Fractional Integral - 2

Calculus Level 5

$\large \int_0^1 \int_0^1 \left \{ \dfrac {x^3}y \right \} \, dx \: dy = \dfrac AB - \dfrac \gamma C$

If the equation above holds true for positive integers $A$ , $B$ and $C$ , with $A$ , $B$ coprime, find $A+B+C$ .

Notations :

1 solution

Aditya Narayan Sharma
Apr 16, 2017

Consider , $\displaystyle I=\int_0^1 \int_0^1 \left \{\dfrac{x^k}{y}\right \} \; dx\;dy$

Set $\displaystyle t=\dfrac{x^k}{y}$

$\displaystyle I = \int_0^1 \overbrace{x^k}^{II} \underbrace{\left(\int_{x^k}^\infty \dfrac{\left \{t\right \}}{t^2}\; dt\right)}_{I}\; dx$

So by IBP we have,

$\displaystyle I =\dfrac{1}{k+1}\int_1^\infty \dfrac{\left \{t\right \}}{t^2}\; dt + \dfrac{k}{k+1}\int_0^1 x^k\; dx$

The first integral is equal to $1-\gamma$ by definition , which makes

$\displaystyle I = \dfrac{2k+1}{(k+1)^2}-\dfrac{\gamma}{k+1}$

For $k=3$ we have,

$\displaystyle I=\int_0^1 \int_0^1 \left \{\dfrac{x^3}{y}\right \} \; dx\;dy = \dfrac{7}{16}-\dfrac{\gamma}{4}$ making $A+B+C=\boxed{27}$

Note for above solution:-

S = Integral(0 to 1) frac part {1/x}dx Let x = 1/t

= integral(1 to infinity ) frac part {t}dt/t^2 = sum(k=1 to infinity) integral(k to k+1) (t-k)dt/t^2 = sum(k=1 to infinity) [ ln((k+1)/k) - 1/(k+1) ]

(OR)

  S = 1 - eulergamma

                          (Q.E.D)

Shivam Sharma - 4 years, 2 months ago