5

trying to integrate $\int_0^1 (1-x)^m x^n dx$, $m$ and $n$ positive integers. I know the answer is a fraction containing gamma functions but don't know how to get there

Mark Viola
  • 179,405
brayton
  • 91

2 Answers2

5

Another approach, that is closely related to the Beta function, is to integrate by parts, assuming $m\gt0$ $$ \begin{align} \int_0^1(1-x)^mx^n\,\mathrm{d}x &=-\frac1{n+1}\int_0^1(1-x)^m\,\mathrm{d}x^{n+1}\\ &=\frac1{n+1}\int_0^1x^{n+1}\,\mathrm{d}(1-x)^m\\ &=\frac{m}{n+1}\int_0^1(1-x)^{m-1}x^{n+1}\,\mathrm{d}x\tag{1} \end{align} $$ We can iterate $(1)$ to get $$ \begin{align} \int_0^1(1-x)^mx^n\,\mathrm{d}x &=\frac{m}{n+1}\frac{m-1}{n+2}\cdots\frac1{n+m}\int_0^1x^{n+m}\,\mathrm{d}x\\ &=\frac{m}{n+1}\frac{m-1}{n+2}\cdots\frac1{n+m}\frac1{n+m+1}\\ &=\frac{n!\,m!}{(n+m+1)!}\\ &=\frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(n+m+2)}\\[6pt] &=\mathrm{B}(n+1,m+1)\tag{2} \end{align} $$ If $m=0$, then $(2)$ does not need iteration.

In fact, the Beta function is often defined as $$ \mathrm{B}(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\,\mathrm{d}t\tag{3} $$

robjohn
  • 345,667
  • @robjon Very straightforward and elegant! Integrate by parts and iterate. It's interesting to see the identity presented from our solutions. Maybe I should post a question asking others for ways to prove the equivalence. Thoughts? – Mark Viola Mar 31 '15 at 00:41
  • @Dr.MV: When I first came across the Beta function, I did just as above to get the values on the integers. Then I extended it to general $x,y$ as in this answer. A question about the various methods to show the equivalence would be interesting, but might meet with resistance from some due to the warning on this page not to ask questions like “I use ______ for ______, what do you use?” – robjohn Mar 31 '15 at 01:28
  • That is interesting. The only time I used the Beta function was in numerically solving an integro-differential equation. The method I used, if memory serves (this was 20 years ago), involved converting the continuous operator equation into a matrix equation. The elements of the matrix were obtained as closed form integrals, including the Beta function. – Mark Viola Mar 31 '15 at 03:36
1

Using the binomial expansion, we can write

$$\begin{align} \int_0^1 (1-x)^m x^n dx&=\int_0^1 \sum_{k=0}^m \binom{m}{k} (-1)^kx^k x^n dx\\ &=\sum_{k=0}^m \binom{m}{k} (-1)^k \int_0^1 x^{k+n} dx\\ &=\sum_{k=0}^m \binom{m}{k} \frac{(-1)^k}{k+n+1} \end{align}$$

Of course, we could write the binomial coefficient in terms of the gamma function as

$$\binom{m}{k}=\frac{\Gamma(m+1)}{\Gamma(k+1)\Gamma(m+1-k)}$$

Mark Viola
  • 179,405