Question Number 109616 by mathmax by abdo last updated on 24/Aug/20
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$
Commented by mathdave last updated on 24/Aug/20
$${this}\:{is}\:{hypergeometric}\:{question}\:{or}\:{question} \\ $$
Answered by mathmax by abdo last updated on 25/Aug/20
![A =∫_0 ^1 (√(1+x^4 ))dx by parts A =[x(√(1+x^4 ))]_0 ^1 −∫_0 ^1 x.((4x^3 )/(2(√(1+x^4 ))))dx =(√2)−2 ∫_0 ^1 (x^4 /( (√(1+x^4 )))) dx =(√2)−2 ∫_0 ^1 ((x^4 +1−1)/( (√(1+x^4 )))) dx =(√2)−2 ∫_0 ^1 (√(1+x^4 ))dx +2 ∫_0 ^1 (dx/((1+x^4 )^(1/2) )) ⇒ 3A =(√2)+2 ∫_0 ^1 (dx/((1+x^4 )^(1/2) )) changement x=t^(1/4) give ∫_0 ^1 (dx/((1+x^4 )^(1/2) )) =(1/4)∫_0 ^1 (t^((1/4)−1) /((1+t)^(1/2) )) dt we know B(p,q) =∫_0 ^1 ((t^(p−1) +t^(q−1) )/((1+t)^(p+q) ))dt =(1/8)∫_0 ^1 ((t^((1/4)−1) +t^((1/4)−1) )/((1+t)^(1/2) )) (p=q =(1/4)) =(1/8)B((1/4),(1/4)) =(1/8).((Γ((1/4))Γ((1/4)))/(Γ((1/2)))) =((Γ^2 ((1/4)))/(8(√π))) ⇒3A =(√2) +2×((Γ^2 ((1/4)))/(8(√π))) =(√2) +((Γ^2 ((1/4)))/(4(√π))) ⇒ A =((√2)/3) +((Γ^2 ((1/4)))/(12(√π)))](https://www.tinkutara.com/question/Q109676.png)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{A}\:=\left[\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}.\frac{\mathrm{4x}^{\mathrm{3}} }{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }}\mathrm{dx} \\ $$$$=\sqrt{\mathrm{2}}−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{4}} }{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }}\:\mathrm{dx}\:=\sqrt{\mathrm{2}}−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{1}−\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }}\:\mathrm{dx} \\ $$$$=\sqrt{\mathrm{2}}−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:\Rightarrow \\ $$$$\mathrm{3A}\:=\sqrt{\mathrm{2}}+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:\:\mathrm{changement}\:\mathrm{x}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\mathrm{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:\mathrm{dt}\:\mathrm{we}\:\mathrm{know}\:\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{p}−\mathrm{1}} +\mathrm{t}^{\mathrm{q}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{p}+\mathrm{q}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:+\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:\left(\mathrm{p}=\mathrm{q}\:=\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{B}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{8}\sqrt{\pi}}\:\Rightarrow\mathrm{3A}\:=\sqrt{\mathrm{2}}\:+\mathrm{2}×\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{8}\sqrt{\pi}}\:=\sqrt{\mathrm{2}}\:+\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\sqrt{\pi}}\:\Rightarrow \\ $$$$\mathrm{A}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\:+\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{12}\sqrt{\pi}} \\ $$
Commented by mathdave last updated on 25/Aug/20
$${how}\:{did}\:{you}\:{get}\:{the}\:\frac{\mathrm{1}}{\mathrm{8}}\:\:{when}\:{it}\:{was}\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt} \\ $$$${has}\:\frac{\mathrm{1}}{\mathrm{2}}\:{didnt}\:{follow}\:{beta}\:{formular}\:{you}\:{stated} \\ $$
Commented by Aziztisffola last updated on 25/Aug/20
$$\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt}=\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt}+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt}\right) \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt}\right) \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} +\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt} \\ $$