Evaluate $\int_{1}^{n} \lfloor x \rfloor^{x- \lfloor x \rfloor} dx$.

Question

$\int_{1}^{n} [x]^{x-[x]} dx$

I tried to approach this with riemann sum method but it seems impossible by that way. Even using other general integration techniques it seems quite complicated .I have no idea about other techniques.

Answer 1

It boils down to splitting down the integral in intervals of length $1$. You get for $i>1$, $$ \int_i^{i+1} [x]^{x-[x]}\,\mathrm{d}x = \int_0^{1} i^x\,\mathrm{d}x = \left[\frac{i^x}{\log(i)}\right]_0^1 = \frac{i-1}{\log i} $$ and for $i=1$, we get $1$. Hence, $$ \int_1^{n} [x]^{x-[x]}\,\mathrm{d}x = \sum_{i=1}^{n-1}\int_i^{i+1} [x]^{x-[x]}\,\mathrm{d}x = 1 + \sum_{i=2}^{n-1}\frac{i-1}{\log i}. $$ You can not simplify this any further.

Answer 2

Splitting this into integrals on $[k,k+1]$, changing variable $x \to x-k$ and noting that the value at the right end is irrelevant since the function is continuous on the interval $[k,k+1)$ we get:

$\int_{1}^{n} [x]^{x-[x]} dx = \sum_{k=1}^{k=n}\int_{0}^{1} k^x dx=1+\sum_{k=2}^{k=n}\frac{k-1}{\log k}$