Question Number 220040 by Nicholas666 last updated on 04/May/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\:\infty} \:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$ \\ $$
Answered by MrGaster last updated on 04/May/25
![x^8 =t⇒x=t^(1/8) ,dx=(1/8)t^(−7/8) dt ∫_0 ^∞ (1/( (√(t+1))))∙(1/( 8))t^(−7/8) dt=(1/8)∫_0 ^∞ t^(−7/8) (1+t)^(−1/2) dt B(a,b)∫_0 ^∞ (t^(a−1) /((1+t)^(a+b) ))dt=((Γ(a)Γ(b))/(Γ(a+b))) a=(1/8),b=(3/8) (1/8)B((1/8),(3/8))=(1/8)∙((Γ((1/8))Γ((3/8)))/(Γ((1/2)))) Γ((1/2))=(√π),Γ((9/8))=(1/8)Γ((1/8)) (1/8)∙((8Γ((9/8))Γ((3/8)))/( (√π)))=((Γ((3/8))Γ((9/8)))/( (√π))) (2):=(1/8)∫_0 ^∞ t^(−(7/8)) (1+t)^(−(1/2)) dt [t=x^8 ] =(1/8)B((1/8),(3/8)) =(1/8)[((Γ((1/8))Γ((3/8)))/(Γ((1/2))))] =(1/8)[((Γ((1/8))Γ((3/8)))/( (√π)))] Γ((9/8))=(1/8)Γ((1/8))⇒Γ((1/8))=8Γ((9/8)) =((Γ((3/8))Γ((9/8)))/( (√π)))](https://www.tinkutara.com/question/Q220060.png)
$${x}^{\mathrm{8}} ={t}\Rightarrow{x}={t}^{\mathrm{1}/\mathrm{8}} ,{dx}=\frac{\mathrm{1}}{\mathrm{8}}{t}^{−\mathrm{7}/\mathrm{8}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\:\sqrt{{t}+\mathrm{1}}}\centerdot\frac{\mathrm{1}}{\:\mathrm{8}}{t}^{−\mathrm{7}/\mathrm{8}} {dt}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} {t}^{−\mathrm{7}/\mathrm{8}} \left(\mathrm{1}+{t}\right)^{−\mathrm{1}/\mathrm{2}} {dt} \\ $$$${B}\left({a},{b}\right)\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{{a}+{b}} }{dt}=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{8}},{b}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}{B}\left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{3}}{\mathrm{8}}\right)=\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi},\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)=\frac{\mathrm{1}}{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\mathrm{8}\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\:\sqrt{\pi}}=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)}{\:\sqrt{\pi}} \\ $$$$\left(\mathrm{2}\right):=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} {t}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}+{t}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt}\:\left[{t}={x}^{\mathrm{8}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}{B}\left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{3}}{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\:\sqrt{\pi}}\right] \\ $$$$\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)=\frac{\mathrm{1}}{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Rightarrow\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)=\mathrm{8}\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{8}}\right)}{\:\sqrt{\pi}} \\ $$