Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 139313 by Dwaipayan Shikari last updated on 25/Apr/21

∫_0 ^∞ (dx/((1+x^5 )^5 ))=((798(√2))/(5^5 (√(5−(√5)))))π

$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{5}} \right)^{\mathrm{5}} }=\frac{\mathrm{798}\sqrt{\mathrm{2}}}{\mathrm{5}^{\mathrm{5}} \sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}\pi \\ $$

Answered by Ar Brandon last updated on 25/Apr/21

I=∫_0 ^∞ (dx/((1+x^5 )^5 )) ,x=u^(1/5) ⇒dx=(1/5)u^(−4/5) du =(1/5)∫_0 ^∞ (u^(−4/5) /((1+u)^5 ))du=(1/5)β((1/5), ((24)/5)) =(1/5)∙((Γ((1/5))Γ(((24)/5)))/(Γ(5)))=(1/5)∙(1/(4!))∙((19)/5)∙((14)/5)∙(9/5)∙(4/5)Γ((1/5))Γ((4/5)) =(1/5)∙(1/(24))∙((19)/5)∙((14)/5)∙(9/5)∙(4/5)∙(π/(sin((π/5))))

$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{5}} \right)^{\mathrm{5}} }\:,\mathrm{x}=\mathrm{u}^{\mathrm{1}/\mathrm{5}} \Rightarrow\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{u}^{−\mathrm{4}/\mathrm{5}} \mathrm{du} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\mathrm{4}/\mathrm{5}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{5}} }\mathrm{du}=\frac{\mathrm{1}}{\mathrm{5}}\beta\left(\frac{\mathrm{1}}{\mathrm{5}},\:\frac{\mathrm{24}}{\mathrm{5}}\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{24}}{\mathrm{5}}\right)}{\Gamma\left(\mathrm{5}\right)}=\frac{\mathrm{1}}{\mathrm{5}}\centerdot\frac{\mathrm{1}}{\mathrm{4}!}\centerdot\frac{\mathrm{19}}{\mathrm{5}}\centerdot\frac{\mathrm{14}}{\mathrm{5}}\centerdot\frac{\mathrm{9}}{\mathrm{5}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\centerdot\frac{\mathrm{1}}{\mathrm{24}}\centerdot\frac{\mathrm{19}}{\mathrm{5}}\centerdot\frac{\mathrm{14}}{\mathrm{5}}\centerdot\frac{\mathrm{9}}{\mathrm{5}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}\centerdot\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)} \\ $$

Commented by Dwaipayan Shikari last updated on 25/Apr/21

Thanks! .But another method?

$${Thanks}!\:.{But}\:{another}\:{method}?\: \\ $$

Commented by Ar Brandon last updated on 25/Apr/21

You can do the partial fraction bro. Haha! 😄

$$\mathrm{You}\:\mathrm{can}\:\mathrm{do}\:\mathrm{the}\:\mathrm{partial}\:\mathrm{fraction}\:\mathrm{bro}.\:\mathrm{Haha}! \\ $$😄

Answered by MJS_new last updated on 25/Apr/21

using Ostrogeadski′s Method we get ∫(dx/((x^5 +1)^5 ))= =((x(1197x^(15) +4256x^(10) +5396x^5 +2712))/(7500(x^5 +1)^4 ))+((399)/(625))∫(dx/(x^5 +1)) and ∫(dx/(x^5 +1))=(1/5)∫(dx/(x+1))−(1/5)∫((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1))dx= =(1/5)∫(dx/(x+1))−(1/5)∫(((1+(√5))x−4)/(2x^2 −(1+(√5))x+2))dx−(1/5)∫(((1−(√5))x−4)/(2x^2 −(1−(√5))x+2))dx now use formulas...

$$\mathrm{using}\:\mathrm{Ostrogeadski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{we}\:\mathrm{get} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{5}} +\mathrm{1}\right)^{\mathrm{5}} }= \\ $$$$=\frac{{x}\left(\mathrm{1197}{x}^{\mathrm{15}} +\mathrm{4256}{x}^{\mathrm{10}} +\mathrm{5396}{x}^{\mathrm{5}} +\mathrm{2712}\right)}{\mathrm{7500}\left({x}^{\mathrm{5}} +\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{399}}{\mathrm{625}}\int\frac{{dx}}{{x}^{\mathrm{5}} +\mathrm{1}} \\ $$$$\mathrm{and}\:\int\frac{{dx}}{{x}^{\mathrm{5}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){x}−\mathrm{4}}{\mathrm{2}{x}^{\mathrm{2}} −\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){x}+\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){x}−\mathrm{4}}{\mathrm{2}{x}^{\mathrm{2}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){x}+\mathrm{2}}{dx} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{formulas}... \\ $$

Commented by Dwaipayan Shikari last updated on 25/Apr/21

Thanks sir!

$${Thanks}\:{sir}! \\ $$

Commented by MJS_new last updated on 25/Apr/21

but it′s a hard path to go...

$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{hard}\:\mathrm{path}\:\mathrm{to}\:\mathrm{go}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com