Symmetric Rearrangement

The question basically revolves around the evaluation of the integral $\displaystyle \int_0^1 \big( 1- x^n \big)^m$ . So, let,
$\displaystyle \begin{array}{l l} I_{n,m} & = \int_0^1 \big( 1- x^n \big)^m \text{ d}x \\ & = \bigg[ \big( 1- x^n \big)^m . x \bigg]_0^1 + mn \int_0^1 \big( 1- x^n \big)^{m-1} . x^n \text{ d}x \\ & = mn \int_0^1 \big( 1- x^n \big)^{m-1} . \big( x^n - 1 +1 \big) \text{ d}x \\ & = mn \bigg( \int_0^1 \big( 1- x^n \big)^{m-1} \text{ d}x - \int_0^1 \big( 1- x^n \big)^m \text{ d}x \bigg) \\ & = mn \Big( I_{n,m-1} - I_{n,m} \Big) \\ \Rightarrow \frac{ I_{n,m}}{I_{n,m-1}} & = \frac{mn}{mn + 1} \\ \end{array}$

Let $\displaystyle n = \frac{1}{p} \Rightarrow \frac{ I_{p,m}}{I_{p,m-1}} = \frac{m}{m + p}$ .Now, the left hand side of the equation is the general term of a telescoping product. So, taking product on both the sides, we get,

$\displaystyle \prod_{m=1}^k \frac{ I_{p,m}}{I_{p,m-1}} = \prod_{m=1}^k \frac{m}{m + p} \Rightarrow \frac{ I_{p,k}}{I_{p,0}} = \frac{k!p!}{(p+k)!}$ .

Trivially, $I_{p,0} = 1$ . Therefore, $\displaystyle I_{p,k} = \frac{k!p!}{(p+k)!}$ . Also it is not difficult to realize that $I_{p,k} = I_{k,p}$ .Coming to the given integral, we have,

$\displaystyle \begin{array}{l l} \int_0^1 \big( 1- x^{1/17} \big)^{23} + \big( 1- x^{1/23} \big)^{17} \text{ d}x & = I_{17,23} + I_{23,17} \\ & = 2 \times I_{17,23} \\ & = 2 \times \frac{17! 23! }{40!} \\ \end{array}$

Compare with the given form to obtain $A \times B \times C \times D = \boxed{31280}$ .

I am going to use property of inverse functions.

Let $f(x) = (1-x^{\frac{1}{17}})^{23}$
Then $f^{-1}(x) = (1-x^{\frac{1}{23}})^{17}$
$\int_{0}^{1}f(x) dx + \int_{f(0)}^{f(1)}f^{-1}(x) dx = 1\cdot f(1) - 0\cdot f(0)$
$\implies \int_{0}^{1}f(x) dx = \int_{0}^{1}f^{-1}(x) dx$

Substitute $x = t^{17}$ and apply by parts repeatedly. The term outside of the integral() is always zero. $\int_{0}^{1}(1-x^{\frac{1}{17}})^{23} dx \\ = 17\int_{0}^{1}t^{16}(1-t)^{23}dt \\ = \frac{17\cdot 16}{24}\int_{0}^{1}t^{15}(1-t)^{24}dt \\ = ... = \frac{17\cdot16\cdot 15... 2\cdot 1}{24\cdot 23...38\cdot 39} \int_{0}^{1}t^{0}(1-t)^{39}dt \\ = \frac{17!\cdot23!}{40!}$

Since $\int_{0}^{1}f(x) dx = \int_{0}^{1}f^{-1}(x) dx$ , the answer is twice the above i.e. $2\cdot \frac{17!\cdot23!}{40!}$

Simply calculate the integral with XPLORE to get I=2.25404613916036e-11. Then multiply by 40! and obtain 1.83911069408686e37. Now assume that B=17 and C=23 (since they must have something to do with exponents of integrand with high probability!) and try I 40!/(B! C!). So determine A=2.00006875852655 which is equal 2 (neglecting rounding errors of the numeric). Results in 2 * 17 * 23 *40 = 31280 much more quickly!

A simple manipulation results in BETA FUNCTIONS!!!

Let u1 = u = x^(1/ 17) and u2 = u = x^(1/ 23) are the crucial start.

The definite integral = Integrate [17 (1 - u)^23 u^16 + 23 (1 - u)^17 u^22] d u from 0 to 1

2 (17!)(23!)/ (40!) for 2 x 17 x 23 x 40 = 31280

However, 34 (16!)(23!)/ (40!) for 500480 and 46 (17!)(22!)/ (40!) for 688160 cold also be answers!

Only by guessing the mind of question maker, we can deduce among ambiguities for 31280. Be careful!

This time, expanding and sum up is not as good as usual numerical method perhaps.

Between 2.25396864937875+E-11 and 1.96181404454876+E-11 before determination, I would think that the former shall most probably be the answer for being more. Computer cannot do very good additions at certain situations. Before further verification, as the numbers {2, 17, 23 and 40} looked good, I entered and found correct!

Answer: 31280