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Prove it!

\[ \Large \displaystyle\int _{ 0 }^{ 1 }{ \left \{ \dfrac { 1 }{ x } \right\} ^{ n } } \, dx=\sum _{ k=1 }^{ \infty }{ \frac { \zeta (k+1)-1 }{ \binom{n+k}{k} } } \]

Prove the equation above for positive integer \(n\).

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This is a part of the set Formidable Series and Integrals

Note by Hummus A
6 months ago

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this is actually a result from a more generalized form,\(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =\frac { k! }{ (n+1)! } \sum _{ i=1 }^{ \infty }{ \frac { (n+i)! }{ (k+i)! } (\zeta (n+i+1)-1) } \)

let's denote the above as

\(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =f(k,n)=\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ t\} ^{ k } }{ { t }^{ n+2 } } dt } \)

converting this to a sum of integrals we have

\(\displaystyle\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ i }^{ i+1 }{ \frac { \{ t\} ^{ k } }{ { t }^{ n+2 } } } dt } =\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { (t-i)^{ k } }{ { t }^{ n+2 } } } \)

u sub \(u=t-i\) we get \(\displaystyle\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ 0 }^{ 1 }{ \frac { { u }^{ k } }{ (i+u)^{ n+2 } } } } du=\int _{ 0 }^{ 1 }{ { u }^{ k }\left( \sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } \right) dy } (1)\) then i tried coming with an integral to represent the sum,after some searching i got \( \frac { 1 }{ (i+u)^{ n+2 } } =\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ { e }^{ -(i+u)u }{ y }^{ n+1 } } dy \) putting that in we get

\(\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } =\frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { e }^{ -(i+u)y }{ y }^{ n+1 }dy } } =\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ { y }^{ n+1 } } { e }^{ -uy }(\displaystyle\sum _{ i=1 }^{ \infty }{ { e }^{ -iy } } )dy=\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 }{ e }^{ -uy } }{ { e }^{ y }-1 } dy } (2)\)

combining (1) and (2) we get \(\frac { 1 }{ (n+1)! } \int _{ 0 }^{ 1 }{ { u }^{ k }\left( \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 }{ e }^{ -uy } }{ { e }^{ y }-1 } dy } \right) du } =\frac { 1 }{ (n+1)! } \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 } }{ { e }^{ y }-1 } } \left( \int _{ 0 }^{ 1 }{ { u }^{ k }{ e }^{ -uy }du } \right) dy\\ \\ \)

by IBP we can evaluate the inner integral as \(k!{ e }^{ -y }\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { { y }^{ i-1 } }{ (k+i)! } } \) so this simplifies to

\( \frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { k! }{ (k+i)! } } \displaystyle\int _{ 0 }^{ \infty }{ \frac { { y }^{ n+i }{ e }^{ -y } }{ { e }^{ y }-1 } dy } \)

evaluating that integral we have

\(\frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { k! }{ (k+i)! } } \displaystyle\int _{ 0 }^{ \infty }{ \frac { { y }^{ n+i }{ e }^{ -y } }{ { e }^{ y }-1 } dy }= \int _{ 0 }^{ \infty }{ { y }^{ n+i }{ e }^{ -2y }\displaystyle\sum _{ j=0 }^{ \infty }{ { e }^{ -jy } } dy } =\sum _{ j=0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { y }^{ n+i }{ e }^{ -(2+j)y }dy } =\displaystyle\sum _{ j=0 }^{ \infty }{ \frac { \Gamma (n+i+1) }{ (2+j)^{ n+i+1 } } } } =(n+i)!(\zeta (n+i+1)-1)\)

hence,

\(\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx } =\frac { k! }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { (n+i)! }{ (k+i)! } (\zeta (n+i+1)-1) }\)

and the result follows

QED

Note:this is the longest \(\LaTeX\) document i've ever done,so it may contain errors Hummus A · 6 months ago

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@Hummus A I, bow to you. Swapnil Das · 5 months, 4 weeks ago

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@Swapnil Das i just happen to really like calculus :) Hummus A · 5 months, 4 weeks ago

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