let f(x) =∫_0 ^∞ (dt/((x−t +t^2 )^3 )) with x>(1/4) 1) calculate f(x) 2) calculate also g(x) =∫_0 ^∞ (dt/((x−t+t^2 )^4 )) 3)find the values of ∫_0 ^∞ (dt/((1−t+t^2 )^3 )) and ∫


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Question Number 65061 by mathmax by abdo last updated on 24/Jul/19

$l e t f (x) = \int_{0}^{\infty} \frac{d t}{{(x - t + t^{2})}^{3}} w i t h x > \frac{1}{4}$ $1) c a l c u l a t e f (x)$ $2) c a l c u l a t e a l s o g (x) = \int_{0}^{\infty} \frac{d t}{{(x - t + t^{2})}^{4}}$ $3) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{d t}{{(1 - t + t^{2})}^{3}} a n d \int_{0}^{\infty} \frac{d t}{{(2 - t + t^{2})}^{4}}$
Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19
$t^2 −t+x=(t−(1/2))^2 +((4x−1)/4)=((4x−1)/4)[(((2t−1)/(√(4x−1))))^2 +1] cause x>(1/4) ⇒ 4x−1>0 f(x)=((4/(4x−1)))^3 ∫_0 ^∞ (dt/([(((2t−1)/(√(4x−1))) )^2 +1]^3 )) let change u=arctan(((2t−1)/(√(4x−1)))) then du=(((2/(√(4x−1))) )/((((2t−1)/(√(4x−1))))^2 +1))dt if t=0 then u=arctan(((−1)/(√(4x−1))))=((−π)/2)+arctan((√(4x−1)))=θ_0 when t=∞ u=(π/2) Then f(x)= ((4/(4x−1)))^3 .(2/(√(4x−1))) ∫_θ_0 ^(π/2) (du/([tan^2 u +1]^2 )) =((4/(4x−1)))^(7/2) .∫_θ_0 ^(π/2) cos^4 u du it is easy to prove that {_(cos^4 u −sin^4 u= cos2u ) ^(cos^4 u +sin^4 u = 1−((sin2u)/2)) then we got cos^4 u = (1/2)−((sin2u)/4)+((cos2u)/2) So f(x)= ((4/(4x−1)))^(7/2) . [(u/2) +((cos2u)/8) +((sin2u)/4) ]_θ_0 ^(π/2) knowing that {_(sin2u=((2tanu)/(1+tan^2 u))) ^(cos2u=((1−tan^2 u)/(1+tan^2 u))) we got f(x)=((4/(4x−1)))^(7/2) .[ (π/4)−(1/8) − ( ((arctan(√(4x−1)))/2) −(π/4) +(1/8).((1−(1/(4x−1)))/(1+(1/(4x−1)))) + (1/4). ((2.((−1)/(√(4x−1))))/(1+(1/(4x−1)))) )] f(x)=((4/(4x−1)))^(7/2) .[ (π/2)−(1/4) −((arctan(√(4x−1)))/2) −((2x−1−2(√(4x−1)))/(16x)) ] 2) (df/dx)=∫_(0 ) ^∞ ((d((1/([x−t+t^2 ]^3 ))))/dx)dt = −3 g(x) so g(x)=((−1)/3) (df/(dx )). 3) by replacing x=1 on the final expression of f(x) we got f(1)=((4/3))^(7/2) [(π/2)−(1/4)−(π/(3×2))−((1−2(√3))/(16))] f(2)=((4/7))^(7/2) .[(π/2)−(1/4)−((arctan(√7))/2) −((3−2(√7))/(16))]$
$t^{2} - t + x = {(t - \frac{1}{2})}^{2} + \frac{4 x - 1}{4} = \frac{4 x - 1}{4} [{(\frac{2 t - 1}{\sqrt{4 x - 1}})}^{2} + 1] c a u s e x > \frac{1}{4} \Rightarrow 4 x - 1 > 0$ $f (x) = {(\frac{4}{4 x - 1})}^{3} \int_{0}^{\infty} \frac{d t}{{[{(\frac{2 t - 1}{\sqrt{4 x - 1}})}^{2} + 1]}^{3}}$ $l e t c h a n g e u = a r c t a n (\frac{2 t - 1}{\sqrt{4 x - 1}}) t h e n d u = \frac{\frac{2}{\sqrt{4 x - 1}}}{{(\frac{2 t - 1}{\sqrt{4 x - 1}})}^{2} + 1} d t$ $i f t = 0 t h e n u = a r c t a n (\frac{- 1}{\sqrt{4 x - 1}}) = \frac{- π}{2} + a r c t a n (\sqrt{4 x - 1}) = θ_{0}$ $w h e n t = \infty u = \frac{π}{2}$ $T h e n f (x) = {(\frac{4}{4 x - 1})}^{3} . \frac{2}{\sqrt{4 x - 1}} \int_{θ_{0}}^{\frac{π}{2}} \frac{d u}{{[{t a n}^{2} u + 1]}^{2}}$ $= {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . \int_{θ_{0}}^{\frac{π}{2}} {c o s}^{4} u d u$ $i t i s e a s y t o p r o v e t h a t {_{{c o s}^{4} u - {s i n}^{4} u = c o s 2 u}^{{c o s}^{4} u + {s i n}^{4} u = 1 - \frac{s i n 2 u}{2}} t h e n w e g o t {c o s}^{4} u = \frac{1}{2} - \frac{s i n 2 u}{4} + \frac{c o s 2 u}{2}$ $S o$ $f (x) = {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . {[\frac{u}{2} + \frac{c o s 2 u}{8} + \frac{s i n 2 u}{4}]}_{θ_{0}}^{\frac{π}{2}}$ $k n o w i n g t h a t {_{s i n 2 u = \frac{2 t a n u}{1 + {t a n}^{2} u}}^{c o s 2 u = \frac{1 - {t a n}^{2} u}{1 + {t a n}^{2} u}} w e g o t$ $f (x) = {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . [\frac{π}{4} - \frac{1}{8} - (\frac{a r c t a n \sqrt{4 x - 1}}{2} - \frac{π}{4} + \frac{1}{8} . \frac{1 - \frac{1}{4 x - 1}}{1 + \frac{1}{4 x - 1}} + \frac{1}{4} . \frac{2 . \frac{- 1}{\sqrt{4 x - 1}}}{1 + \frac{1}{4 x - 1}})]$ $f (x) = {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . [\frac{π}{2} - \frac{1}{4} - \frac{a r c t a n \sqrt{4 x - 1}}{2} - \frac{2 x - 1 - 2 \sqrt{4 x - 1}}{16 x}]$ $2) \frac{d f}{d x} = \int_{0}^{\infty} \frac{d (\frac{1}{{[x - t + t^{2}]}^{3}})}{d x} d t = - 3 g (x)$ $s o g (x) = \frac{- 1}{3} \frac{d f}{d x} .$ $3) b y r e p l a c i n g x = 1 o n t h e f i n a l e x p r e s s i o n o f f (x) w e g o t$ $f (1) = {(\frac{4}{3})}^{\frac{7}{2}} [\frac{π}{2} - \frac{1}{4} - \frac{π}{3 \times 2} - \frac{1 - 2 \sqrt{3}}{16}]$ $f (2) = {(\frac{4}{7})}^{\frac{7}{2}} . [\frac{π}{2} - \frac{1}{4} - \frac{a r c t a n \sqrt{7}}{2} - \frac{3 - 2 \sqrt{7}}{16}]$
Commented by mathmax by abdo last updated on 25/Jul/19

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