How to solve this challenging integral?

Question

How do you integrate $\int_0^1 \frac{(1-x)^{a+2}x^{b+1}}{a+2}dx$, where $a,b$ are constants? I've tried integration by parts, u-sub, and I have had no luck so far. The answer is supposed to be $\frac{(a+1)!(b+1)!}{(a+b+4)!}$.

Answer 1

Binomial Method:

Call $a+2 = N$, $b+1 = M$ then:

$$\frac{1}{a+2}\int_0^1 (1+x)^N x^M\ dx$$

Expand the first term in the binomial way:

$$\frac{1}{a+2}\int_0^1 \sum_{k =0}^N \binom{N}{k} x^{N-k} x^M\ dx$$

$$\frac{1}{a+2}\sum_{k =0}^N \binom{N}{k}\int_0^1 x^{N-k+M}\ dx$$

$$\frac{1}{a+2}\sum_{k =0}^N \binom{N}{k}\frac{1}{N-k+M+1}$$

The sum is well known even if it expressed in terms of the Hypergeometric function, and the result is

$$\frac{1}{a+2}\frac{\, _2F_1(-M-n-1,-n;-M-n;-1)}{M+n+1}$$

Where $M = b+1$ and $N = a+2$. Now substitute back.

The above series can be rewritten in terms of the Beta function here below

In any case, we end up with that.

Answer 2

By using the Euler Beta integral, defined as

$$\mathcal{B}(A, B) = \int_0^1 x^{A-1} (1-x)^{B-1}\ dt$$

Setting $A-1 = a+2\ $ and $\ B-1 = b+1\ $, you can easily get the result which is:

$$\frac{1}{a+2}\frac{\Gamma (a+3) \Gamma (b+2)}{\Gamma (a+b+5)}$$

The result in terms of Euler Gamma function is obtained by making use of the general relation:

$$\mathcal{B}(A, B) = \frac{\Gamma(A)\Gamma(A)}{\Gamma(A+A)}$$

To get your final result, remember that Euler Gamma is such that for $A, B \in \mathbb{N}$:

$$\Gamma(A) = (A-1)!$$

Indeed you have

$$\Gamma(A)\Gamma(B) = (A-1)!(B-1)!$$

$$\Gamma(A + B) = (A + B - 1)!$$

In your case then, expanding the whole:

$$\frac{1}{a-2}\frac{(a-2)!(b-1)!}{(a + b + 5 - 1)!} = \frac{(a-1)!(b-1)!}{(a + b + 4)!}$$