Who's up to the challenge?10

Calculus Level 5

0 ( ( x 5 + 1 ) 1 5 x ) d x = A C B ( D F , 1 G ) \displaystyle\int _{ 0 }^{ \infty }{ \left( ({ x }^{ 5 }+1)^{ \frac { 1 }{ 5 } }-x \right)} \, dx =\dfrac { A }{ C } B\left(\dfrac { D }{ F } ,\frac { 1 }{ G } \right)\\

Given that the equation holds true where B ( x , y ) B(x,y) is the beta function , and A , C , D , F , G A,C,D,F,G positive integers gcd ( A , C ) = gcd ( D , F ) = 1 \gcd(A,C) = \gcd(D,F) = 1 . Find min { A + C + D + F + G } \text{min}\{A+C+D+F+G\}


this is a part of Who's up to the challenge?


The answer is 24.

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1 solution

Neelesh Vij
Jul 14, 2017

Hint:

I ( a ) = 0 ( ( x 5 + a ) 1 / 5 x ) d x \displaystyle I(a) = \int_{0}^{\infty} \left ( (x^5 +a)^{1/5} -x \right ) dx

Then proceed by differentiating wrt a a

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