Question Number 65061 by mathmax by abdo last updated on 24/Jul/19
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({x}−{t}\:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{with}\:\:\:{x}>\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{also}\:\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\left({x}−{t}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{values}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{2}−{t}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$ \\ $$
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))]](https://www.tinkutara.com/question/Q65088.png)
$$\:\:{t}^{\mathrm{2}} −{t}+{x}=\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{4}{x}−\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{4}{x}−\mathrm{1}}{\mathrm{4}}\left[\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)^{\mathrm{2}} +\mathrm{1}\right]\:\:{cause}\:{x}>\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:\mathrm{4}{x}−\mathrm{1}>\mathrm{0} \\ $$$$\:{f}\left({x}\right)=\left(\frac{\mathrm{4}}{\mathrm{4}{x}−\mathrm{1}}\right)^{\mathrm{3}} \int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\left[\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\:\right)^{\mathrm{2}} +\mathrm{1}\right]^{\mathrm{3}} } \\ $$$$\:\:{let}\:{change}\:{u}={arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)\:\:\:\:{then}\:\:\:{du}=\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\:}{\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\:\:{if}\:\:\:{t}=\mathrm{0}\:\:\:{then}\:\:{u}={arctan}\left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\right)=\frac{−\pi}{\mathrm{2}}+{arctan}\left(\sqrt{\mathrm{4}{x}−\mathrm{1}}\right)=\theta_{\mathrm{0}} \\ $$$${when}\:{t}=\infty\:\:\:{u}=\frac{\pi}{\mathrm{2}} \\ $$$${Then}\:\:{f}\left({x}\right)=\:\left(\frac{\mathrm{4}}{\mathrm{4}{x}−\mathrm{1}}\right)^{\mathrm{3}} .\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}\:\int_{\theta_{\mathrm{0}} } ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{du}}{\left[{tan}^{\mathrm{2}} {u}\:+\mathrm{1}\right]^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\left(\frac{\mathrm{4}}{\mathrm{4}{x}−\mathrm{1}}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} .\int_{\theta_{\mathrm{0}} } ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{4}} {u}\:{du} \\ $$$$\:\:\:\:{it}\:{is}\:{easy}\:{to}\:{prove}\:\:\:\:\:{that}\:\left\{_{{cos}^{\mathrm{4}} {u}\:−{sin}^{\mathrm{4}} {u}=\:{cos}\mathrm{2}{u}\:\:} ^{{cos}^{\mathrm{4}} {u}\:+{sin}^{\mathrm{4}} {u}\:=\:\mathrm{1}−\frac{{sin}\mathrm{2}{u}}{\mathrm{2}}} {then}\:\:{we}\:{got}\:\:{cos}^{\mathrm{4}} {u}\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{{sin}\mathrm{2}{u}}{\mathrm{4}}+\frac{{cos}\mathrm{2}{u}}{\mathrm{2}}\right. \\ $$$${So}\:\: \\ $$$${f}\left({x}\right)=\:\left(\frac{\mathrm{4}}{\mathrm{4}{x}−\mathrm{1}}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} .\:\left[\frac{{u}}{\mathrm{2}}\:+\frac{{cos}\mathrm{2}{u}}{\mathrm{8}}\:+\frac{{sin}\mathrm{2}{u}}{\mathrm{4}}\:\right]_{\theta_{\mathrm{0}} } ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:{knowing}\:{that}\:\:\left\{_{{sin}\mathrm{2}{u}=\frac{\mathrm{2}{tanu}}{\mathrm{1}+{tan}^{\mathrm{2}} {u}}} ^{{cos}\mathrm{2}{u}=\frac{\mathrm{1}−{tan}^{\mathrm{2}} {u}}{\mathrm{1}+{tan}^{\mathrm{2}} {u}}} \:\:\:{we}\:\:{got}\right. \\ $$$${f}\left({x}\right)=\left(\frac{\mathrm{4}}{\mathrm{4}{x}−\mathrm{1}}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} .\left[\:\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}\:−\:\left(\:\frac{{arctan}\sqrt{\mathrm{4}{x}−\mathrm{1}}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}.\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{1}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{1}}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}.\:\frac{\mathrm{2}.\frac{−\mathrm{1}}{\:\sqrt{\mathrm{4}{x}−\mathrm{1}}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{1}}}\:\right)\right] \\ $$$$\:\:{f}\left({x}\right)=\left(\frac{\mathrm{4}}{\mathrm{4}{x}−\mathrm{1}}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} .\left[\:\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{{arctan}\sqrt{\mathrm{4}{x}−\mathrm{1}}}{\mathrm{2}}\:−\frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{2}\sqrt{\mathrm{4}{x}−\mathrm{1}}}{\mathrm{16}{x}}\:\right] \\ $$$$\:\:\: \\ $$$$\left.\mathrm{2}\right)\:\:\:\frac{{df}}{{dx}}=\int_{\mathrm{0}\:\:\:\:} ^{\infty} \:\frac{{d}\left(\frac{\mathrm{1}}{\left[{x}−{t}+{t}^{\mathrm{2}} \right]^{\mathrm{3}} }\right)}{{dx}}{dt}\:=\:−\mathrm{3}\:{g}\left({x}\right)\: \\ $$$${so}\:\:{g}\left({x}\right)=\frac{−\mathrm{1}}{\mathrm{3}}\:\frac{{df}}{{dx}\:}. \\ $$$$\left.\mathrm{3}\right)\:\:{by}\:{replacing}\:\:{x}=\mathrm{1}\:\:\:{on}\:{the}\:{final}\:{expression}\:{of}\:{f}\left({x}\right)\:{we}\:{got} \\ $$$${f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} \left[\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\pi}{\mathrm{3}×\mathrm{2}}−\frac{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{16}}\right] \\ $$$${f}\left(\mathrm{2}\right)=\left(\frac{\mathrm{4}}{\mathrm{7}}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} .\left[\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{{arctan}\sqrt{\mathrm{7}}}{\mathrm{2}}\:−\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{16}}\right] \\ $$
Commented by mathmax by abdo last updated on 25/Jul/19
$${thank}\:{you}\:{sir}\:. \\ $$