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I am trying to solve the following integral:

$$\int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx$$

My intuition is that this integral is equal to zero but I am unsure as to which direction to take to prove this. I was thinking binomal expansion but I believe there must be a better way, possibly using summation notation instead.

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  • $\begingroup$ Well some ideas: always good to exploit symmetry. Also what happens if you replace $700$ and $300$ by very much smaller (different) numbers? [Does it matter that they are both even?] $\endgroup$ May 16 '20 at 21:20
  • $\begingroup$ Set $x=-t$..... $\endgroup$
    – imranfat
    May 16 '20 at 21:23
  • $\begingroup$ @MarkBennet : It doesn't matter if they are even or odd, nor whether they are integers. See my answer below. $\endgroup$ May 16 '20 at 21:23
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    $\begingroup$ @MichaelHardy Quite. I was suggesting thoughts and questions which might possibly arise, and might be explored, rather than giving a direct answer. I put first the first thing to do (and if you work that out, I agree the other things are redundant). I put the "use small values" piece in, because that can help (eg be more obviously sketchable) and then the odd/even thing because if you don;'t know what's going on, you sometimes have to take a little care in building a small model of the problem. $\endgroup$ May 17 '20 at 5:12
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\begin{align} \text{Let } u & = 1-x \\ \text{and consequently } x & = 1-u \\ du & = -dx \end{align} As $x$ goes from $0$ to $1,$ $u$ goes from $1$ to $0.$

This substitution shows that this integral is $-1$ times this integral. So it is $0.$

Appendix by the original poster:

Working with the second integral, his substitution shows that:

$$ \int_0^1x^{300}(1-x)^{700}dx = \int_1^0(1-u)^{300}u^{700}(-du) = \int_0^1u^{700}(1-u)^{300}du $$

Thus \begin{align} & \int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx \\[8pt] = {} & \int_0^1 x^{700}(1-x)^{300} \, dx - \int_0^1(1-u)^{300}u^{700}\, du = 0 \end{align}

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    $\begingroup$ Doesn't $u$ go from $1$ to $0$, not $1$ to $u$? $\endgroup$
    – jeremy909
    May 16 '20 at 21:35
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Let $1/2 - x \to x^\prime$ to see that the integral is $0$.

Convert the integrand to

$$(x^\prime /2)^{700} (x^\prime /2)^{300} - (x^\prime /2)^{300}(x^\prime /2)^{700}$$

and put in the right limits...

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  • $\begingroup$ Can you explain this solution further? I went with the answer but this seems like an interesting method $\endgroup$ May 16 '20 at 21:35
  • $\begingroup$ @S.Miller : I think possibly he has in mind something about like this: $$\begin{align} & u = 2\left( \tfrac 1 2 - x \right) = 1 - 2x \\ {} \\ & x = \frac {1-u} 2 \quad \text{ and } \quad 1-x = \frac{1+u}2 \\ {} \\ & \text{As $x$ goes from 0 to $\tfrac 1 2,$} \\ & \text{$u$ goes from 1 to 0.} \\ {} \\ & x^{700} = \left( \frac {1-u} 2 \right)^{700} \\ {} \\ & (1-x)^{300} = \left( \frac{1+u} 2 \right)^{300} \end{align}$$ $$ \int_0^{1/2} x^{700} (1-x)^{300} \, dx = \frac 1 {2^{1000}} \int_1^0 (1-u)^{700} (1+u)^{300} \, \left( \frac{-du} 2 \right) $$ $$\text{and so on.}$$ $\endgroup$ May 16 '20 at 23:36
  • $\begingroup$ Yep............ $\endgroup$ May 16 '20 at 23:37
  • $\begingroup$ @S.Miller : Note that this $\quad\uparrow\quad$ is $\displaystyle \int_0^{1/2},$ not $\displaystyle \int_0^1.$ You'd also need to look at $\displaystyle \int_0^{1/2}$ of the other term, which has $700$ and $300$ interchanged and has a minus sign, and also at $\displaystyle \int_{1/2}^1$ of both functions. $\qquad$ $\endgroup$ May 16 '20 at 23:39
  • $\begingroup$ @S.Miller : One thing that might suggest this method is that when $x=1/2$ then the value of the function being integrated is $0,$ so the polynomial must be divisible by $x - \tfrac 1 2. \qquad$ $\endgroup$ May 16 '20 at 23:44
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Well, solving a more general case we have:

$$\mathcal{I}_\beta\left(\text{n},\text{k}\right):=\int_0^\beta\left(x^\text{n}\left(\beta-x\right)^\text{k}-x^\text{k}\left(\beta-x\right)^\text{n}\right)\space\text{d}x\tag1$$

Let $\text{u}=\beta-x$, so we get $-\text{du}=\text{d}x$, so:

$$\mathcal{I}_\beta\left(\text{n},\text{k}\right)=\int_\beta^0-\left(\left(\beta-\text{u}\right)^\text{n}\text{u}^\text{k}-\left(\beta-\text{u}\right)^\text{k}\text{u}^\text{n}\right)\space\text{du}=$$ $$\int_0^\beta\left(\text{u}^\text{k}\left(\beta-\text{u}\right)^\text{n}-\text{u}^\text{n}\left(\beta-\text{u}\right)^\text{k}\right)\space\text{du}=$$ $$\int_0^\beta\left(x^\text{k}\left(\beta-x\right)^\text{n}-x^\text{n}\left(\beta-x\right)^\text{k}\right)\space\text{d}x\tag2$$

So, we get:

$$\mathcal{I}_\beta\left(\text{n},\text{k}\right)+\mathcal{I}_\beta\left(\text{n},\text{k}\right)=0\tag3$$

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First let $$f(x)=x^{700}(1-x)^{300}-x^{300}(1-x)^{700}$$ and see that $$\begin{aligned} f\left(\frac{1}{2}+x\right) &= \left(\frac{1}{2}+x\right)^{700}\left(\frac{1}{2}-x\right)^{300}-\left(\frac{1}{2}+x\right)^{300}\left(\frac{1}{2}-x\right)^{700} \\ f\left(\frac{1}{2}-x\right) &= \left(\frac{1}{2}-x\right)^{700}\left(\frac{1}{2}+x\right)^{300}-\left(\frac{1}{2}-x\right)^{300}\left(\frac{1}{2}+x\right)^{700} \end{aligned}$$ Sum these to get $$f\left(\frac{1}{2}+x\right)+f\left(\frac{1}{2}-x\right)=0$$

$$f\left(\frac{1}{2}+x\right)=-f\left(\frac{1}{2}-x\right)$$ So $f$ is odd with respect to the point $x_0=1/2$ which means $$\begin{aligned} \int_0^1 f(x)\,dx&=\int_0^{1/2}f(x)\,dx+\int_{1/2}^1f(x)\,dx \\ &=-\mathcal{J}'+\mathcal{J'}\\ &=0. \end{aligned}$$

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  • $\begingroup$ The location of the very last sentence-ending punctuation in your answer was very strange. I fixed it. $\endgroup$ May 16 '20 at 23:07
  • $\begingroup$ @MichaelHardy Thank you. $\endgroup$
    – dromastyx
    May 16 '20 at 23:09
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We know that the Beta Function is defined by $$\beta(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,\mathrm dx$$ Applying the King rule of integration $$\int_a^b f(x)\,\mathrm dx=\int_a^b f(a+b-x)\,\mathrm dx$$ we can write $$\beta(m,n)=\int_0^1(1-x)^{m-1}x^{n-1}\,\mathrm dx=\beta(n,m)$$

Our integral is of the form $$\begin{align}I&=\int_0^1x^{m-1}(1-x)^{n-1}\,\mathrm dx-\int_0^1x^{n-1}(1-x)^{m-1}\,\mathrm dx\\&=\beta(m,n)-\beta(n,m)\\&=\beta(m,n)-\beta(m,n)\\&=0\end{align}$$

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