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Question Number 220038 by Nicholas666 last updated on 04/May/25
∫^( 1) _( 0) (1/( (√(x^8 + 1)))) dx
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$\: \\ $$
Answered by MrGaster last updated on 04/May/25
=∫_0 ^1 (1+x^8 )^(−(1/2)) dx =∫_0 ^1 Σ_(k=1) ^∞ (((−1)^k ((1/2))_k )/(k!))x^(8k) dx =Σ_(k=0) ^∞ (((−1)^k ((1/2))_k )/(k!))∫_0 ^1 x^(8k) dx =Σ_(k=1) ^∞ (((−1)^k ((1/2))_k )/(k!(k!+1))) =(1/8)Σ_(k=0) ^∞ ((((1/8))_k ((1/2))_k )/(((9/8))_k )) (((−1)^k )/(k!)) =(1/8)∙((Γ((1/8))Γ((9/8)−(1/2)))/(Γ((9/8))Γ((1/8)−(1/2)+(9/8)))) = _2 F_1 ((1/8),(1/2);(9/8);−1) (2): ∫^( 1) _( 0) (1/( (√(x^8 + 1)))) dx=^(t=x^8 ) (1/8)∫_0 ^1 t^(−(7/8)) (1+t)^(−(1/2)) dt=(1/8)∙8∙((Γ((1/8))Γ((9/8)−(1/8)))/(Γ((9/8))))∙ _2 F_1 ((1/2),(1/8);(9/8);−1) =((Γ((1/8))Γ((7/8)))/(Γ((9/8))))∙ _2 F_1 ((1/8),(1/2);(9/8);−1)=^(Γ(s)Γ(1−s)=(π/(sin(πs)))) (π/(sin((π/8))))∙ _2 F_1 ((1/8),(1/2);(9/8);−1) ⇒ _2 F_1 ((1/8),(1/2);(9/8);−1)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{8}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{{k}!}{x}^{\mathrm{8}{k}} {dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{{k}!}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{8}{k}} {dx} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{{k}!\left({k}!+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{8}}\right)_{{k}} \left(\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}} {\right)}}{\left(\frac{\mathrm{9}}{\mathrm{8}}\right)_{{k}} }\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{8}}\right)} \\ $$$$=\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{9}}{\mathrm{8}};−\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right):\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx}\overset{{t}={x}^{\mathrm{8}} } {=}\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}+{t}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt}=\frac{\mathrm{1}}{\mathrm{8}}\centerdot\mathrm{8}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}\right)}{\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)}\centerdot\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{8}};\frac{\mathrm{9}}{\mathrm{8}};−\mathrm{1}\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)}\centerdot\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{9}}{\mathrm{8}};−\mathrm{1}\right)\overset{\Gamma\left({s}\right)\Gamma\left(\mathrm{1}−{s}\right)=\frac{\pi}{\mathrm{sin}\left(\pi{s}\right)}} {=}\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)}\centerdot\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{9}}{\mathrm{8}};−\mathrm{1}\right) \\ $$$$\Rightarrow\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{9}}{\mathrm{8}};−\mathrm{1}\right) \\ $$

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