Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 184031 by SEKRET last updated on 02/Jan/23

∫_1 ^∞ ((x^7 −x)/(x^(10) −1))dx = ?

$$\:\int_{\mathrm{1}} ^{\infty} \:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{7}} −\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}^{\mathrm{10}} −\mathrm{1}}\boldsymbol{\mathrm{dx}}\:=\:? \\ $$

Answered by ARUNG_Brandon_MBU last updated on 02/Jan/23

∫_1 ^∞ ((x^7 −x)/(x^(10) −1))dx=∫_0 ^1 (((1/x^7 )−(1/x))/((1/x^(10) )−1))∙(1/x^2 )dx=∫_0 ^1 ((x−x^7 )/(1−x^(10) ))dx =(1/(10))∫_0 ^1 ((x^(−(4/5)) −x^(−(1/5)) )/(1−x))dx=(1/(10))(ψ((4/5))−ψ((1/5))) =(π/(10))cot((π/5))=(π/(10))(√(1+(2/( (√5)))))

$$\int_{\mathrm{1}} ^{\infty} \frac{{x}^{\mathrm{7}} −{x}}{{x}^{\mathrm{10}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}}{{x}^{\mathrm{7}} }−\frac{\mathrm{1}}{{x}}}{\frac{\mathrm{1}}{{x}^{\mathrm{10}} }−\mathrm{1}}\centerdot\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{x}^{\mathrm{7}} }{\mathrm{1}−{x}^{\mathrm{10}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{−\frac{\mathrm{4}}{\mathrm{5}}} −{x}^{−\frac{\mathrm{1}}{\mathrm{5}}} }{\mathrm{1}−{x}}{dx}=\frac{\mathrm{1}}{\mathrm{10}}\left(\psi\left(\frac{\mathrm{4}}{\mathrm{5}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{10}}\mathrm{cot}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\pi}{\mathrm{10}}\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}} \\ $$

Commented by SEKRET last updated on 02/Jan/23

beatiful solution. thanks sir

$$\boldsymbol{\mathrm{beatiful}}\:\boldsymbol{\mathrm{solution}}.\:\:\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{sir}} \\ $$

Commented by ARUNG_Brandon_MBU last updated on 02/Jan/23

����

Commented by manxsol last updated on 02/Jan/23

what function is ψ ?

$${what}\:{function}\:{is}\:\:\psi\:\:? \\ $$

Commented by ARUNG_Brandon_MBU last updated on 02/Jan/23

digamma Γ ′(x)=Γ(x)ψ(x) ψ(α)=−γ+∫_0 ^1 ((1−t^(α−1) )/(1−t))dt ψ(x)−ψ(1−x)=−πcot(πx)

$$\mathrm{digamma} \\ $$$$\Gamma\:'\left({x}\right)=\Gamma\left({x}\right)\psi\left({x}\right) \\ $$$$\psi\left(\alpha\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{\alpha−\mathrm{1}} }{\mathrm{1}−{t}}{dt} \\ $$$$\psi\left({x}\right)−\psi\left(\mathrm{1}−{x}\right)=−\pi\mathrm{cot}\left(\pi{x}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com