Prove it!

∫01{1x}n dx=∑k=1∞ζ(k+1)−1(n+kk) \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 nn.

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

We will proceed to generalize the above integral into

∫01{1x}kxndx\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx }

Denote

f(k,n)=∫01{1x}kxndxf(k,n)=\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} ^{ k }{ x }^{ n }dx }

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

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

This is simply a sum of consecutive integrals:

f(k,n)=∑i=1∞∫ii+1{t}ktn+2dt=∑i=1∞∫ii+1(t−i)ktn+2 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−iu=t-i we obtain: f(k,n)=∑i=1∞∫01uk(i+u)n+2duf(k,n)=\displaystyle\sum _{ i=1 }^{ \infty }{ \displaystyle\int _{ 0 }^{ 1 }{ \frac { { u }^{ k } }{ (i+u)^{ n+2 } } } } du =∫01uk(∑i=1∞1(i+u)n+2)dy             (1)=\int _{ 0 }^{ 1 }{ { u }^{ k }\left( \sum _{ i=1 }^{ \infty }{ \frac { 1 }{ (i+u)^{ n+2 } } } \right) dy } \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(1)}

Given that ∫0∞e−kxxn dx=k−n−1n!\int_{0}^{\infty}{e^{-kx}x^n \ dx}=k^{-n-1}n! We can see that 1(i+u)n+2=1(n+1)!∫0∞e−(i+u)yyn+1dy \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:

∑i=1∞1(i+u)n+2=1(n+1)!∑i=1∞∫0∞e−(i+u)yyn+1dy\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 } }

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

=1(n+1)!∫0∞yn+1e−uyey−1dy                 (2)=\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)=1(n+1)!∫01uk(∫0∞yn+1e−uyey−1dy)duf(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 } =1(n+1)!∫0∞yn+1ey−1(∫01uke−uydu)dy=\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∑i=1∞yi−1(k+i)!k!{ e }^{ -y }\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { { y }^{ i-1 } }{ (k+i)! } } . Therefore,

f(k,n)=1(n+1)!∑i=1∞k!(k+i)!∫0∞yn+ie−yey−1dyf(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:

1(n+1)!∑i=1∞k!(k+i)!∫0∞yn+ie−yey−1dy\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 }

=1(n+1)!∑i=1∞k!(k+i)!∫0∞yn+ie−2y∑j=0∞e−jydy=\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 }

=1(n+1)!∑i=1∞k!(k+i)!∑j=0∞∫0∞yn+ie−(2+j)ydy=\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 }}

=1(n+1)!∑i=1∞k!(k+i)!∑j=0∞Γ(n+i+1)(2+j)n+i+1=\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 } } }

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

Hence,

∫01{1x}kxndx=k!(n+1)!∑i=1∞(n+i)!(k+i)!(ζ(n+i+1)−1)\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=0n=0 gives us our original integral ∫01{1x}n dx=∑k=1∞ζ(k+1)−1(n+kk) \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} } } }

QED\textbf{QED}

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

Hamza A - 5 years, 2 months ago

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

찬홍 민 - 4 years, 8 months ago

I, bow to you.

Swapnil Das - 5 years, 2 months ago

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i just happen to really like calculus :)

Hamza A - 5 years, 2 months ago
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