# "Easy looking" indefinite integral to be performed

In my work, I am confronted with the following indefinite integral, which looks like it should be performed easily, although I (nor Maple) can find it. I would appreciate any help finding it explicitly. $$\int{\left(1-\frac{1}{x}\right)}^p\,{\rm{d}}x,$$ where $p$ is a real number, and where we consider $x<0$.

• What is the domain of x? If x >1, then 1-1/x <0, which means that raising it to a non-integer power gets very messy. Apr 2 '18 at 18:56
• For what it's worth, applying the substitution $u = 1 - \frac 1x$ lets us rewrite this integral as $$\int \frac{u^p}{(1-u)^2}\,du$$ Apr 2 '18 at 19:02
• This integral leads to a hypergeometric series Apr 2 '18 at 19:03
• It doesn't look easy.
– user65203
Apr 2 '18 at 19:20
• To @Acccumulation, I added that I am interested to $x<0$. Apr 2 '18 at 20:32

For positive integer $p$,

$$\int\left(1-x^{-1}\right)^pdx=\sum_{k=0}^p\int\binom pk(-x)^{-k}dx=x-p\log x+\sum_{k=2}^p\binom pk\frac{(-x)^{1-k}}{k-1}$$

which doesn't seem to have a simpler form.

For fractional $p$, you will obtain an infinite series by means of the generalized binomial theorem.

Call the integral $J$. The substitution $s = 1/x$ gives you $$J = - \int (1-s)^p s^{-2}\; ds$$

For $0 < |s| < 1$ the integrand has the binomial series expansion

$$\sum_{k=0}^\infty {p \choose k} (-1)^k s^{k-2}$$ We integrate the $k=0$ and $k=1$ terms separately, and obtain

\eqalign{J &= \frac{1}{s} + p \ln(s) + \sum_{k=2}^\infty (-1)^{k+1} {p \choose k} \frac{s^{k-1}}{k-1} + c\cr &= \frac{1}{s} + p \ln(s) -{p\choose 2}s\; {\mbox{_3F_2}(1,1,2-p;\,2,3;\,s)} +c } where ${}_3F_2$ is a generalized hypergeometric function.