let f(x) =∫_0 ^∞ (dt/((x^2 +t^2 )^2 )) with x>0 1) find a explicit form of (x) 2)find also g(x) =∫_0 ^∞ (dt/((x^2 +t^2 )^3 )) 3)find the values of integrals ∫_0 ^∞ (dt/((t^2 +3)^2 )) and ∫_0 ^∞ (dt/((t^2 +3)^3 )) 4) calculate U_θ =∫_0 ^∞ (dt/((t^2 +cos^2 θ)^2 )) with 0<θ<(π/2) 5) find f^((n)) (x) and f^((n)) (0) 6) developp f at integr serie


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Question Number 67008 by mathmax by abdo last updated on 21/Aug/19

$l e t f (x) = \int_{0}^{\infty} \frac{d t}{{(x^{2} + t^{2})}^{2}} w i t h x > 0$ $1) f i n d a e x p l i c i t f o r m o f (x)$ $2) f i n d a l s o g (x) = \int_{0}^{\infty} \frac{d t}{{(x^{2} + t^{2})}^{3}}$ $3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{2}} a n d \int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{3}}$ $4) c a l c u l a t e U_{θ} = \int_{0}^{\infty} \frac{d t}{{(t^{2} + {c o s}^{2} θ)}^{2}} w i t h 0 < θ < \frac{π}{2}$ $5) f i n d f^{(n)} (x) a n d f^{(n)} (0)$ $6) d e v e l o p p f a t i n t e g r s e r i e$
Commented by mathmax by abdo last updated on 25/Aug/19
$1) f(x) =∫_0 ^∞ (dt/((x^2 +t^2 )^2 )) ⇒2f(x) =∫_(−∞) ^(+∞) (dt/((t^2 +x^2 )^2 )) let =_(t=xz) ∫_(−∞) ^(+∞) ((xdz)/(x^4 (z^2 +1)^2 )) =(1/x^3 ) ∫_(−∞) ^(+∞) (dz/((z^2 +1)^2 )) let ϕ(z)=(1/((z^2 +1)^2 )) we have ϕ(z) =(1/((z−i)^2 (z+i)^2 )) resi_ dus tbeorem hive ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(z+i)^(−2) }^((1)) =lim_(z→i) −2(z+i)^(−3) =−2(2i)^(−3) =((−2)/((2i)^3 )) =((−2)/(−8i)) =(1/(4i)) ⇒ ∫_(−∞) ^(+∞ ) ϕ(z)dz =2iπ×(1/(4i)) =(π/2) ⇒2f(x)=(π/(2x^3 )) ⇒f(x) =(π/(4x^3 )) (x>0) 2) we have f^′ (x) =−∫_0 ^∞ ((2(2x)(x^(2 ) +t^2 ))/((x^2 +t^2 )^4 ))dx =−4x ∫_0 ^∞ (dx/((x^2 +t^2 )^3 )) =−4x g(x) ⇒g(x) =−(1/(4x))f^′ (x) f^′ (x) =(π/4)(x^(−3) )^′ =(π/4)(−3)x^(−4) =((−3π)/(4x^4 )) ⇒ g(x) =−(1/(4x))×((−3π)/(4x^4 )) =((3π)/(16x^5 ))$
$1) f (x) = \int_{0}^{\infty} \frac{d t}{{(x^{2} + t^{2})}^{2}} \Rightarrow 2 f (x) = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + x^{2})}^{2}} l e t$ $=_{t = x z} \int_{- \infty}^{+ \infty} \frac{x d z}{x^{4} {(z^{2} + 1)}^{2}} = \frac{1}{x^{3}} \int_{- \infty}^{+ \infty} \frac{d z}{{(z^{2} + 1)}^{2}} l e t φ (z) = \frac{1}{{(z^{2} + 1)}^{2}}$ $w e h a v e φ (z) = \frac{1}{{(z - i)}^{2} {(z + i)}^{2}} {r e s i}_{} d u s t b e o r e m h i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} {{(z - i)}^{2} φ (z)}^{(1)} = {l i m}_{z \to i} {{(z + i)}^{- 2}}^{(1)}$ $= {l i m}_{z \to i} - 2 {(z + i)}^{- 3} = - 2 {(2 i)}^{- 3} = \frac{- 2}{{(2 i)}^{3}} = \frac{- 2}{- 8 i} = \frac{1}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{1}{4 i} = \frac{π}{2} \Rightarrow 2 f (x) = \frac{π}{2 x^{3}} \Rightarrow f (x) = \frac{π}{4 x^{3}} (x > 0)$ $2) w e h a v e f^{^{'}} (x) = - \int_{0}^{\infty} \frac{2 (2 x) (x^{2} + t^{2})}{{(x^{2} + t^{2})}^{4}} d x = - 4 x \int_{0}^{\infty} \frac{d x}{{(x^{2} + t^{2})}^{3}}$ $= - 4 x g (x) \Rightarrow g (x) = - \frac{1}{4 x} f^{^{'}} (x)$ $f^{^{'}} (x) = \frac{π}{4} {(x^{- 3})}^{^{'}} = \frac{π}{4} (- 3) x^{- 4} = \frac{- 3 π}{4 x^{4}} \Rightarrow$ $g (x) = - \frac{1}{4 x} \times \frac{- 3 π}{4 x^{4}} = \frac{3 π}{16 x^{5}}$
Commented by mathmax by abdo last updated on 25/Aug/19

$3) \int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{2}} = f (\sqrt{3}) = \frac{π}{4 {(\sqrt{3})}^{3}} = \frac{π}{12 \sqrt{3}} = \frac{π \sqrt{3}}{36}$ $\int_{0}^{\infty} \frac{d t}{{(t^{2} + 3)}^{3}} = g (\sqrt{3}) = \frac{3 π}{16 {(\sqrt{3})}^{5}} = \frac{3 π}{16 \times 9 \times \sqrt{3}} = \frac{π}{48 \sqrt{3}} = \frac{π \sqrt{3}}{144}$
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