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Question Number 201266 by Mathspace last updated on 02/Dec/23
calculate ∫_1 ^∞ (dx/( (√(1+x^3 ))))
$${calculate}\:\int_{\mathrm{1}} ^{\infty} \frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }} \\ $$
Answered by witcher3 last updated on 03/Dec/23
(1/x^3 )=y ⇔(1/3)∫_0 ^1 (y^(−(5/6)) /( (√(1+y))))dy (1+y)^(−(1/2)) =1+Σ_(n≥1) (−(1/2))....(−(1/2)−n+1).(x^n /(n!)) (y^(−(1/6)) /( (√(1+y))))=Σ_(n≥0) ((Γ((1/2)+n)(−y)^n y^(−(5/6)) )/(Γ((1/2))n!)) ⇔(1/3)∫_0 ^1 Σ_(n≥0) ((Γ((1/2)+n))/(Γ((1/2))n!))(−1)^n .y^(−(5/6)+n) dy=(1/3)Σ_(n≥0) (−1)^n ((Γ((1/2)+n))/(Γ((1/2))n!)).(1/(n+(1/6))) =2Σ_(n≥0) ((((Γ(n+(1/2)))/(Γ((1/2)))).((Γ(n+(1/6)))/(Γ((1/6)))))/((Γ((7/6)+n))/(Γ((7/6)))))(((−1)^n )/(n!)) =2 _2 F_1 ((1/2),(1/6),(7/6),−1)≈1,89
$$\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{y} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{y}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\:\sqrt{\mathrm{1}+\mathrm{y}}}\mathrm{dy} \\ $$$$\left(\mathrm{1}+\mathrm{y}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{1}+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)….\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{n}+\mathrm{1}\right).\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$\frac{\mathrm{y}^{−\frac{\mathrm{1}}{\mathrm{6}}} }{\:\sqrt{\mathrm{1}+\mathrm{y}}}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{n}\right)\left(−\mathrm{y}\right)^{\mathrm{n}} \mathrm{y}^{−\frac{\mathrm{5}}{\mathrm{6}}} }{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{n}!} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{n}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{n}!}\left(−\mathrm{1}\right)^{\mathrm{n}} .\mathrm{y}^{−\frac{\mathrm{5}}{\mathrm{6}}+\mathrm{n}} \mathrm{dy}=\frac{\mathrm{1}}{\mathrm{3}}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{n}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{n}!}.\frac{\mathrm{1}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$=\mathrm{2}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\frac{\Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}.\frac{\Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{6}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}}{\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{6}}+\mathrm{n}\right)}{\Gamma\left(\frac{\mathrm{7}}{\mathrm{6}}\right)}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\mathrm{2}\:\:\:_{\mathrm{2}} \mathrm{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{7}}{\mathrm{6}},−\mathrm{1}\right)\approx\mathrm{1},\mathrm{89} \\ $$
Answered by Calculusboy last updated on 03/Dec/23
Solution: ∫_1 ^∞ (1+x^3 )^(−(1/2)) dx let x^(−3) +1=t^2 x=(t^2 −1)^(−(1/3)) dx=−(1/3)(t^2 −1)^(−(4/3)) ∙2tdt=−(2/3)(t^2 −1)^(−(4/3)) tdt when x=∞ t=1 and when x=1 t=∞ I=∫_∞ ^1 ((−(2/3)(t^2 +1)^(−(4/3)) t dt)/([1+(t^2 −1)^(−1) ]^(1/2) )) ⇔ I=(2/3)∫_1 ^∞ (((t^2 −1)^(−(4/3)) t dt)/(((t^2 )^(1/2) )/((t^2 −1)^(1/2) ))) I=(2/3)∫_1 ^∞ (dt/((t^2 −1)^(5/6) )) =1.8947
$$\boldsymbol{{Solution}}:\:\int_{\mathrm{1}} ^{\infty} \left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{3}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{dx}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{x}}^{−\mathrm{3}} +\mathrm{1}=\boldsymbol{{t}}^{\mathrm{2}} \:\:\:\:\boldsymbol{{x}}=\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \:\:\:\:\:\boldsymbol{{dx}}=−\frac{\mathrm{1}}{\mathrm{3}}\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{1}\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} \centerdot\mathrm{2}\boldsymbol{{tdt}}=−\frac{\mathrm{2}}{\mathrm{3}}\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{1}\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} \boldsymbol{{tdt}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\infty\:\:\boldsymbol{{t}}=\mathrm{1}\:\:\boldsymbol{{and}}\:\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{1}\:\:\boldsymbol{{t}}=\infty \\ $$$$\boldsymbol{{I}}=\int_{\infty} ^{\mathrm{1}} \:\frac{−\frac{\mathrm{2}}{\mathrm{3}}\left(\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{1}\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} \boldsymbol{{t}}\:\boldsymbol{{dt}}}{\left[\mathrm{1}+\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{1}\right)^{−\mathrm{1}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} }\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{1}} ^{\infty} \:\frac{\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{1}\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} \boldsymbol{{t}}\:\boldsymbol{{dt}}}{\frac{\left(\boldsymbol{{t}}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{1}} ^{\infty} \:\frac{\boldsymbol{{dt}}}{\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{6}}} }\:\:=\mathrm{1}.\mathrm{8947} \\ $$

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