Let's do some calculus! (48)

Lemma:

$\int_0^1 {\left\{ \dfrac 1k \right\}}^k \mathrm{d} x = \sum_{j=1}^{\infty} \dfrac{\zeta (j+1) - 1}{\dbinom{k+j}{j}} = \sum_{j=1}^{\infty} \dfrac{\zeta (j+1) - 1}{\dbinom{k+j}{k}}$

$*$ This implies that our integral is $\displaystyle \sum_{j=1}^{\infty} \dfrac{\zeta (j+1) - 1}{\dbinom{2017+j}{j}}$ , which gives $A+B+C = \boxed{2019}$ .

Proof:

Let

$I = \int_0^1 {\left\{ \dfrac 1k \right\}}^k \mathrm{d} x$

Make the substitution $t = \dfrac 1x \implies \mathrm{d} t = - \dfrac{1}{x^2} \mathrm{d} x$ and $\displaystyle \int_0^1 \mapsto \int_{\infty}^1$ . So we get

$\begin{aligned} I &= \int_0^1 {\left\{ \dfrac 1k \right\}}^k \cdot (-x^2) \cdot \left( - \dfrac{1}{x^2} \right) \mathrm{d} x \\ &= - \int_{\infty}^1 \dfrac{ {\left\{ t \right\}}^k }{t^2} \mathrm{d} t \\ &= \int_1^{\infty} \dfrac{ {\left\{ t \right\}}^k }{t^2} \mathrm{d} t \\ &= \int_1^{\infty} \dfrac{ {(t - \left\lfloor t \right\rfloor ) }^k }{t^2} \mathrm{d} t \\ &= \sum_{n=1}^{\infty} \int_n^{n+1} \dfrac{ (t - n)^k }{t^2} \mathrm{d} t \end{aligned}$

Make another substitution $y = t-n \implies \mathrm{d} y = \mathrm{d}t$ and $\displaystyle \int_n^{n+1} \mapsto \int_0^1$ . So we have

$\begin{aligned} I &= \sum_{n=1}^{\infty} \int_0^1 \dfrac{y^k}{ (y+n)^2 } \mathrm{d} y \\ &= \int_0^1 y^k \left( \sum_{n=1}^{\infty} \dfrac{1}{ (y+n)^2 } \right) \mathrm{d} y \end{aligned}$

Note that

$\dfrac{1}{ (y+n)^2 } = \dfrac{\Gamma(2)}{ (y+n)^2 } = \dfrac{1}{(y+n)^2} \cdot \int_0^{\infty} s e^{-s} \mathrm{d} s = \int_0^{\infty} u e^{-(n+y)u} \mathrm{d} u$

Thus

$\sum_{n=1}^{\infty} \dfrac{1}{ (y+n)^2 } = \sum_{n=1}^{\infty} \int_0^{\infty} u e^{-(n+y)u} \mathrm{d} u = \int_0^{\infty} \sum_{n=1}^{\infty} u e^{-(n+y)u} \mathrm{d} u = \int_0^{\infty} \dfrac{u e^{-uy}}{e^u - 1} \mathrm{d} u$

And so

$\begin{aligned} I &= \int_0^1 y^k \left( \int_0^{\infty} \dfrac{u e^{-uy}}{e^u - 1} \mathrm{d} u \right) \mathrm{d} y \\ &= \int_0^{\infty} \dfrac{u}{e^u -1} \left( \int_0^1 y^k e^{-uy} \mathrm{d} y \right) \mathrm{d} u \end{aligned}$

Let

$L_k = \int_0^1 y^k e^{-uy} \mathrm{d} y$

integrating by parts gives

$L_k = - \dfrac{e^{-u}}{u} + \left( \dfrac ku \right) L_{k-1}$

Let $M_k = \dfrac{u^k}{k!} L_k$ . Note that

$\begin{aligned} & \dfrac{u^k}{k!} \cdot L_k = - \dfrac{e^{-u}}{u} \cdot \dfrac{u^k}{k!} + \left( \dfrac{u^{k-1}}{(k-1)!} \right) L_{k-1} \\ \implies & M_k = - \dfrac{e^{-u}}{u} \cdot \dfrac{u^k}{k!} + M_{k-1} \end{aligned}$

Thus we have obtained a recursive pattern which gives us

$M_k = - \dfrac{e^{-u}}{u} \left( \sum_{r=0}^{k-1} \dfrac{u^{k-r}}{(k-r)!} \right) + M_0$

Here

$M_0 = \dfrac{u^0}{0!} L_0 = \int_0^1 e^{-uy} \mathrm{d} y = \dfrac{1-e^{-u}}{u}$

Hence

$\begin{aligned} M_k &= - \dfrac{e^{-u}}{u} \left( \sum_{r=0}^{k-1} \dfrac{u^{k-r}}{(k-r)!} \right) + \dfrac{1-e^{-u}}{u} \\ &= - \dfrac{e^{-u}}{u} \left( \sum_{r=0}^{k-1} \dfrac{u^{k-r}}{(k-r)!} \right) + \dfrac{e^{-u} \cdot e^u -e^{-u}}{u} \\ &= \dfrac{e^{-u}}{u} \left( e^u - \left( 1 + \sum_{r=0}^{k-1} \dfrac{u^{k-r}}{(k-r)!} \right) \right) \\ &= \dfrac{e^{-u}}{u} \sum_{j=1}^{\infty} \dfrac{u^{k+j}}{(k+j)!} \end{aligned}$

This gives

$L_k = \dfrac{k!}{u^k} \cdot \dfrac{e^{-u}}{u} \sum_{j=1}^{\infty} \dfrac{u^{k+j}}{(k+j)!} = k! e^{-u} \sum_{j=1}^{\infty} \dfrac{u^{j-1}}{(k+j)!}$

So

$I = \int_0^{\infty} \dfrac{u}{e^u -1} \left( k! e^{-u} \sum_{j=1}^{\infty} \dfrac{u^{j-1}}{(k+j)!} \right) \mathrm{d} u = \sum_{j=1}^{\infty} \dfrac{k!}{(k+j)!} \int_0^{\infty} \dfrac{u^j e^{-u}}{e^u -1} \mathrm{d} u$

Let

$\begin{aligned} T &= \int_0^{\infty} \dfrac{u^j e^{-u}}{e^u -1} \mathrm{d} u \\ &= \int_0^{\infty} u^j e^{-2u} \sum_{r=0}^{\infty} e^{-ru} \mathrm{d} u \\ &= \sum_{r=0}^{\infty} \int_0^{\infty} u^j e^{-(r+2)u} \mathrm{d}u \\ &= \sum_{r=0}^{\infty} \dfrac{\Gamma(j+1)}{ (r+2)^{j+1} } \\ &= \sum_{r=0}^{\infty} \dfrac{ j! }{ (r+2)^{j+1} } \\ &= j! \left( \zeta(j+1) -1 \right) \end{aligned}$

Finally, we get

$\begin{aligned} I &= \sum_{j=1}^{\infty} \dfrac{k! j!}{(k+j)!} \left( \zeta(j+1) -1 \right) \\ &= \sum_{j=1}^{\infty} \dfrac{ \zeta(j+1) -1 }{ \frac{(k+j)!}{k! j!} } \\ &= \sum_{j=1}^{\infty} \dfrac{ \zeta(j+1) -1 }{ \dbinom{k+j}{j} } \text{ or } \sum_{j=1}^{\infty} \dfrac{ \zeta(j+1) -1 }{ \dbinom{k+j}{k} } \end{aligned}$

$\mathbf{Q.E.D.}$