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

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.

• 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?] May 16 '20 at 21:20
• Set $x=-t$..... May 16 '20 at 21:23
• @MarkBennet : It doesn't matter if they are even or odd, nor whether they are integers. See my answer below. May 16 '20 at 21:23
• @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. May 17 '20 at 5:12

\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}

• Doesn't $u$ go from $1$ to $0$, not $1$ to $u$? May 16 '20 at 21:35

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...

• Can you explain this solution further? I went with the answer but this seems like an interesting method May 16 '20 at 21:35
• @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} $$\int_0^{1/2} x^{700} (1-x)^{300} \, dx = \frac 1 {2^{1000}} \int_1^0 (1-u)^{700} (1+u)^{300} \, \left( \frac{-du} 2 \right)$$ $$\text{and so on.}$$ May 16 '20 at 23:36
• Yep............ May 16 '20 at 23:37
• @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. $\qquad$ May 16 '20 at 23:39
• @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 $x - \tfrac 1 2. \qquad$ May 16 '20 at 23:44

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:

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

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}

• The location of the very last sentence-ending punctuation in your answer was very strange. I fixed it. May 16 '20 at 23:07
• @MichaelHardy Thank you. May 16 '20 at 23:09

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}