# 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 $n$.

Notations:

• $\{ \cdot \}$ denotes the fractional part function.

• $\zeta(\cdot)$ denotes the Riemann zeta function.

• $\binom MN$ denotes the binomial coefficient, $\binom MN = \frac{M!}{N!(M-N)!}$.

This is a part of the set Formidable Series and Integrals

Note by Hamza A
3 years, 10 months ago

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We will proceed to generalize the above integral into

$\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx }$

Denote

$f(k,n)=\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx }$

After the substitution $\frac{1}{t}=x$ we obtain:

$f(k,n)=\displaystyle\int _{ 1 }^{ \infty }{ \frac { \{ t\} ^{ k } }{ { t }^{ n+2 } } dt }$

This is simply a sum of consecutive integrals:

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

By the u-substitution $u=t-i$ we obtain: $f(k,n)=\displaystyle\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ 0 }^{ 1 }{ \frac { { u }^{ k } }{ (i+u)^{ n+2 } } } } du$ $=\int _{ 0 }^{ 1 }{ { u }^{ k }\left( \sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } \right) dy } \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(1)}$

Given that $\int_{0}^{\infty}{e^{-kx}x^n \ dx}=k^{-n-1}n!$ We can see that $\frac { 1 }{ (i+u)^{ n+2 } } =\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ { e }^{ -(i+u)y }{ y }^{ n+1 } } dy$ Proceeding to substitute the above expression we obtain:

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

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

$=\frac { 1 }{ (n+1)! } \displaystyle\int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 }{ e }^{ -uy } }{ { e }^{ y }-1 } dy } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(2)}$

Substituting (2) into (1) : $f(k,n)=\frac { 1 }{ (n+1)! } \int _{ 0 }^{ 1 }{ { u }^{ k }\left( \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 }{ e }^{ -uy } }{ { e }^{ y }-1 } dy } \right) du }$ $=\frac { 1 }{ (n+1)! } \int _{ 0 }^{ \infty }{ \frac { { y }^{ n+1 } }{ { e }^{ y }-1 } } \left( \int _{ 0 }^{ 1 }{ { u }^{ k }{ e }^{ -uy }du } \right) dy$

It is easy to show through integration by parts that the inner integral is $k!{ e }^{ -y }\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { { y }^{ i-1 } }{ (k+i)! } }$ . Therefore,

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

After evaluating that integral in the expression we obtain:

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

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

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

$=\frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { k! }{ (k+i)! } }\displaystyle\sum _{ j=0 }^{ \infty }{ \frac { \Gamma (n+i+1) }{ (2+j)^{ n+i+1 } } }$

$=\frac { 1 }{ (n+1)! } \displaystyle\sum _{ i=1 }^{ \infty }{ \frac { k! }{ (k+i)! } }(n+i)!(\zeta (n+i+1)-1)$

Hence,

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

Plugging in $n=0$ gives us our original integral $\Large \boxed{\displaystyle\int _{ 0 }^{ 1 }{ \left \{ \dfrac { 1 }{ x } \right\} ^{ n } } \, dx=\sum _{ k=1 }^{ \infty }{ \frac { \zeta (k+1)-1 }{ \binom{n+k}{k} } } }$

$\textbf{QED}$

Note:this is the longest $\LaTeX$ document i've ever done,so it may contain errors

- 3 years, 10 months ago

How long did you take to prove this identity? It's just awesome!

- 3 years, 3 months ago

I, bow to you.

- 3 years, 10 months ago

i just happen to really like calculus :)

- 3 years, 10 months ago