Greatest Integer Function Integral

Let $I=\int _{ 1 }^{ \infty }{ \frac { \left\lfloor x \right\rfloor }{ { x }^{ 5 } } dx }$ .

$\Longrightarrow 8I=8\int _{ 1 }^{ 2 }{ \frac { 1 }{ { x }^{ 5 } } dx } +8\int _{ 2 }^{ 3 }{ \frac { 2 }{ { x }^{ 5 } } dx } +8\int _{ 3 }^{ 4 }{ \frac { 3 }{ { x }^{ 5 } } dx } +...$

$=8\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { n }{ { x }^{ 5 } } dx } }$

$=8\sum _{ n=1 }^{ \infty }{ { \left[ \frac { n }{ -4{ x }^{ 4 } } \right] }_{ x=n }^{ x=n+1 } } \\$

$=2\sum _{ n=1 }^{ \infty }{ (\frac { 1 }{ { n }^{ 3 } } -\frac { n }{ { (n+1) }^{ 4 } } ) }$ .

Now note that $\frac { n }{ { (n+1) }^{ 4 } } =\frac { (n+1)-1 }{ { (n+1) }^{ 4 } } \\ =\frac { n+1 }{ { (n+1) }^{ 4 } } -\frac { 1 }{ { (n+1) }^{ 4 } } \\ =\frac { 1 }{ { (n+1) }^{ 3 } } -\frac { 1 }{ { (n+1) }^{ 4 } }$ .

So the sum is now equal to $2\sum _{ n=1 }^{ \infty }{ (\frac { 1 }{ { n }^{ 3 } } -\frac { 1 }{ { (n+1) }^{ 3 } } +\frac { 1 }{ { (n+1) }^{ 4 } } ) } \\ =2\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 3 } } } -2\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (n+1) }^{ 3 } } } +2\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (n+1) }^{ 4 } } } \\ =2\zeta (3)-2(\frac { 1 }{ { 2 }^{ 3 } } +\frac { 1 }{ { 3 }^{ 3 } } +\frac { 1 }{ { 4 }^{ 3 } } +...)+2(\frac { 1 }{ { 2 }^{ 4 } } +\frac { 1 }{ { 3 }^{ 4 } } +\frac { 1 }{ { 4 }^{ 4 } } +...)\\ =2\zeta (3)-2(\zeta (3)-1)+2(\zeta (4)-1)\\ =2\zeta (3)-2\zeta (3)+2+2\zeta (4)-2\\ =2\zeta (4)\\ =2\cdot \frac { { \pi }^{ 4 } }{ 90 } =\frac { { \pi }^{ 4 } }{ 45 }$ .

$a=4, b=45$ and $a+b=\boxed {49}$ .

Relevant wiki: Riemann Zeta Function

Similar solution with @Anthony Muleta 's

$\begin{aligned} I & = \int_1^\infty \frac {\lfloor x \rfloor}{x^5}\ dx \\ & = 8 \sum_{k=1}^\infty \int_k^{k+1} \frac k{x^5}\ dx \\ & = 8 \sum_{k=1}^\infty \frac k{4x^4}\ \bigg|_{k+1}^k \\ & = 2 \sum_{k=1}^\infty \left(\frac k{k^4} - \frac k{(k+1)^4} \right) \\ & = 2 \sum_{k=1}^\infty \left(\frac 1{k^3} - \frac {k+1-1}{(k+1)^4} \right) \\ & = 2 \sum_{k=1}^\infty \left(\frac 1{k^3} - \frac 1{(k+1)^3} + \frac 1{(k+1)^4} \right) \\ & = 2 \left(1 + \sum_{\color{#3D99F6}k=1}^\infty \frac 1{\color{#3D99F6}(k+1)^4} \right) \\ & = 2 \left(1 + \sum_{\color{#D61F06}k=2}^\infty \frac 1{\color{#D61F06}k^4} \right) \\ & = 2 \left(1 + \sum_{\color{#3D99F6}k=1}^\infty \frac 1{k^4} \color{#3D99F6} - 1 \right) \\ & = 2 \color{#3D99F6} \sum_{k=1}^\infty \frac 1{k^4} & \small \color{#3D99F6} \text{Riemann zeta function }\zeta(n) = \sum_{k=1}^\infty \frac 1{k^n} \\ & = 2 {\color{#3D99F6} \zeta (4)} = 2 \times {\color{#3D99F6} \frac {\pi^4}{90}} = \frac {\pi^4}{45} \end{aligned}$

Therefore, $a+b = 4+45 = \boxed{49}$ .