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$$\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$$

I have tried integration by parts using calling $f (x)=(x-1)(x-2)... (x-2016)$ but it doesn't help. I can't think of anything else. Please help.

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    $\begingroup$ Here's a tutorial in MathJax $\endgroup$ Jun 5, 2017 at 14:12
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    $\begingroup$ I've taken the liberty of formatting the equation. $\endgroup$ Jun 5, 2017 at 14:15
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    $\begingroup$ You would need $2016$ steps to plough through integration by parts! Fortunately there are some shortcuts :) Nice question (+1). $\endgroup$ Jun 5, 2017 at 14:26

3 Answers 3

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Note that

$$\scriptsize\begin{align} \int_0^2 x(x-1)(x-2)dx&=\int_{-1}^1 (x-1)x(x+1)dx&=0\\ \int_0^4 x(x-1)(x-2)(x-3)(x-4)dx&=\int_{-2}^2 (x-2)(x-1)x(x+1)(x+2)dx&=0\\ &\vdots\\ \int_0^{2016} x(x-1)(x-2)(x-3)\cdots (x-2016)dx&=\int_{-1008}^{1008} (x-1008)(x-1007)\cdots x\cdots (x+1007)(x+1008)dx&=\color{red}0\\ \end{align}$$

The result follows from the antisymmetric nature of the curve $\scriptsize(x-a)(x-a+1)\cdots x\cdots (x+a-1)(x+a)$.

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  • $\begingroup$ Nice use of script size, and good answer..+1 $\endgroup$
    – user284001
    Jun 5, 2017 at 14:27
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    $\begingroup$ @Bacon - Thanks! Had to get it to fit, for typographical symmetry :) $\endgroup$ Jun 5, 2017 at 14:29
  • $\begingroup$ The $+$ signs in the LHS integrands should be $-$ signs. $\endgroup$ Jun 5, 2017 at 14:35
  • $\begingroup$ @JohnBentin - Yes, thanks! (It still works if the limits are from $-2n$ to 0 though! :)) $\endgroup$ Jun 5, 2017 at 14:40
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Hint Shift the variable of integration so that the interval is centered at $0$, that is, set $$x = u + 1008 , \qquad dx = du .$$

Alternatively, substitute $$x = 2016 - v, \qquad dx = -dv ,$$ and compare the result with the given integral.

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This Definite integral can be evaluated using the property $$\int_a^b f(x) dx= \int_a^b f(a+b-x) dx$$

Lets say that $$I= \int_0^{2016} x(x-1)(x-2) \cdots (x-2016) dx$$

Applying the property above, $$I= \int_0^{2016} (2016-x)(2015-x) \cdots (-x) $$ $$\implies I= - \int_0^{2016} x(x-1)(x-2) \cdots (x-2016) dx$$

Adding our initial equation to the one we have obtained, we can see that $$ 2I=0 \implies I=0$$

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  • $\begingroup$ Ok thanks all. It was easy, i couldn't see it. $\endgroup$
    – reco
    Jun 5, 2017 at 14:27

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