Interesting Integral

We will use poly-gamma function for this.

${ \psi }_{ 2 }\left( n+1 \right) =\int _{ 0 }^{ 1 }{ \frac { { x }^{ n }{ \left( lnx \right) }^{ 2 } }{ x-1 } }$

Now, we substitute $n=0$

${ \psi }_{ 2 }\left( 1 \right) =\int _{ 0 }^{ 1 }{ \frac { { \left( lnx \right) }^{ 2 } }{ x-1 } }$

Now, we use the relation: ${ \psi }_{ n }\left( z \right) ={ \left( -1 \right) }^{ n+1 }n!\zeta \left( n+1,z \right)$

Here $\zeta \left( n+1,z \right)$ is Hurwitz zeta function.

Now, we substitute $n=2$ and $z=1$ .

${ \psi }_{ 2 }\left( 1 \right) =-2\zeta \left( 3,1 \right)$

Now, we use the property: $\zeta \left( n+1,1 \right) =\zeta \left( n+1 \right)$

Therefore we get: ${ \psi }_{ 2 }\left( 1 \right) =-2\zeta \left( 3 \right)$

For the proof part of the question:

$I=\int _{ 1 }^{ \infty }{ \frac { \left\lfloor x \right\rfloor }{ { x }^{ a+1 } } dx } \\ =\int _{ 1 }^{ 2 }{ \frac { 1 }{ { x }^{ a+1 } } dx } +\int _{ 2 }^{ 3 }{ \frac { 2 }{ { x }^{ a+1 } } dx } +\int _{ 3 }^{ 4 }{ \frac { 3 }{ { x }^{ a+1 } } dx } +...$

Now converting it into a summation and simplifying, we get:

$I=\frac { \zeta \left( a \right) }{ a }$

Therefore, as far as the question is concerned, $a=3$

---

Using Harsh's hint: $\frac { 1 }{ 1-x } =\sum _{ n=0 }^{ \infty }{ { x }^{ n } }$

Now,

$I=\sum _{ n=0 }^{ \infty }{ { \int _{ 0 }^{ 1 }{ { x }^{ n }{ \left( lnx \right) }^{ 2 }dx } } }$

we substitute: $x={ e }^{ -\frac { t }{ n } }$

We get: $I=\sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( 3 \right) }{ { n }^{ 3 } } } \\ \therefore I=2\zeta \left( 3 \right)$

It was foolish of me not to think in a straight way.

Hint :

Use $\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \cdots$ .