Who's up to the challenge? 31

Calculus Level 5

$\large \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \dfrac { 1 }{ { x }^{ 1729 } } -\dfrac { 1 }{ (1-x)^{ 1729 } } \right\} { x }^{ 6 }(1-x)^{ 6 } \, dx } =\dfrac { a}{ b }$

The equation above holds true for coprime positive integers $a$ and $b$ . Find $a+b$ .

Bonus : Generalize it.

Notation : $\{ \cdot \}$ denotes the fractional part function .

This problem is a part of " Who's up to the challenge? "

The answer is 24025.

1 solution

Mark Hennings
Apr 5, 2016

If $f$ is a function such that $f(1-x) = -f(x)$ then, since $\{u\} + \{-u\} = 1$ for all $u$ , we have $\begin{array}{rcl} \displaystyle \int_0^1 \{ f(x) \} x^n(1-x)^n\,dx & = & \displaystyle \frac12\int_0^1 \big[ \{f(x)\} + \{f(1-x)\}\big] x^n(1-x)^n\,dx \\ & = & \displaystyle \frac12\int_0^1 x^n(1-x)^n\,dx \; = \; \tfrac12B(n+1,n+1) \\ & = & \displaystyle \frac{(n!)^2}{2(2n+1)!} \end{array}$ With $f(x) = x^{-1729} - (1-x)^{-1729}$ and $n=6$ , the integral equals $\tfrac{(6!)^2}{2 \times 13!} \; = \; \tfrac{1}{24024}$ , and so the answer is $\boxed{24025}$ .

exactly the same way i did it!

Hamza A - 5 years, 2 months ago

How did you equate this {f(x)}=({f(x)}+{f(1-x)})/2 in first step ??? You can't equate them as f(x)=-f(1-x) so {f(x)} = 1-{f(1-x)}

Rohan Chidrewar - 5 years, 2 months ago

The second term comes from a $y =1-x$ substitution, with $y$ then replaced by $x$ ...

Mark Hennings - 5 years, 2 months ago

Who's up to the challenge? 31

This problem is a part of " Who's up to the challenge? "

The answer is 24025.

1 solution

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