Let's do some calculus! (47)

Lemma:

$\int_0^1 \left\{ \dfrac{1}{\sqrt[k] x} \right\} \, \mathrm{d}x = \dfrac{k}{k-1} - \zeta(k)$

$*$ This implies that the value of our integral is $\dfrac{2017}{2016} - \zeta(2017)$ , which gives $a+b+c = \boxed{6050}$ .

Proof:

Let

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

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

$I = \int_0^1 \left\{ \dfrac{1}{\sqrt[k] x} \right\} \cdot (-x^2) \cdot \left( - \dfrac{1}{x^2} \right) \, \mathrm{d}x = - \int_{\infty}^1 \dfrac{ \left\{ \sqrt[k] t \right\} }{t^2} \, \mathrm{d} t = \int_1^{\infty} \dfrac{ \left\{ \sqrt[k] t \right\} }{t^2} \, \mathrm{d} t$

Again make a substitution $u = t^{\frac 1k}$ , this gives $\mathrm{d} u = \dfrac 1k \cdot t^{\frac{1-k}{k}} \mathrm{d} t$ and $\displaystyle \int_1^{\infty} \mapsto \int_1^{\infty}$ . We get

$\begin{aligned} I &= \int_1^{\infty} \dfrac{ \left\{ \sqrt[k] t \right\} }{t^2} \, \mathrm{d} t \\ &= \int_1^{\infty} \dfrac{ \left\{ \sqrt[k] t \right\} }{t^2} \cdot k t^{\frac{k-1}{k}} \cdot \left( \dfrac 1k t^{\frac{1-k}{k}} \right) \, \mathrm{d} t \\ &= k \int_1^{\infty} \dfrac{ \left\{ u \right\} }{ u^{2k} } \cdot u^{k-1} \, \mathrm{d} u \\ &= k \int_1^{\infty} \dfrac{\left\{ u \right\} }{ u^{k+1} } \mathrm{d} u \\ &= k \int_1^{\infty} \dfrac{u - \left\lfloor u \right\rfloor}{ u^{k+1} } \mathrm{d} u \\ &= k \int_1^{\infty} \dfrac{1}{u^k} \mathrm{d} u - k \int_1^{\infty} \dfrac{ \left\lfloor u \right\rfloor }{u^{k+1}} \mathrm{d} u \\ &= \dfrac{k}{1-k} \cdot \dfrac{1}{u^{k-1}} { {\huge|}_1^{\infty}} - k \sum_{n=1}^{\infty} \int_n^{n+1} \dfrac{n}{u^{k+1}} \mathrm{d} u \\ &= \dfrac{k}{1-k} \cdot (-1) - k \left( - \dfrac 1k \right) \sum_{n=1}^{\infty} n \cdot \left[ \dfrac{1}{u^k} \right] { {\huge|}_n^{n+1}} \\ &= \dfrac{k}{k-1} + \sum_{n=1}^{\infty} \left[ \dfrac{n}{ (n+1)^k } - \dfrac{n}{n^k} \right] \\ &= \dfrac{k}{k-1} + \left( \dfrac{1}{2^k} - \dfrac{1}{1^k} + \dfrac{2}{3^k} - \dfrac{2}{2^k} + \dfrac{3}{4^k} - \dfrac{3}{3^k} + \cdots \right) \\ &= \dfrac{k}{k-1} + \left( - \dfrac{1}{1^k} - \dfrac{1}{2^k} - \dfrac{1}{3^k} - \cdots \right) \\ &= \dfrac{k}{k-1} - \zeta(k) \end{aligned}$

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