Fractions, don't leave me alone!

Calculus Level 5

Let $k\gt 0$ , then evaluate

$\large \int\limits_0^1 \left\{\left(\dfrac{1}{x}\right)^k-\left(\dfrac{1}{1-x}\right)^k\right\}x^3 (1-x)^3 \, dx.$

Which of the options best represent the answer?

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

\dfrac{1}{280}

A function of

k

that is not linear A linear function of

k

\dfrac{3}{4}

1 solution

Aditya Narayan Sharma
Apr 16, 2017

The key point is $\{x\}+\{-x\}=1$

Denoting the integral by I , if we apply the substitution $x=1-y$ then

$\displaystyle I=\int\limits_0^1\left(1-\left\{\left(\dfrac{1}{y}\right)^k-\left(\dfrac{1}{1-y}\right)^k\right\}\right)y^3 (1-y)^3\;dy$

where we have used the property $\{x\}+\{-x\}=1$

$\displaystyle I=\int\limits_0^1 y^{m}(1-y)^m\; dy-I$

$\displaystyle I = \dfrac{\beta(m+1,m+1)}{2}=\dfrac{(m!)^2}{2(2m+1)!}$

Put $m=3$ to get $\displaystyle I_3 = \dfrac{1}{280}$

Here , $\displaystyle \beta(m+1,m+1)=\int\limits_0^1 y^{m}(1-y)^m\; dy$ is the Beta Function

Nice method. Did the same way. I think k should be greater than 0 as k=0 will make the integral 0.

Indraneel Mukhopadhyaya - 4 years, 1 month ago

Did the same as your

Aditya Kumar - 4 years ago

This problem is a good one for you to try. It's been long since season 4 of integration contest has been held. Have you thought of starting it?

Aditya Narayan Sharma - 4 years ago