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Question Number 205625 by Lindemann last updated on 25/Mar/24

∫_0 ^1 (√(1−x^4 ))dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$

Answered by mathzup last updated on 25/Mar/24

I=∫_0 ^1 (1−x^4 )^(1/2) dx (x=t^(1/4) ) =(1/4)∫_0 ^1 (1−t)^(1/2) t^((1/4)−1) dt we know B(p,q)=∫_0 ^1 t^(p−1) (1−t)^(q−1) dt (p>0 et q>0) ⇒I=(1/4)∫_0 ^1 t^((1/4)−1) (1−t)^((3/2)−1) dt =(1/4)B((1/4),(3/2))=(1/4)((Γ((1/4))Γ((3/2)))/(Γ((1/4)+(3/2)))) but Γ((3/2))=Γ(1+(1/2))=(1/2)Γ((1/2))=((√π)/2) Γ((1/4)+(3/2))=Γ((1/4)+1+(1/2)) =Γ((3/4)+1)=(3/4)Γ((3/4)) ⇒ I=(1/4)((Γ((1/4)).((√π)/2))/((3/4)Γ((3/4))))=((√π)/6)×((Γ((1/4)))/(Γ((3/4)))) Γ((1/4)).Γ((3/4))=Γ((1/4))Γ(1−(1/4))=(π/(sin((π/4)))) =(√2)π

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}\:\:\:\:\:\left({x}={t}^{\frac{\mathrm{1}}{\mathrm{4}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dt} \\ $$$${we}\:{know}\:{B}\left({p},{q}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{p}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{q}−\mathrm{1}} {dt}\:\left({p}>\mathrm{0}\:{et}\:{q}>\mathrm{0}\right) \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{B}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$${but}\:\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:\Rightarrow \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\frac{\sqrt{\pi}}{\mathrm{2}}}{\frac{\mathrm{3}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}=\frac{\sqrt{\pi}}{\mathrm{6}}×\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\sqrt{\mathrm{2}}\pi \\ $$

Commented by Frix last updated on 26/Mar/24

Something went wrong. (√2)π>4 but obviously ∫_0 ^1 (√(1−x^4 ))dx<1

$$\mathrm{Something}\:\mathrm{went}\:\mathrm{wrong}. \\ $$$$\sqrt{\mathrm{2}}\pi>\mathrm{4}\:\mathrm{but}\:\mathrm{obviously}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}<\mathrm{1} \\ $$

Commented by Lindemann last updated on 26/Mar/24

exactly. it is possible to do this without the beta function ?

$${exactly}. \\ $$$${it}\:{is}\:{possible}\:{to}\:{do}\:{this}\:{without}\:{the} \\ $$$${beta}\:{function}\:? \\ $$

Commented by mr W last updated on 26/Mar/24

I=((√π)/6)×((Γ((1/4)))/(Γ((3/4))))≈0.87401918 ≠(√2)π

$${I}=\frac{\sqrt{\pi}}{\mathrm{6}}×\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\approx\mathrm{0}.\mathrm{87401918}\:\neq\sqrt{\mathrm{2}}\pi \\ $$

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