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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...

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

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