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According to WolframAlpha, $$\int (x^2+1)^n dx = x \cdot _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)$$ and $$_2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{n=0}^{\infty} \frac{1}{3}(-n)\frac{(-x^2)^n}{n!}.$$

Can you tell which are correct terms of the sum of solution of the integral $\int (x^2+1)^n dx $? these $$(n=0): x, (n=1): 1/3x(x^2+3), (n=2): 1/3 x(x^4+x^2), \dots$$

or (by using $_2F_1(a,b;c;z)= \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$(1)), these which I got $$(n=0): 0, (n=1): 1/3x(x^2), (n=2): 1/3 x(-x^4+x^2), \dots$$ If mine are incorrect, can you compute some terms of the sum (1)?

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    I edited your question, but I'm not 100% sure that I captured the correct meaning. Specifically, your original question did not specifically refer to WolframAlpha, except for an almost hidden link, but that seems to be the origin of the question. Could you please take a look? The last part could certainly use a further edit. – Mark McClure May 30 '13 at 11:25
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    What's your question exactly? – Mhenni Benghorbal May 30 '13 at 11:52

1 Answers1

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You are getting the wrong answer. The answer should be

$$ x\,{\mbox{$_2$F$_1$}\left(\frac{1}{2},-n;\,\frac{3}{2};\,-{x}^{2}\right)}. $$

Now, you won't have problems. For $n=0,1,2,3$, we get the corresponding answers

$$ x,\, \frac{x}{3} \left( {x}^{2}+3 \right), \, \frac{x}{15} \left( 3\,{x}^{4}+10\,{x}^{2}+15 \right),\, \, \frac{x}{35} \left( 5\,{x}^{6}+21\,{x}^{4}+35\,{x}^{2}+35 \right). $$

Added: You should use different index for the sum as

$$ _2F_1(\frac{1}{2},-n;\frac{1}{3};-x^2)= \sum_{k=0}^{\infty} \frac{(1/2)_k(-n)_k}{(3/2)_k}\frac{(-x^2)^k}{k!}. $$

Now, you need this identity

$$ (a)_m = \frac{\Gamma(m+a)}{\Gamma(a)}= \frac{(m+a-1)!}{(a-1)!}. $$

Mhenni Benghorbal
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  • @MhBe: See my edit. is it true that $2F_1 (1/2 ,−n;1/3 ;−x^2 )=\sum{n=0}^{\infty} 1/3(−n)(−x^2 )^n/n! $? If yes, can you compute how you get eg. $x/3(x^2+3)$? –  May 30 '13 at 12:32
  • @alvoutila: You got the wrong hypergeometric function. The third entry in the hypergeometric function should be $\frac{3}{2}$ instead of $\frac{1}{3}$. – Mhenni Benghorbal May 30 '13 at 13:12
  • Yes, It was just typo. I meant $_2F_1(1/2,-n;3/2;-x^2)=...$. Can you give some terms of this sum that I put in first comment? –  May 30 '13 at 13:15