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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 \(k-b>-1\).

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This is a part of the set Formidable Series and Integrals

Note by Hummus A
5 months, 3 weeks 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

\(\boxed{\displaystyle\int _{ 0 }^{ 1 }{ ]\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }\frac { { y }^{ a } }{ x^{ b } } dxdy } } =\frac { 1 }{ a-b+2 } \left( \frac { 1 }{ k-b+1 } +\frac { k! }{ (a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (k+n)! } \left( \zeta (a+n+1)-1 \right) } \right)}\) Hummus A · 5 months, 3 weeks ago

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