can-solve-dx-x-17-1-via-elementary-calculus-

Question Number 77422 by john santu last updated on 06/Jan/20

can solve \int \frac{dx}{x^{17} - 1} via

elementary calculus ?

Commented by aliesam last updated on 07/Jan/20

\int \frac{1}{x^{17} - 1} d x

= - \int \frac{1}{1 - x^{17}} (\frac{x}{x^{17}}) (\frac{x^{17}}{x}) d x

= - \int {(1 - x)}^{17 - 1} {(x)}^{1 - 17} {(x)}^{17 - 1} d x

= - \int {(1 - x)}^{17 - 1} {(x^{17})}^{\frac{1}{17} - 1} {(x)}^{17 - 1} d x

= - x_{2} F_{1} (1, \frac{1}{17}; \frac{18}{17}; x^{17}) + c

Answered by mind is power last updated on 06/Jan/20

X^{17} - 1 = \underset{k = 0}{\prod^{16}} (x - e^{\frac{2 ik π}{17}})

\int \frac{dx}{x^{17} - 1} = \int \frac{dx}{\underset{k = 0}{\prod^{16}} (x - e^{\frac{2 ik π}{17}})} = \int {\underset{k = 0}{\sum^{16}} \frac{a_{k}}{x - e^{\frac{2 ik π}{17}}}} dx

a_{k} = \frac{1}{{17 e}^{\frac{2 ik π}{17} . 16}} = \frac{e^{- \frac{2 ik π}{17}}}{17}

\Rightarrow \int \frac{dx}{x^{17} - 1} = \int \underset{k = 0}{\sum^{16}} \frac{e^{\frac{- 2 ik π}{17}}}{17 (x - e^{\frac{2 ik π}{17}})} . dx

we can see that

\underset{k = 0}{\sum^{16}} \frac{e^{- \frac{2 ik π}{17}}}{x - e^{\frac{2 ik π}{17}}} = (\frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} (\frac{e^{- \frac{2 ik π}{17}}}{x - e^{\frac{2 ik π}{17}}} + \frac{e^{\frac{- 2 i π (17 - k)}{17}}}{x - e^{\frac{2 i (17 - k) π}{17}}}))

= \frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} (\frac{e^{- \frac{2 ik π}{17}}}{x - e^{\frac{2 ik π}{17}}} + \frac{e^{\frac{2 ik π}{17}}}{x - e^{\frac{- 2 ik π}{17}}}) = \frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} \frac{2 (xcos (\frac{2 k π}{17}) - \cos (\frac{2 k π}{17}))}{x^{2} - 2 \cos (\frac{2 π k}{17}) x + 1}

\int 2 \frac{xcos (\frac{2 k π}{17}) - \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} dx = \int {\cos (\frac{2 k π}{17}) \frac{2 x - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} + \frac{{2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1}} dx

since ∣ \cos (\frac{2 k π}{17}) ∣< 1, \forall k \in {1, \dots . . 8}

\int \cos (\frac{2 k π}{17}) \frac{2 x - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 xcos (\frac{2 π k}{17}) + 1} = \cos (\frac{2 k π}{17}) \ln (x^{2} - 2 xcos (\frac{2 k π}{17}) x + 1)

\int \frac{{2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} = ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) \int \frac{dx}{(x - \cos {(\frac{2 k π}{17})}^{2} + \sin^{2} (\frac{2 k π}{17})}

= ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) \arctan (\frac{x}{\sin (\frac{2 k π}{17})} - \cot (\frac{2 k π}{17}))

we Get

\int \frac{dx}{x^{17} - 1} = \frac{1}{17} \int {\frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} \cos (\frac{2 k π}{17}) \frac{2 x - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} + ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) . \frac{1}{x^{2} - 2 xcos (\frac{2 k π}{17}) + 1}} dx

= \frac{1}{17} {\ln ∣ x - 1 ∣ + \underset{k = 1}{\sum^{8}} [\cos (\frac{2 k π}{17}) \ln (x^{2} - 2 xcos (\frac{2 π k}{17}) + 1) + ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) \arctan (\frac{x}{\sin (\frac{2 k π}{17})} - \cot (\frac{2 k π}{17}))]} + c

c \in R constant

Commented by john santu last updated on 06/Jan/20

thanks you sir