Evaluate $\int_{0}^1 \prod_{k=2}^n \lfloor kx \rfloor dx$ Let $n\ge2$ be an integer , then $$\int_{0}^1 \prod_{k=2}^n \lfloor kx \rfloor dx=\text{ ?}, $$  where $\lfloor \space \rfloor$ is the "floor-function"
 A: This is not an answer, but too long for a comment. Here is the result of a numerical experiment done with Maple:
$$seq(int(mul(floor(k*x), k = 2 .. n), x = 0 .. 1), n = 2 .. 15) $$ outputs $$1/2,5/6, 13/6, 15/2, 166/5, 888/5, 5617/5, 57339/7, 473958/7, 48185388/77,$$  $$ 493082052/77, 10264563168/143, 125213779200/143, 1651372471104/143.$$ Then $$plots:-pointplot([seq([n, evalf(ln(int(mul(floor(k*x), k = 2 .. n), x = 0 .. 1)))],$$ $$ n = 2 .. 15)], color = red, symbolsize = 15) $$ 
In my opinion, the plot is close to a straight line. Therefore, this suggests the asymptotic dependence of the form $A+B\exp(k)$
A: A partial answer and insight to the nature of the integral.
Let $$I=\int^1_0dx\prod^n_{k=2}\lfloor kx\rfloor $$
Make the substitution $u=1/x$ therefore $dx=-1/u^2du$ and the integral becomes
$$I=\int^\infty_1du\frac1{u^2}\left[\prod^n_{k=2}\lfloor k/u\rfloor\right]$$
Then integrate by parts you will see it evaluates to (integrand obtained by considering derivatives of products).
$$I=n!+\int^\infty_1\frac1u\left[\sum^n_{j=2}\prod^n_{\substack{i=2\\i\ne j}}\lfloor i/u\rfloor\right]du$$
Then by manipulation we get 
$$I=n!+\sum_{j=2}^n\int^\infty_1du\frac{1}{u\lfloor j/u\rfloor}\prod^n_{i=2}\lfloor i/u\rfloor$$
Now focus on the other integral. Let 
$$I_2=\int^\infty_1du\frac1{u\lfloor j/u\rfloor}\prod^n_{i=2}\lfloor i/u\rfloor$$
Now let $x=1/u$ again then $du=-dx/x^2$ and $$I_2=\int_0^1dx\frac1{x\lfloor jx \rfloor}\prod^n_{i=2}\lfloor ix\rfloor$$
I tried to manipulate this further however that factor will not go away.
So now I am going to look at the asymptotics of the integral. Clearly $I(n)\in\omega(n!)$. The summation term may give us a more precise approximation however. Through inspection it can be seen that $\sum I_2(n)\in O(n!)$. Therefore $I(n)$ can be thought of as $I(n)\in\Theta(n!)$. For approximation low values of $n$ can be though of as $I(n)\approx\frac23(n-2)!$. This can easily be made more accurate.
I would be surprised if a closed form for this integral exists. You should also note that I tried Riemann sums however they end up being messy and difficult to manipulate.
A: 
In the following paragraph, the letter k has a different connotation than it does in the question.

Let $F_n$ be the Farey sequence of n. It always has $2k+1$ elements for $n\ge2$ $($which is clearly the case here$)$, and its middle element is always $\dfrac12$, the first and last being $0$ and $1$ respectively. Now, we are interested in the sub-sequence starting at $F_n(k+1)=\dfrac12$ and ending at $F_n(2k+1)=1.~$ $($The number of terms in each Farey sequence can be found here$)$. Now, the product vanishes on $\Big(0,\frac12\Big)$, so the integral can be written as $\displaystyle\sum_{j=1}^kC_j\bigg[F_n(k+j+1)-F_n(k+j)\bigg]$, since the product is a constant on each such sub-interval, whose two extremities are consecutive Farey numbers, the value of each such constant being $C_j=\displaystyle\prod_{i=2}^n\Big\lfloor i\cdot F_n(k+j)\Big\rfloor$.
