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# Fractional parts, again!

$\large\displaystyle\int _{ 0 }^{ 1 }{ \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \dfrac { x }{ y } \right\} ^{ k }\dfrac { { y }^{ a } }{ { x }^{ b } } \, dx \; dy } } =\dfrac { 1 }{ a-b+2 } \left( \dfrac { 1 }{ k-b+1 } +\dfrac { k! }{ (a+1)! } \sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (k+n)! } \left( \zeta (a+n+1)-1 \right) } \right)$

Prove the equation above with $$k$$ being a real number $$\ge1$$ and $$a,b$$ nonnegative integers such that $$a-b>-2$$ and $$k-b>-1$$.

Notations:

This is a part of the set Formidable Series and Integrals

Note by Hummus A
8 months ago

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rearranging the integral we get

$$\displaystyle\int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ b } } \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }{ y }^{ a }dxdy } }$$

then we let $$\frac { x }{ y } =u$$ and the integral converts into

$$\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx$$

now we integrate by parts to get

$$\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx=\left( \frac { { x }^{ a+2-b } }{ a+2-b } \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right)\Big|_{x=0}^{x=1} +\frac { 1 }{ a+2-b } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ k-b }dx } =\\ \frac { 1 }{ a+2-b } \displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } +\frac { 1 }{ (a+2-b)(k-b+1) }$$

i evaluated that integral here

putting that in we get

$$\boxed{\displaystyle\int _{ 0 }^{ 1 }{ ]\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }\frac { { y }^{ a } }{ x^{ b } } dxdy } } =\frac { 1 }{ a-b+2 } \left( \frac { 1 }{ k-b+1 } +\frac { k! }{ (a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (k+n)! } \left( \zeta (a+n+1)-1 \right) } \right)}$$ · 8 months ago