dx-1-x-97-7-

Question Number 129672 by Dwaipayan Shikari last updated on 17/Jan/21

\int \frac{d x}{{(1 + x^{97})}^{7}}

Answered by mindispower last updated on 17/Jan/21

u = x^{97} \Rightarrow d x = \frac{u^{- \frac{96}{97}}}{97} d u

= \int \frac{u^{- \frac{96}{97}}}{97 {(1 + u)}^{7}} d u \dots E

\int \frac{x^{s}}{1 + x} d x = \int \frac{x^{s} - x^{s + 1}}{1 - x^{2}} d x \dots F

x^{2} = t

= \frac{1}{2} \int \frac{1}{1 - t} (t^{2 s - \frac{1}{2}} - t^{2 s + \frac{3}{2}}) d t

= \frac{1}{2} \int (\frac{1 - t^{2 s + \frac{3}{2}}}{1 - t} - \frac{1 - t^{2 s - \frac{1}{2}}}{1 - t})

= \frac{1}{2} (Ψ (2 s + \frac{5}{2}) - Ψ (2 s + \frac{1}{2}))

i i n t e g r a t e b y p a r t 6 t i m e . . E & F

g i v e u s a n s w r

Commented by Dwaipayan Shikari last updated on 17/Jan/21

G r e a t s i r! C a n i t b e f r a n s f o r m e d i n t o H y p e r g e o m e t r i c

f u n c t i o n s ?

Commented by mindispower last updated on 18/Jan/21

y e s

\frac{d^{6}}{{d u}^{6}} . \frac{1}{(1 + u)} = \frac{6!}{{(1 + u)}^{7}} = \frac{d^{6}}{{d u}^{6}} \sum_{k ⩾ 0} {(- 1)}^{k} u^{k}

= \sum_{k ⩾ 6} {(- 1)}^{k} k (k - 1) \dots (k - 6) . u^{k - 6}

= \sum_{k ⩾ 0} {(- 1)}^{k} . (k + 6) \dots \dots k . u^{k}

\int \frac{d x}{{(1 + x^{97})}^{7}} = \int \sum_{k ⩾ 0} {(- x^{97})}^{k} . \frac{(k + 6) \dots k}{1} d x, \forall ∣ x ∣< 1

= \sum_{k ⩾ 0} \frac{{(- 1)}^{k} x^{97 k + 1} . k \dots (k + 6)}{(97 k + 1)} + c

= x . (1 + \sum_{k ⩾ 1} \frac{{(- x^{97})}^{k} . (k + 6)!}{(k - 1)! . (k + \frac{1}{97)})}

= x (1 + \sum_{k ⩾ 0} \frac{{(- x)}^{97 k + 97} . (k + 7)!}{k! . (k + \frac{98}{97})})

= x (1 + x^{97} . \sum_{k ⩾ 0} \frac{{(- x^{97})}^{k}}{k!} . \frac{{(8)}_{k} . {(\frac{1}{97})}_{k}}{{(\frac{98}{97})}_{k}})

= x + x^{98} ._{2} F_{1} (8, \frac{1}{97}; \frac{98}{97}; - x^{97}) + c