# Evaluation of Indefinite Integral resulting in Hypergeometric Function

I am attempting to derive the result:

$$\int \left(1+x^n\right)^{-1/m}dx= x\,_2F_1\left(\frac 1m,\frac 1n;1+\frac 1n;-x^n\right)$$

First, I start off with the binomial expansion of the integrand to get:

$$\int\sum_{k=0}^{\infty}\frac{(1/m)_k}{k!}\left(-x^n\right)^kdx$$

Then, I pull out a $-1$ and interchange summation and integration:

$$\sum_k\int (-1)^k\frac{(1/m)_k}{k!}x^{nk}dx$$

$$\sum_k (-1)^k\frac{(1/m)_k}{k!}\frac{x^{nk+1}}{nk+1}$$

$$x\sum_k \frac{(1/m)_k}{k!}\frac{(-x^n)^k}{nk+1}$$

So, I'm just right there, but not sure how to express this series in the general form of a hypergeometric series with the two additional Pochhammer symbols. I even typed this last series into Mathematica, and it returned the Hyper. Function. What am I missing, or how do I transform this last series into the desired form?

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Yes! I had just figured that out – TylerHG Jul 21 '14 at 17:18
More generally, if the ratio of two consequtive terms is a rational function of $k$, the series can always be reduced to a generalized hypergeometric one. – Start wearing purple Jul 21 '14 at 17:21
You can use the identity $\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma}{\gamma+k}$ to simplify the sum. – achille hui Jul 21 '14 at 17:34
Interesting @O.L. I'm finding that this function and series comes up everywhere – TylerHG Jul 21 '14 at 18:14
What are n,m? integers or real numbers? – user198810 Dec 9 '14 at 7:18

$$\frac{(1/n)_k}{(1+1/n)_k}=\frac{\frac1n(\frac1n+1)(\frac1n+2)\cdots(\frac1n+k-1)}{(1+\frac1n)(1+\frac1n+1)(1+\frac1n+2)\cdots(\frac1n+1+k-2)(\frac1n+1+k-1)} =\frac{1}{nk+1}$$