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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 at 17:18
1  
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. –  O.L. Jul 21 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 at 17:34
    
Interesting @O.L. I'm finding that this function and series comes up everywhere –  TylerHG Jul 21 at 18:14

2 Answers 2

Deriving from integral approach is easier: \begin{align} \int(1+x^n)^{-\frac{1}{m}}~dx &=\int_0^x(1+t^n)^{-\frac{1}{m}}~dt+C\\ &=\int_0^{x^n}(1+t)^{-\frac{1}{m}}~d(t^\frac{1}{n})+C\\ &=\dfrac{1}{n}\int_0^{x^n}t^{\frac{1}{n}-1}(1+t)^{-\frac{1}{m}}~dt+C\\ &=\dfrac{1}{n}\int_0^1(x^nt)^{\frac{1}{n}-1}(1+x^nt)^{-\frac{1}{m}}~d(x^nt)+C\\ &=\dfrac{x}{n}\int_0^1t^{\frac{1}{n}-1}(1+x^nt)^{-\frac{1}{m}}~dt+C\\ &=x~_2F_1\left(\dfrac{1}{m},\dfrac{1}{n};1+\dfrac{1}{n};-x^n\right)+C \end{align}

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Ah cool! I didn't think of doing it this way. Thank you –  TylerHG Jul 21 at 17:11
up vote 1 down vote accepted

I was in the end able to derive the correct expression as...

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

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