A calculus problem by Atul Kumar Ashish

Let $I=\int _{ 1 }^{ 5 }{ \frac { \left\{ x \right\} }{ \left\lfloor x \right\rfloor } dx } =\sum _{ k=1 }^{ 4 }{ \int _{ k }^{ k+1 }{ \frac { \left\{ x \right\} }{ \left\lfloor x \right\rfloor } dx } }$

when $x\in \left( k,k+1 \right) \Rightarrow \begin{cases} \left\{ x \right\} =x-k \\ \left\lfloor x \right\rfloor =k \end{cases}$

so $I=\sum _{ k=1 }^{ 4 }{ \int _{ k }^{ k+1 }{ \frac { x-k }{ k } dx } }$

$I=\quad \sum _{ k=1 }^{ 4 }{ \frac { 1 }{ k } \left( \frac { { x }^{ 2 } }{ 2 } -kx \right) { | }_{ k }^{ k+1 } } =\sum _{ k=1 }^{ 4 }{ \frac { 1 }{ k } \left( \frac { 1 }{ 2 } \right) }$

$I=\frac { 1 }{ 2 } \cdot \left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } \right)$

$I=\frac { 25 }{ 24 }$

$\begin{aligned} \int_1^5 \frac {\{x\}}{\lfloor x \rfloor} dx & = \sum_{k=1}^4 \frac 1k \int_0^1 x \ dx \\ & = \left(\frac 11 + \frac 12 + \frac 13 + \frac 14\right) \frac 12 \\ & = \boxed{\frac {25}{24}} \end{aligned}$

The above definite integral can be expressed as:

$\int_{1}^{5} \frac{\{x\}}{\lfloor x \rfloor} dx = \int_{1}^{5} \frac{x - \lfloor x \rfloor}{\lfloor x \rfloor} dx = \int_{1}^{2} \frac{x}{1} dx + \int_{2}^{3} \frac{x}{2} dx + \int_{3}^{4} \frac{x}{3} dx + \int_{4}^{5} \frac{x}{4} dx - \int_{1}^{5} 1 dx$ ;

or $\frac{x^2}{2}|^{2}_{1} + \frac{x^2}{4}|^{3}_{2} + \frac{x^2}{6}|^{4}_{3} + \frac{x^2}{8}|^{5}_{4} - x|^{5}_{1} = \frac{3}{2} + \frac{5}{4} + \frac{7}{6} + \frac{9}{8} - 4 = \boxed{\frac{25}{24}}$ .