Fractional Party! 1

Calculus Level 5

∫ 0 1 ∫ 0 1 { x y } 3 { y x } 3 d x d y = A − π B C − π B + 2 D − ζ ( E ) \large \int_0^1 \int _{ 0 }^{ 1 }{ { \left\{ \dfrac { x }{ y } \right\} }^{ 3 }{ \left\{ \dfrac { y }{ x } \right\} }^{ 3 } \, dx \; dy } =A-\dfrac { { \pi }^{ B } }{ C } -\dfrac { { \pi }^{ B+2 } }{ D } -\zeta \left( E \right)

The equation above holds true for positive integers A , B , C , D A,B,C,D and E E . Find A + B + C + D + E A+B+C+D+E .

Notations :


The answer is 390.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Hamza A
Apr 5, 2016

this is a result of this ultimate generalized form

∫ 0 1 ∫ 0 1 { x y } a { y x } b d x d y = a ! 2 ( b + 1 ) ! ∑ n = 1 ∞ ( a + n ) ! ( b + n ) ! ( ζ ( a + n + 1 ) − 1 ) + b ! 2 ( a + 1 ) ! ∑ n = 1 ∞ ( b + n ) ! ( a + n ) ! ( ζ ( b + n + 1 ) − 1 ) \displaystyle\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ a }\left\{ \frac { y }{ x } \right\} ^{ b }dxdy } } =\frac { a! }{ 2(b+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (b+n)! } (\zeta (a+n+1)-1) } +\frac { b! }{ 2(a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (b+n)! }{ (a+n)! } } (\zeta (b+n+1)-1)

we let x y = u \frac { x }{ y } =u then the integral converts to

∫ 0 1 x ∫ x ∞ { u } a { 1 u } b d u u 2 \displaystyle\int _{ 0 }^{ 1 }{ x\int _{ x }^{ \infty }{ \{ u\} ^{ a }\left\{ \frac { 1 }{ u } \right\} ^{ b }\frac { du }{ { u }^{ 2 } } } }

integrating by parts we get

∫ 0 1 x ∫ x ∞ { u } a { 1 u } b d u u 2 = ( x 2 2 ∫ x ∞ { u } a { 1 u } b d u u 2 ) ∣ x = 0 x = 1 + 1 2 ∫ 0 1 x a { 1 x } b d x = 1 2 ∫ 0 1 x b { 1 x } a d x + 1 2 ∫ 0 1 x a { 1 x } b d x \displaystyle\int _{ 0 }^{ 1 }{ x\int _{ x }^{ \infty }{ \{ u\} ^{ a }\left\{ \frac { 1 }{ u } \right\} ^{ b }\frac { du }{ { u }^{ 2 } } } } =\left( \frac { { x }^{ 2 } }{ 2 } \displaystyle\int _{ x }^{ \infty }{ \{ u\} ^{ a }\left\{ \frac { 1 }{ u } \right\} ^{ b }\frac { du }{ { u }^{ 2 } } } \right)\Big|_{x=0}^{x=1} +\frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a }\left\{ \frac { 1 }{ x } \right\} ^{ b }dx } =\\ \frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ b }\left\{ \frac { 1 }{ x } \right\} ^{ a }dx } +\frac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ a }\left\{ \frac { 1 }{ x } \right\} ^{ b }dx }

see this for the evaluation of the above integrals

putting that in we get

∫ 0 1 ∫ 0 1 { x y } a { y x } b d x d y = a ! 2 ( b + 1 ) ! ∑ n = 1 ∞ ( a + n ) ! ( b + n ) ! ( ζ ( a + n + 1 ) − 1 ) + b ! 2 ( a + 1 ) ! ∑ n = 1 ∞ ( b + n ) ! ( a + n ) ! ( ζ ( b + n + 1 ) − 1 ) \displaystyle\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { x }{ y } \right\} ^{ a }\left\{ \frac { y }{ x } \right\} ^{ b }dxdy } } =\frac { a! }{ 2(b+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (a+n)! }{ (b+n)! } (\zeta (a+n+1)-1) } +\frac { b! }{ 2(a+1)! } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { (b+n)! }{ (a+n)! } } (\zeta (b+n+1)-1)

putting a = b = 3 a=b=3 we get ∫ 0 1 { x y } 3 { y x } 3 d x = 1 − π 2 24 − π 4 360 − ζ ( 3 ) \boxed{\displaystyle\int _{ 0 }^{ 1 }{ { \left\{ \frac { x }{ y } \right\} }^{ 3 }{ \left\{ \frac { y }{ x } \right\} }^{ 3 }dx } =1-\frac { { \pi }^{ 2 } }{ 24 } -\frac { { \pi }^{ 4 } }{ 360 } -\zeta \left( 3 \right)}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes this was the correct method!

Aditya Kumar - 5 years, 2 months ago

I think its not correct answer it should be 1 − 1 4 ( ζ ( 2 ) + ζ ( 3 ) + ζ ( 4 ) ) 1 -\frac{1}{4}\left(\zeta(2)+\zeta(3)+\zeta(4)\right) giving us 1 − π 2 24 − ζ ( 3 ) 4 − π 4 360 1-\frac{\pi^2}{24}-\frac{\zeta(3)}{4} -\frac{\pi^4}{360}

Naren Bhandari - 1 year, 1 month ago

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