1) let f(x) =∫_0 ^(+∞) (dt/(t^3 +x^3 )) with x>0 calculate f(x) 2) find also g(x) =∫_0 ^∞ (dt/((t^3 +x^3 )^2 )) 3) find the values of integrals ∫_0 ^∞ (dt/(t^3 +1)) and ∫_0 ^∞ (dt/((t^3 +1)^2 )) 4) give f^((n)) (x) at form of integrals.


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Question Number 65288 by mathmax by abdo last updated on 27/Jul/19

$1) l e t f (x) = \int_{0}^{+ \infty} \frac{d t}{t^{3} + x^{3}} w i t h x > 0$ $c a l c u l a t e f (x)$ $2) f i n d a l s o g (x) = \int_{0}^{\infty} \frac{d t}{{(t^{3} + x^{3})}^{2}}$ $3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d t}{t^{3} + 1} a n d \int_{0}^{\infty} \frac{d t}{{(t^{3} + 1)}^{2}}$ $4) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l s .$
Commented by mathmax by abdo last updated on 30/Jul/19
$1) f(x)=∫_0 ^∞ (dt/(t^3 +x^3 ))dt ⇒f(x) =∫_0 ^∞ (dt/((t+x)(t^2 −xt +x^2 ))) let decompose F(t) =(1/((t+x)(t^2 −xt +x^2 ))) F(t) =(a/(t+x)) +((bt +c)/(t^2 −xt +x^2 )) (x>0) a =lim_(t→−x) (t+x)F(t) =(1/(3x^2 )) lim_(t→+∞) tF(t) =0 =a+b ⇒b =−(1/(3x^2 )) ⇒F(t)=(1/(3x^2 (t+x))) +((−(1/(3x^2 ))t+c)/(t^2 −xt+x^2 )) F(0) =(1/x^3 ) =(1/(3x^3 )) +(c/x^2 ) ⇒c =0 ⇒1=(1/3) +xc ⇒xc =(2/3) ⇒c =(2/(3x)) ⇒ F(t) =(1/(3x^2 (t+x))) +((−(t/(3x^2 ))+(2/(3x)))/(t^2 −xt+x^2 )) =(1/(3x^2 (t+x)))−(1/(3x^2 )) ((t−2x)/(t^2 −xt +x^2 )) ⇒ f(x) =(1/(3x^2 )){ ∫_0 ^∞ (dt/(t+x)) −∫_0 ^∞ ((t−2x)/(t^2 −xt +x^2 ))} =(1/(3x^2 )){ ∫_0 ^∞ (dt/(t+x)) −(1/2)∫_0 ^∞ ((2t−x+x)/(t^2 −xt +x^2 )) +∫_0 ^∞ ((2x)/(t^2 −xt+x^2 ))dt} =(1/(3x^2 )){ [ln∣((t+x)/(√(t^2 −xt+x^2 )))∣]_0 ^(+∞) +((3x)/2)∫_0 ^∞ (dt/(t^2 −xt +x^2 ))} =(1/(2x))∫_0 ^∞ (dt/(t^2 −xt +x^2 )) and ∫_0 ^∞ (dt/(t^2 −xt +x^2 )) =∫_0 ^∞ (dt/(t^2 −2(x/2)t +(x^2 /4)+x^2 −(x^2 /4))) =∫_0 ^∞ (dt/((t−(x/2))^2 +(3/4)x^2 )) =_(t−(x/2)=((√3)/2)xu) ∫_(−(1/(√3))) ^(+∞) (1/((3/4)x^2 (1+u^2 )))((√3)/2) xdu =(4/(3x^2 )) ((√3)/2) x ∫_(−(1/(√3))) ^(+∞) (du/(1+u^2 )) =(2/(x(√3)))[arctan(u)]_(−(1/(√3))) ^(+∞) =(2/(x(√3))){(π/2) +arctan((1/(√3)))} =(2/(x(√3))){π −arctan((√3)) =(2/(x(√3))){π−(π/3)} =(2/(x(√3)))((2π)/3) =((4π)/(3x(√3))) ⇒f(x)=(1/(2x)) ((4π)/(3x(√3))) =((2π)/(3x^2 (√3))) ⇒ f(x) =((2π)/(3(√3)x^2 )) (x>0)$
$1) f (x) = \int_{0}^{\infty} \frac{d t}{t^{3} + x^{3}} d t \Rightarrow f (x) = \int_{0}^{\infty} \frac{d t}{(t + x) (t^{2} - x t + x^{2})}$ $l e t d e c o m p o s e F (t) = \frac{1}{(t + x) (t^{2} - x t + x^{2})}$ $F (t) = \frac{a}{t + x} + \frac{b t + c}{t^{2} - x t + x^{2}} (x > 0)$ $a = {l i m}_{t \to - x} (t + x) F (t) = \frac{1}{3 x^{2}}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = - \frac{1}{3 x^{2}} \Rightarrow F (t) = \frac{1}{3 x^{2} (t + x)} + \frac{- \frac{1}{3 x^{2}} t + c}{t^{2} - x t + x^{2}}$ $F (0) = \frac{1}{x^{3}} = \frac{1}{3 x^{3}} + \frac{c}{x^{2}} \Rightarrow c = 0 \Rightarrow 1 = \frac{1}{3} + x c \Rightarrow x c = \frac{2}{3} \Rightarrow c = \frac{2}{3 x} \Rightarrow$ $F (t) = \frac{1}{3 x^{2} (t + x)} + \frac{- \frac{t}{3 x^{2}} + \frac{2}{3 x}}{t^{2} - x t + x^{2}} = \frac{1}{3 x^{2} (t + x)} - \frac{1}{3 x^{2}} \frac{t - 2 x}{t^{2} - x t + x^{2}} \Rightarrow$ $f (x) = \frac{1}{3 x^{2}} {\int_{0}^{\infty} \frac{d t}{t + x} - \int_{0}^{\infty} \frac{t - 2 x}{t^{2} - x t + x^{2}}}$ $= \frac{1}{3 x^{2}} {\int_{0}^{\infty} \frac{d t}{t + x} - \frac{1}{2} \int_{0}^{\infty} \frac{2 t - x + x}{t^{2} - x t + x^{2}} + \int_{0}^{\infty} \frac{2 x}{t^{2} - x t + x^{2}} d t}$ $= \frac{1}{3 x^{2}} {{[l n ∣ \frac{t + x}{\sqrt{t^{2} - x t + x^{2}}} ∣]}_{0}^{+ \infty} + \frac{3 x}{2} \int_{0}^{\infty} \frac{d t}{t^{2} - x t + x^{2}}}$ $= \frac{1}{2 x} \int_{0}^{\infty} \frac{d t}{t^{2} - x t + x^{2}} a n d$ $\int_{0}^{\infty} \frac{d t}{t^{2} - x t + x^{2}} = \int_{0}^{\infty} \frac{d t}{t^{2} - 2 \frac{x}{2} t + \frac{x^{2}}{4} + x^{2} - \frac{x^{2}}{4}}$ $= \int_{0}^{\infty} \frac{d t}{{(t - \frac{x}{2})}^{2} + \frac{3}{4} x^{2}} =_{t - \frac{x}{2} = \frac{\sqrt{3}}{2} x u} \int_{- \frac{1}{\sqrt{3}}}^{+ \infty} \frac{1}{\frac{3}{4} x^{2} (1 + u^{2})} \frac{\sqrt{3}}{2} x d u$ $= \frac{4}{3 x^{2}} \frac{\sqrt{3}}{2} x \int_{- \frac{1}{\sqrt{3}}}^{+ \infty} \frac{d u}{1 + u^{2}} = \frac{2}{x \sqrt{3}} {[a r c t a n (u)]}_{- \frac{1}{\sqrt{3}}}^{+ \infty}$ $= \frac{2}{x \sqrt{3}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{3}})} = \frac{2}{x \sqrt{3}} {π - a r c t a n (\sqrt{3}) = \frac{2}{x \sqrt{3}} {π - \frac{π}{3}}$ $= \frac{2}{x \sqrt{3}} \frac{2 π}{3} = \frac{4 π}{3 x \sqrt{3}} \Rightarrow f (x) = \frac{1}{2 x} \frac{4 π}{3 x \sqrt{3}} = \frac{2 π}{3 x^{2} \sqrt{3}} \Rightarrow$ $f (x) = \frac{2 π}{3 \sqrt{3} x^{2}} (x > 0)$
Commented by mathmax by abdo last updated on 30/Jul/19

$2) w e h a v e f (x) = \int_{0}^{\infty} \frac{d t}{t^{3} + x^{3}} \Rightarrow f^{^{'}} (x) = - \int_{0}^{\infty} \frac{3 x^{2}}{{(t^{3} + x^{3})}^{2}} d t$ $= - 3 x^{2} \int_{0}^{\infty} \frac{d t}{{(t^{3} + x^{3})}^{2}} = - 3 x^{2} g (x) \Rightarrow g (x) = - \frac{1}{3 x^{2}} f^{^{'}} (x)$ $f (x) = \frac{2 π}{3 \sqrt{3} x^{2}} \Rightarrow f^{^{'}} (x) = \frac{2 π}{3 \sqrt{3}} (- \frac{2 x}{x^{4}}) = - \frac{4 π}{3 \sqrt{3} x^{3}} \Rightarrow$ $g (x) = - \frac{1}{3 x^{2}} \times (- \frac{4 π}{3 \sqrt{3} x^{3}}) = \frac{4 π}{9 \sqrt{3} x^{5}}$
Commented by mathmax by abdo last updated on 30/Jul/19

$3) \int_{0}^{\infty} \frac{d t}{t^{3} + 1} = f (1) = \frac{2 π}{3 \sqrt{3}}$
Commented by mathmax by abdo last updated on 30/Jul/19

$\int_{0}^{\infty} \frac{d t}{{(t^{3} + 1)}^{2}} = g (1) = \frac{4 π}{9 \sqrt{3}}$
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