Prove it!

01{1x}ndx=k=1ζ(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.

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This is a part of the set Formidable Series and Integrals

Note by Hamza A
3 years, 6 months ago

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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=1ii+1{t}ktn+2dt=i=1ii+1(ti)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=tiu=t-i we obtain: f(k,n)=i=101uk(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=11(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 0ekxxn dx=kn1n!\int_{0}^{\infty}{e^{-kx}x^n \ dx}=k^{-n-1}n! We can see that 1(i+u)n+2=1(n+1)!0e(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=11(i+u)n+2=1(n+1)!i=10e(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)!0yn+1euy(i=1eiy)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)!0yn+1euyey1dy                 (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(0yn+1euyey1dy)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)!0yn+1ey1(01ukeuydu)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!eyi=1yi1(k+i)!k!{ e }^{ -y }\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { { y }^{ i-1 } }{ (k+i)! } } . Therefore,

f(k,n)=1(n+1)!i=1k!(k+i)!0yn+ieyey1dyf(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=1k!(k+i)!0yn+ieyey1dy\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=1k!(k+i)!0yn+ie2yj=0ejydy=\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=1k!(k+i)!j=00yn+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=1k!(k+i)!j=0Γ(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=1k!(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}ndx=k=1ζ(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 - 3 years, 6 months ago

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

찬홍 민 - 3 years ago

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I, bow to you.

Swapnil Das - 3 years, 6 months ago

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i just happen to really like calculus :)

Hamza A - 3 years, 6 months ago

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