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 kk being a real number ≥1\ge1 and a,ba,b nonnegative integers such that a−b>−2a-b>-2 and k−b>−1k-b>-1.

Notations:

This is a part of the set Formidable Series and Integrals

#Calculus

Note by Hamza A
5 years, 2 months ago

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Comments

After rearranging the integral we obtain:

∫011xb∫01{xy}kyadxdy\displaystyle\int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ b } } \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ k }{ y }^{ a }dxdy } }

We now let xy=u\frac { x }{ y } =u. In doing so the integral is converted to:

∫01xa+1−b(∫x∞{u}kua+2du)dx\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx

We proceed to integrate by parts to get:

∫01xa+1−b(∫x∞{u}kua+2du)dx\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a+1-b } } \left( \displaystyle\int _{ x }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } \right) dx =(xa+2−ba+2−b∫x∞{u}kua+2du)∣x=0x=1+1a+2−b∫01xk−bdx=\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 } =1a+2−b∫1∞{u}kua+2du+1(a+2−b)(k−b+1)=\\ \frac { 1 }{ a+2-b } \displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ u\} ^{ k } }{ { u }^{ a+2 } } du } +\frac { 1 }{ (a+2-b)(k-b+1) }

See the evaluation of that integral here

Plugging in this result we get:

∫01∫01{xy}kyaxbdxdy=1a−b+2(1k−b+1+k!(a+1)!∑n=1∞(a+n)!(k+n)!(ζ(a+n+1)−1))\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)}

Hamza A - 5 years, 2 months ago
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