Who's up to the challenge? 72

Calculus Level 5

$\large \int _{ 0 }^{ 1 }{ { x }^{ 3 }\sqrt { 1-{ x }^{ 7 } } dx } =\frac { A\Gamma \left( \frac { B }{ C } \right) \sqrt { \pi } }{ D\Gamma \left( \frac { E }{ F } \right) }$

If the equation above holds true for positive integers $A$ , $B$ , $C$ , $D$ , $E$ and $F$ with $\text{gcd}(A,D)$ $=\text{gcd}(B,C)$ $=\text{gcd}(E,F)=1$ and $\dfrac{B}{C},\dfrac{E}{F}<1$ , find $A+B+C+D+E+F$ .

The answer is 55.

1 solution

Chew-Seong Cheong
Jun 11, 2016

$\begin{aligned} I & = \int_0^1 x^3\sqrt{1-\color{#3D99F6}{x^7}} dx \quad \quad \small \color{#3D99F6}{\text{Let }u^2 = x^7 \implies x = u^\frac27 \implies dx = \frac27 u^{-\frac57} du} \\ & = \frac27 \int_0^1 u^\frac67 u^{-\frac57} (1-u^2)^\frac12 du \\ & = \frac27 \int_0^1 u^\frac17 (1-u^2)^\frac12 du \\ & = \frac17 B \left(\frac47, \frac32 \right) \\ & = \frac{\Gamma \left(\frac47 \right)\Gamma \left(\frac32 \right)}{7\Gamma \left(\frac{29}{14} \right)} \\ & = \frac{\Gamma \left(\frac47 \right) \cdot \frac12 \sqrt{\pi}}{7 \cdot \frac{15}{14} \cdot \frac 1{14} \Gamma \left(\frac{1}{14} \right)} \\ & = \frac{14\Gamma \left(\frac47 \right) \sqrt{\pi}}{15\Gamma \left(\frac{1}{14} \right)} \end{aligned}$

$\implies A+B+C+D+E+F = 14+4+7+15+1+14 = \boxed{55}$

Gamma(15/14) can also be expanded as Gamma(1+ 1/14)=1/14*Gamma(1/14)

Kushal Bose - 5 years ago

Yes, I have lodged a report.

Chew-Seong Cheong - 5 years ago

Who's up to the challenge? 72

The answer is 55.

1 solution

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