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Question Number 77422 by john santu last updated on 06/Jan/20
can solve ∫ (dx/(x^(17) −1)) via   elementary calculus?
$$\mathrm{can}\:\mathrm{solve}\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{17}} −\mathrm{1}}\:\mathrm{via}\: \\ $$$$\mathrm{elementary}\:\mathrm{calculus}? \\ $$
Commented by aliesam last updated on 07/Jan/20
∫(1/(x^(17) −1)) dx  = −∫(1/(1−x^(17) )) ((x/x^(17) ))((x^(17) /x)) dx  =−∫(1−x)^(17−1)  (x)^(1−17)  (x)^(17−1)  dx  =−∫(1−x)^(17−1)  (x^(17) )^((1/(17))−1) (x)^(17−1)  dx  =−x_2 F_1 (1, (1/(17)) ; ((18)/(17)) ;x^(17) )+c
$$\int\frac{\mathrm{1}}{{x}^{\mathrm{17}} −\mathrm{1}}\:{dx} \\ $$$$=\:−\int\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{17}} }\:\left(\frac{{x}}{{x}^{\mathrm{17}} }\right)\left(\frac{{x}^{\mathrm{17}} }{{x}}\right)\:{dx} \\ $$$$=−\int\left(\mathrm{1}−{x}\right)^{\mathrm{17}−\mathrm{1}} \:\left({x}\right)^{\mathrm{1}−\mathrm{17}} \:\left({x}\right)^{\mathrm{17}−\mathrm{1}} \:{dx} \\ $$$$=−\int\left(\mathrm{1}−{x}\right)^{\mathrm{17}−\mathrm{1}} \:\left({x}^{\mathrm{17}} \right)^{\frac{\mathrm{1}}{\mathrm{17}}−\mathrm{1}} \left({x}\right)^{\mathrm{17}−\mathrm{1}} \:{dx} \\ $$$$=−{x}_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\:\frac{\mathrm{1}}{\mathrm{17}}\:;\:\frac{\mathrm{18}}{\mathrm{17}}\:;{x}^{\mathrm{17}} \right)+{c} \\ $$
Answered by mind is power last updated on 06/Jan/20
X^(17) −1=Π_(k=0) ^(16) (x−e^((2ikπ)/(17)) )  ∫(dx/(x^(17) −1))=∫(dx/(Π_(k=0) ^(16) (x−e^((2ikπ)/(17)) )))=∫{Σ_(k=0) ^(16) (a_k /(x−e^((2ikπ)/(17)) ))}dx  a_k =(1/(17e^(((2ikπ)/(17)).16) ))=((e^(−((2ikπ)/(17)))  )/(17))  ⇒∫(dx/(x^(17) −1))=∫Σ_(k=0) ^(16) (e^((−2ikπ)/(17)) /(17(x−e^((2ikπ)/(17)) ))).dx  we can see that  Σ_(k=0) ^(16) (e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))=((1/(x−1))+Σ_(k=1) ^8 ((e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))+(e^((−2iπ(17−k))/(17)) /(x−e^((2i(17−k)π)/(17)) ))))  =(1/(x−1))+Σ_(k=1) ^8 ((e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))+(e^((2ikπ)/(17)) /(x−e^((−2ikπ)/(17)) )))=(1/(x−1))+Σ_(k=1) ^8 ((2(xcos(((2kπ)/(17)))−cos(((2kπ)/(17)))))/(x^2 −2cos(((2πk)/(17)))x+1))  ∫2((xcos(((2kπ)/(17)))−cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))dx=∫{cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))+((2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))}dx  since ∣cos(((2kπ)/(17)))∣<1,∀k∈{1,.....8}   ∫cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2xcos(((2πk)/(17)))+1))=cos(((2kπ)/(17)))ln(x^2 −2xcos(((2kπ)/(17)))x+1)  ∫((2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))=(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))∫(dx/((x−cos(((2kπ)/(17)))^2 +sin^2 (((2kπ)/(17)))))  =(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))arctan((x/(sin(((2kπ)/(17)))))−cot(((2kπ)/(17))))  we Get  ∫(dx/(x^(17) −1))=(1/(17))∫{(1/(x−1))+Σ_(k=1) ^8 cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))+(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17)))).(1/(x^2 −2xcos(((2kπ)/(17)))+1))}dx  =(1/(17)){ln∣x−1∣+Σ_(k=1) ^8 [cos(((2kπ)/(17)))ln(x^2 −2xcos(((2πk)/(17)))+1)+(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))arctan((x/(sin(((2kπ)/(17)))))−cot(((2kπ)/(17))))]}+c  c∈R constant
$$\mathrm{X}^{\mathrm{17}} −\mathrm{1}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{16}} {\prod}}\left(\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} \right) \\ $$$$\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{17}} −\mathrm{1}}=\int\frac{\mathrm{dx}}{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{16}} {\prod}}\left(\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} \right)}=\int\left\{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{16}} {\sum}}\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }\right\}\mathrm{dx} \\ $$$$\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{1}}{\mathrm{17e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}.\mathrm{16}} }=\frac{\mathrm{e}^{−\frac{\mathrm{2ik}\pi}{\mathrm{17}}} \:}{\mathrm{17}} \\ $$$$\Rightarrow\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{17}} −\mathrm{1}}=\int\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{16}} {\sum}}\frac{\mathrm{e}^{\frac{−\mathrm{2ik}\pi}{\mathrm{17}}} }{\mathrm{17}\left(\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} \right)}.\mathrm{dx} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that} \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{16}} {\sum}}\frac{\mathrm{e}^{−\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }=\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}\left(\frac{\mathrm{e}^{−\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }+\frac{\mathrm{e}^{\frac{−\mathrm{2i}\pi\left(\mathrm{17}−\mathrm{k}\right)}{\mathrm{17}}} }{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2i}\left(\mathrm{17}−\mathrm{k}\right)\pi}{\mathrm{17}}} }\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}\left(\frac{\mathrm{e}^{−\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }+\frac{\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{17}}} }{\mathrm{x}−\mathrm{e}^{\frac{−\mathrm{2ik}\pi}{\mathrm{17}}} }\right)=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}\frac{\mathrm{2}\left(\mathrm{xcos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{2}\pi\mathrm{k}}{\mathrm{17}}\right)\mathrm{x}+\mathrm{1}} \\ $$$$\int\mathrm{2}\frac{\mathrm{xcos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{x}+\mathrm{1}}\mathrm{dx}=\int\left\{\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\frac{\mathrm{2x}−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{x}+\mathrm{1}}+\frac{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{x}+\mathrm{1}}\right\}\mathrm{dx} \\ $$$$\mathrm{since}\:\mid\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mid<\mathrm{1},\forall\mathrm{k}\in\left\{\mathrm{1},…..\mathrm{8}\right\}\: \\ $$$$\int\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\frac{\mathrm{2x}−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\frac{\mathrm{2}\pi\mathrm{k}}{\mathrm{17}}\right)+\mathrm{1}}=\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{x}+\mathrm{1}\right) \\ $$$$\int\frac{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{x}+\mathrm{1}}=\left(\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right)\int\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right.} \\ $$$$=\left(\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right)\mathrm{arctan}\left(\frac{\mathrm{x}}{\mathrm{sin}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}−\mathrm{cot}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right) \\ $$$$\mathrm{we}\:\mathrm{Get} \\ $$$$\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{17}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{17}}\int\left\{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\frac{\mathrm{2x}−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{x}+\mathrm{1}}+\left(\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right).\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)+\mathrm{1}}\right\}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{17}}\left\{\mathrm{ln}\mid\mathrm{x}−\mathrm{1}\mid+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}\left[\mathrm{cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\frac{\mathrm{2}\pi\mathrm{k}}{\mathrm{17}}\right)+\mathrm{1}\right)+\left(\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)−\mathrm{2cos}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right)\mathrm{arctan}\left(\frac{\mathrm{x}}{\mathrm{sin}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)}−\mathrm{cot}\left(\frac{\mathrm{2k}\pi}{\mathrm{17}}\right)\right)\right]\right\}+\mathrm{c} \\ $$$$\mathrm{c}\in\mathbb{R}\:\mathrm{constant} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by john santu last updated on 06/Jan/20
thanks you sir
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

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