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Question Number 42810 by maxmathsup by imad last updated on 02/Sep/18

find ∫_0 ^∞ (x^5 /(1+x^7 ))dx .

$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{5}} }{\mathrm{1}+{x}^{\mathrm{7}} }{dx}\:\:. \\ $$

Commented by prof Abdo imad last updated on 03/Sep/18

changement x=t^(1/7) give I = ∫_0 ^∞ (t^(5/7) /(1+t)) (1/7) t^((1/7)−1) dt =(1/7) ∫_0 ^∞ (t^((6/7)−1) /(1+t)) dt =(1/7) (π/(sin(((6π)/7)))) = (π/(7sin((π/7)))) .

$${changement}\:\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{7}}} \:\:\:{give}\: \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{5}}{\mathrm{7}}} }{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{7}}\:{t}^{\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{7}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\frac{\mathrm{6}}{\mathrm{7}}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{7}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)}\:=\:\frac{\pi}{\mathrm{7}{sin}\left(\frac{\pi}{\mathrm{7}}\right)}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18

x^7 =tan^2 α x=(tanα)^(2/7) dx=(2/7)tan^((−5)/7) α sec^2 α dα ∫_0 ^(Π/2) (((tanα)^((10)/7) )/(sec^2 α))×(2/7)(tanα)^((−5)/7) ×sec^2 αdα (2/7)∫_0 ^(Π/2) (tanα)^(5/7) αdα (2/7)∫_0 ^(Π/2) (sinα)^(5/7) (cosα)^((−5)/7) dα using formula ∫_0 ^(Π/2) sin^(2p−1) αcos^(2q−1) α dα =((⌈p ×⌈q)/(2⌈(p+q))) here 2p−1=(5/7) 2p=((12)/7) p=(6/7) 2q−1=−(5/7) 2q=(2/7) q=(1/7) =(2/7)∫_0 ^(Π/2) (sinα^ )^(2×(6/7)−1) (cosα)^(((2×1)/7)−1) dα (2/7)×((⌈((6/7))×⌈((1/7)))/(2⌈((6/7)+(1/7))))=(1/7)×⌈((1/7))×⌈(1−(1/7))=(1/7)×(Π/(sin((Π/7)))) ⌈(p)×⌈(1−p)=(Π/(sin(pΠ)))

$${x}^{\mathrm{7}} ={tan}^{\mathrm{2}} \alpha\:\:\:{x}=\left({tan}\alpha\right)^{\frac{\mathrm{2}}{\mathrm{7}}} \\ $$$${dx}=\frac{\mathrm{2}}{\mathrm{7}}{tan}^{\frac{−\mathrm{5}}{\mathrm{7}}} \alpha\:{sec}^{\mathrm{2}} \alpha\:{d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \frac{\left({tan}\alpha\right)^{\frac{\mathrm{10}}{\mathrm{7}}} }{{sec}^{\mathrm{2}} \alpha}×\frac{\mathrm{2}}{\mathrm{7}}\left({tan}\alpha\right)^{\frac{−\mathrm{5}}{\mathrm{7}}} ×{sec}^{\mathrm{2}} \alpha{d}\alpha \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \left({tan}\alpha\right)^{\frac{\mathrm{5}}{\mathrm{7}}} \alpha{d}\alpha \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \left({sin}\alpha\right)^{\frac{\mathrm{5}}{\mathrm{7}}} \left({cos}\alpha\right)^{\frac{−\mathrm{5}}{\mathrm{7}}} {d}\alpha \\ $$$${using}\:{formula} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \alpha{cos}^{\mathrm{2}{q}−\mathrm{1}} \alpha\:{d}\alpha\:=\frac{\lceil{p}\:×\lceil{q}}{\mathrm{2}\lceil\left({p}+{q}\right)} \\ $$$${here}\:\mathrm{2}{p}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{7}}\:\:\mathrm{2}{p}=\frac{\mathrm{12}}{\mathrm{7}}\:\:\:{p}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\mathrm{2}{q}−\mathrm{1}=−\frac{\mathrm{5}}{\mathrm{7}}\:\:\:\mathrm{2}{q}=\frac{\mathrm{2}}{\mathrm{7}}\:\:\:{q}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \left({sin}\alpha^{} \right)^{\mathrm{2}×\frac{\mathrm{6}}{\mathrm{7}}−\mathrm{1}} \left({cos}\alpha\right)^{\frac{\mathrm{2}×\mathrm{1}}{\mathrm{7}}−\mathrm{1}} {d}\alpha \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}×\frac{\lceil\left(\frac{\mathrm{6}}{\mathrm{7}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}\right)}{\mathrm{2}\lceil\left(\frac{\mathrm{6}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}}\right)}=\frac{\mathrm{1}}{\mathrm{7}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}\right)×\lceil\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{7}}×\frac{\Pi}{{sin}\left(\frac{\Pi}{\mathrm{7}}\right)} \\ $$$$ \\ $$$$\lceil\left({p}\right)×\lceil\left(\mathrm{1}−{p}\right)=\frac{\Pi}{{sin}\left({p}\Pi\right)} \\ $$

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