# How this integral was evaluated using hypergeometric series?

s Hello everyone,i can't solve the following integral using the hypergeometric series:

$$\int [x(a-x)]^{n}dx$$

Wolframalpha showed this

but i don't understand how Wolframalpha came to this result.

Thanks in advance!

My steps are:

$$\int [x(a-x)]^{n}dx =\int [(xa)^{n}][1-(x/a)]^{n} dx=\int [(xa)^{n}] \sum_{k=0}^\infty\frac{[(\Gamma(k-n))/\Gamma(-n)]}{k!}\frac{x^{k}}{a^{k}}dx=\sum_{k=0}^\infty\frac{[(\Gamma(k-n))/\Gamma(-n)]a^{n-k}}{k!}\frac{x^{k+n+1}}{k+n+1}$$

How can i continue to bring the whole form of the hypergeometric function?

• Probably just binomial expansion... – Simply Beautiful Art Apr 6 '17 at 14:06
• The hypergeometric functions reduce to polynomials if the orders are off by integers from what I recall. This is just repackaging binomial expansion. – Cameron Williams Apr 6 '17 at 14:17

## 1 Answer

We can always turn an indefinite integral into a definite one by imposing limits (and then adding the arbitrary constant on at the end), so let's evaluate $$I(x) = \int_0^x t^n (a-t)^n \, dt = a^n \int_0^x t^n (1-t/a)^n \, dt.$$ This has the right derivative, so it will work. Change variables to $t=xu$, $dt=x \, du$ and the limits change to $0$ and $1$, so $$I(x) = a^n x^{n+1} \int_0^1 u^n (1-(x/a)u)^n \, du.$$ Now we recognise the form of the hypergeometric integral, $$F(a,b;c;z) = \frac{1}{B(b,c-b)} \int_0^1 u^{b-1} (1-u)^{c-b-1} (1-zu)^{-a} \, du,$$ and it follows that $$I(x) =a^n x^{n+1} B(n+1,1)F(-n,n+1,n+2,x/a).$$ $B(n+1,1)= 1/(n+1)$, so the answer agrees with Wolfram|Alpha's when we simplify the group in front of the $F$.

Note also that for $a$ or $b$ a nonpositive integer, the hypergeometric function is a polynomial, which agrees with what we expect from expanding the brackets and integrating term by term if $n$ is a nonnegative integer.

• Thanks very very much for your help ! Good job – C.C.12 Apr 6 '17 at 14:34