N integrations

Let's first focus on the first integral, $\int_0^n \{ x_1 + ... + x_n \} dx_1$ Leaving the other variables fixed, the values the function takes when $0 \leq x_1 \leq 1$ are the same values it takes when $k \leq x_1 < k+1$ for any integer $k$ . Thus, $\int_0^n \{ x_1 + ... + x_n \} dx_1 = n \int_0^1 \{ x_1 + ... + x_n \} dx_1$ . But we can do this with respect to all the variables, giving us our first simplification: $\int_0^n ... \int_0^n \{ x_1 + ... + x_n \} dx_1 ... d_n = n^n \int_0^1 ... \int_0^1 \{x_1 + ... + x_n \} dx_1 ... dx_n$

Let's now focus again on the first integral, $\int_0^1 \{ x_1 + ... + x_n \} dx_1$ . Remembering that with respect to $x_1$ , all the other variables are constant, let's study the integral $\int_0^1 \{x + c \} dx$ . Remembering that $\{ \cdot \}$ is periodic with it taking its whole range of values in $[0,1)$ , shifting it by a constant just changes the order in which those same values are attained. But we are integrating over the entirety of $[0,1)$ so it doesn't matter. Thus, $\int_0^1 \{x + c \} dx = \int_0^1 \{ x \} dx = \int_0^1 x dx = \frac{1}{2}$ . With this result we can compute: $\int_0^1 ... \int_0^1 \{x_1 + ... + x_n \} dx_1 ... dx_n = \int_0^1 ... \int_0^1 \frac{1}{2} dx_2 ... dx_n = \frac{1}{2}$ giving us the final result: $I_n = \frac{n^n}{2}$