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I am trying to evaluate the following integral: $$ \int_0^1 \frac{n(n-1)(n-2)}{2} y^{3} \left( 1 - y \right)^{n-3} \, dy $$ Here is my solution. Let $I$ be the integral we are trying to evaluate. \begin{align*} I &= \int_0^1 \frac{n(n-1)(n-2)}{2} y^{3} \left( 1 - y \right)^{n-3} \, dy \end{align*} Now to evaluate this integral we use the substitution $u = 1 - y$ which gives us $y = 1 - u$ and $du = -dy$. \begin{align*} I &= - \int_0^1 \frac{n(n-1)(n-2)}{2} (1-u)^3 u^{n-3} \, du \\ I &= \int_1^0 \frac{n(n-1)(n-2)}{2} (1-u)^3 u^{n-3} \, du \\ \end{align*} Observe that: \begin{align*} (1-u)^3 &= (1-u)(1-u)^2 = (u-1)^2(1-u) = (u^2 - 2u + 1)(1-u) \\ (1-u)^3 &= -u^3 + 3u^2 - 3u + 1 \end{align*} Now back to the integral: \begin{align*} I &= \int_1^0 \frac{n(n-1)(n-2)}{2} ( -u^3 + 3u^2 - 3u + 1) u^{n-3} \, du \\ I &= \int_0^1 \frac{n(n-1)(n-2)}{2} ( u^3 - 3u^2 + 3u - 1) u^{n-3} \, du \\ I &= \int_0^1 \frac{n(n-1)(n-2)}{2} ( u^n - 3u^{n-1} + 3u^{n-2} - u^{n-3} ) \, du \\ I &= \frac{n(n-1)(n-2)}{2} \left( \frac{u^{n+1}}{n+1} - \frac{3u^n}{n} + \frac{3u^{n-1}}{n-1} - \frac{u^{n-2}}{n_2} \right) \bigg|_0^1 \\ I &= \frac{n(n-1)(n-2)}{2} \left( \frac{1}{n+1} - \frac{3}{n} + \frac{3}{n-1} - \frac{1}{n-2} \right) \\ I &= \frac{n(n-1)(n-2) -3(n+1)(n-1)(n-2) + 3(n+1)(n)(n-2) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{n^3 - 3n^2 + 2n -3(n^2-1)(n-2) + 3(n+1)(n)(n-2) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{n^3 - 3n^2 + 2n -3(n^3-2n^2-n+2) + 3(n+1)(n)(n-2) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{-2n^3 + 3n^2 + 5n - 6 + 3(n^3-n^2 - 2n) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{-2n^3 + 3n^2 + 5n - 6 + 3n^3-3n^2 - 6n - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{n^3 + 3n^2 + 5n - 6 - 3n^2 - 6n - (n+1)(n^2-n)}{2(n+1)} \\ I &= \frac{n^3 + 5n - 6 - 6n - (n+1)(n^2-n)}{2(n+1)} \\ I &= \frac{n^3 - n - 6 - (n^3 -n^2 + n^2 - n)}{2(n+1)} \\ I &= \frac{n^3 - n - 6 - n^3 + n}{2(n+1)} = \frac{-6}{2(n+1)} \\ I &= \frac{-3}{n+1} \\ \end{align*} I have reason to believe that the value of this integral is: $$ \frac{3}{n+1} $$ Where did I go wrong in evaluating this integral?

Michael Hardy
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Bob
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    Why do you drag these $n$ factors all along ?? This is calling for problems. –  Nov 17 '19 at 16:31
  • They are needed because of the $y^{n-3}$ term. – Bob Nov 17 '19 at 16:57
  • I don't see why. –  Nov 17 '19 at 18:42
  • They're a constant factor wrt $y$, meaning they don't really affect the "behavior" of the integral, and so you can pull them out of the integrand and ignore them until the end. If you have to divide by $n - 2$ or something at some point, it's fine to do it without these factors. – Bladewood Nov 18 '19 at 01:10
  • You can also notice that if $n>2$, then $I$ must be positive in your first two lines and must be negative in the third line, which is a big hint to examine that step. – aschepler Nov 18 '19 at 01:27
  • At the URL below you see it confirmed that your guess about the value of the integral is correct. (And clearly the number you get should be positive.) https://math.stackexchange.com/questions/86542/prove-binomnk-1-n1-int-01xk1-xn-kdx-for-0-leq-k-le/86578#86578 – Michael Hardy Nov 18 '19 at 03:06

2 Answers2

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You forgot to change the bounds in your third equation. It should be $$I=-\int_{\color{red}{1}}^{\color{red}{0}}\frac{n(n-1)(n-2)}{2}(1-u)^{3}u^{n-3}\,du$$

MathIsFun7225
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This is a complete Beta integral and we immediately have

$$\frac{n(n-1)(n-2)}{2} \int_0^1 y^{3} \left( 1 - y \right)^{n-3} \, dy =\frac{n(n-1)(n-2)}{2}\frac{3!(n-3)!}{(n+1)!}.$$

Simplify.

Alternatively:

Indeed,

$$y^{n-3}-3y^{n-2}+3y^{n-1}-y^{n}$$ integrates to

$$\frac1{n-2}-\frac3{n-1}+\frac3{n}-\frac1{n+1}=\frac6{(n+1)n(n-1)(n-2)}.$$