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Say I have some polynomial $p(x)$ and want to express its $n$th integral, is there a closed form for this?

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It doesn't have a unique $n^{th}$ integral; this is only well-defined up to a polynomial of degree $n-1$. –  Qiaochu Yuan Sep 11 '11 at 3:45
    
I don't see your point. Take for instance $p(x) = x$, so its degree is 1. We have the 2nd integral as $\frac{x^{2}}{6}$ however.. edit: that is, plus a constant –  Pedro Sep 11 '11 at 3:49
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The $n^{\rm th}$ integral of $x^r$ is $$ \frac{r!}{(r+n)!} x^{r+n} \ \ (+ \ \text{arbitrary poly of degree } n-1). $$ You can then use linearity to add the integrals of individual terms. –  Srivatsan Sep 11 '11 at 3:50
    
On the other hand, $$\underbrace{\int_0^x\int_0^{t_{n-1}}\cdots\int_0^{t_1}}_{n} t^k\;\mathrm dt\cdots\mathrm dt_{n-2}\mathrm dt_{n-1}=\frac1{(n-1)!}\int_0^x t^k (x-t)^{n-1}\mathrm dt=\frac{k!}{(n+k)!}x^{n+k}$$ –  J. M. Sep 11 '11 at 3:53
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Oh, now I see. I forgot to integrate the constant along with it –  Pedro Sep 11 '11 at 3:53

1 Answer 1

Noticing that for $d\leqslant p$, we have that the $d$-th derivative of $x^p$ is $$x^{p-d}\frac{p!}{(p-d)!},$$ we get what Srivatsan wrote in his comment, that is, the $n$-th integral of $x^r$ is $\frac{r!}{(r+n)!}x^{r+n}$ (take $p-d=r$ and $d=n$) up to a polynomial of degree $\leqslant n-1$ (as at each integration, we need to integrate the constant of the previous one). Then the result follows by linearity.

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