2})}^{3}} w i t h x > 0$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $1) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(t^{2} + 3)}^{3}} a n d \int_{0}^{\infty} \frac{d t}{{(t^{2} + 4)}^{3}}$ $2) f i n d t h e v a l u e o f A (θ) = \int_{0}^{\infty} \frac{d t}{{(t^{2} + {s i n}^{2} θ)}^{3}} w i t h 0 < θ < π .$
Commented by maxmathsup by imad last updated on 17/Apr/19
$1) we have f(x)=(1/2) ∫_(−∞) ^(+∞) (dt/((t^2 +x^2 )^3 )) let consider the complex function ϕ(z) =(1/((z^2 +x^2 )^3 )) poles of ϕ? we have ϕ(z) =(1/((z−ix)^3 (z+ix)^3 )) so the poles of are +^− ix (triples) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ix) Res(ϕ,ix) =lim_(z→ix) (1/((3−1)!)){ (z−ix)^3 ϕ(z)}^((2)) =lim_(z→ix) (1/2){ (1/((z+ix)^3 ))}^((2)) we have {(1/((z+ix)^3 ))}^((1)) =−((3(z+ix)^2 )/((z+ix)^6 )) =−(3/((z+ix)^4 )) ⇒{(1/((z+ix)^3 ))}^((2)) =−3 ((−4(z+ix)^3 )/((z+ix)^8 )) =((12)/((z+ix)^5 )) ⇒Res(ϕ,ix) =lim_(z→ix) (6/((z+ix)^5 )) =(6/((2ix)^5 )) =(6/(2^5 x^5 i)) =(6/(32ix^5 )) =(3/(16ix^5 )) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (3/(16 x^5 i)) =((3π)/(8 x^5 )) ⇒ f(x) =((3π)/(16 x^5 )) ( with x>0) 2) ∫_0 ^∞ (dt/((t^2 +3)^3 )) =f((√3)) = ((3π)/(16 ((√3))^5 )) =((3π)/(16 ((√3))^4 (√3))) =((3π)/(9×16 (√3))) ∫_0 ^∞ (dt/((t^2 +4)^3 )) =f(2) =((3π)/(16 (2)^5 )) =((3π)/(16×32))$
$1) w e h a v e f (x) = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + x^{2})}^{3}} l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{1}{{(z^{2} + x^{2})}^{3}} p o l e s o f φ ?$ $w e h a v e φ (z) = \frac{1}{{(z - i x)}^{3} {(z + i x)}^{3}} s o t h e p o l e s o f a r e \bar{+} i x (t r i p l e s) \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i x)$ $R e s (φ, i x) = {l i m}_{z \to i x} \frac{1}{(3 - 1)!} {{(z - i x)}^{3} φ (z)}^{(2)}$ $= {l i m}_{z \to i x} \frac{1}{2} {\frac{1}{{(z + i x)}^{3}}}^{(2)} w e h a v e$ ${\frac{1}{{(z + i x)}^{3}}}^{(1)} = - \frac{3 {(z + i x)}^{2}}{{(z + i x)}^{6}} = - \frac{3}{{(z + i x)}^{4}} \Rightarrow {\frac{1}{{(z + i x)}^{3}}}^{(2)}$ $= - 3 \frac{- 4 {(z + i x)}^{3}}{{(z + i x)}^{8}} = \frac{12}{{(z + i x)}^{5}} \Rightarrow R e s (φ, i x) = {l i m}_{z \to i x} \frac{6}{{(z + i x)}^{5}}$ $= \frac{6}{{(2 i x)}^{5}} = \frac{6}{2^{5} x^{5} i} = \frac{6}{32 {i x}^{5}} = \frac{3}{16 {i x}^{5}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{3}{16 x^{5} i} = \frac{3 π}{8 x^{5}} \Rightarrow$ $f (x) = \frac{3 π}{16 x^{5}} (w i t h x > 0)$ $2) \int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{3}} = f (\sqrt{3}) = \frac{3 π}{16 {(\sqrt{3})}^{5}} = \frac{3 π}{16 {(\sqrt{3})}^{4} \sqrt{3}} = \frac{3 π}{9 \times 16 \sqrt{3}}$ $\int_{0}^{\infty} \frac{d t}{{(t^{2} + 4)}^{3}} = f (2) = \frac{3 π}{16 {(2)}^{5}} = \frac{3 π}{16 \times 32}$
Commented by maxmathsup by imad last updated on 17/Apr/19

$3) \int_{0}^{\infty} \frac{d t}{{(t^{2} + {s i n}^{2} θ)}^{3}} = f (s i n θ) = \frac{3 π}{16 {s i n}^{5} θ}$
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