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Question Number 131547 by liberty last updated on 06/Feb/21
J = ∫_0 ^( ∞) ((x^8 −1)/(x^(10) +1)) dx ?
$$\:\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{x}^{\mathrm{8}} −\mathrm{1}}{\mathrm{x}^{\mathrm{10}} +\mathrm{1}}\:\mathrm{dx}\:? \\ $$
Answered by rs4089 last updated on 06/Feb/21
0
$$\mathrm{0} \\ $$
Answered by liberty last updated on 06/Feb/21
consider I=∫_0 ^∞ (x^8 /(x^(10) +1)) dx replace x by (1/x) I=∫_∞ ^( 0) ((((1/x))^8 )/(((1/x))^(10) +1)).(−(1/x^2 ))dx I= −∫_∞ ^0 (x^2 /(1+x^(10) )).((dx/x^2 ))=∫_0 ^∞ (dx/(x^(10) +1)) so we have J = ∫_0 ^∞ (x^8 /(x^(10) +1))dx−∫_0 ^∞ (dx/(x^(10) +1)) J=I−I = 0
$$\mathrm{consider}\:\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{x}^{\mathrm{10}} +\mathrm{1}}\:\mathrm{dx}\: \\ $$$$\mathrm{replace}\:\mathrm{x}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{I}=\int_{\infty} ^{\:\mathrm{0}} \:\frac{\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{8}} }{\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{10}} +\mathrm{1}}.\left(−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$\mathrm{I}=\:−\int_{\infty} ^{\mathrm{0}} \:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{10}} }.\left(\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} }\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{10}} +\mathrm{1}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{x}^{\mathrm{10}} +\mathrm{1}}\mathrm{dx}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{10}} +\mathrm{1}} \\ $$$$\mathrm{J}=\mathrm{I}−\mathrm{I}\:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 06/Feb/21
J =∫_0 ^∞ (x^8 /(1+x^(10) ))dx−∫_0 ^∞ (dx/(1+x^(10) )) but ∫_0 ^∞ (x^8 /(1+x^(10) ))dx=_(x^(10) =t) (1/(10))∫_0 ^∞ (t^(8/(10)) /(1+t))t^((1/(10))−1) =(1/(10))∫_0 ^∞ (t^((9/(10))−1) /(1+t))dt =(1/(10)).(π/(sin(((9π)/(10))))) also ∫_0 ^∞ (dx/(1+x^(10) )) =_(x^(10) =t) (1/(10))∫_0 ^∞ (t^((1/(10))−1) /(1+t))dt =(1/(10)).(π/(sin((π/(10))))) ⇒J =(π/(10)){(1/(sin(((9π)/(10)))))−(1/(sin((π/(10)))))} but sin(((9π)/(10)))=sin((π/(10))) ⇒ J=0
$$\mathrm{J}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{1}+\mathrm{x}^{\mathrm{10}} }\mathrm{dx}−\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{10}} }\:\:\mathrm{but}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{1}+\mathrm{x}^{\mathrm{10}} }\mathrm{dx}=_{\mathrm{x}^{\mathrm{10}} =\mathrm{t}} \:\:\frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\frac{\mathrm{8}}{\mathrm{10}}} }{\mathrm{1}+\mathrm{t}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\frac{\mathrm{9}}{\mathrm{10}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{10}}.\frac{\pi}{\mathrm{sin}\left(\frac{\mathrm{9}\pi}{\mathrm{10}}\right)}\:\:\mathrm{also}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{10}} }\:=_{\mathrm{x}^{\mathrm{10}} =\mathrm{t}} \:\:\frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}.\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{10}}\right)}\:\Rightarrow\mathrm{J}\:=\frac{\pi}{\mathrm{10}}\left\{\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{9}\pi}{\mathrm{10}}\right)}−\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\pi}{\mathrm{10}}\right)}\right\}\:\mathrm{but}\:\mathrm{sin}\left(\frac{\mathrm{9}\pi}{\mathrm{10}}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{10}}\right)\:\Rightarrow \\ $$$$\mathrm{J}=\mathrm{0} \\ $$

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