Floor Integral

First necessary realization is that we can convert this from an integral to a summation as all values of x are being floored. Thus we can rewrite this as:

$\sum_{x=1}^{\infty} (x + 2 + \frac{1}{x})(\frac{1}{(x - 1)!}) dx$

We can simplify this summation by multiplying the top and bottom by x:

$S = \sum_{x=1}^{\infty} (x^{2} + 2x + 1)(\frac{1}{x!}) dx$

$S = \sum_{x=1}^{\infty} (x+1)^{2}(\frac{1}{x!}) dx$

Writing out the first few terms, we get:

$S = 4 + \frac{9}{2} + \frac{16}{6} + ...$

A similar function is $e^{x}$ :

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...$

If we wish to get squared terms on the top we can multiply by x and take the derivative:

$xe^{x} = x + x^{2} + \frac{x^{3}}{2!} + \frac{x^{4}}{3!} + ...$

$e^{x} + xe^{x} = 1 + 2x + \frac{3x^{2}}{2!} + \frac{4x^{3}}{3!} + ...$

We can repeat this step to square each value:

$xe^{x} + x^{2}e^{x} = x + 2x^{2} + \frac{3x^{3}}{2!} + \frac{4x^{4}}{3!} + ...$

$e^{x} + xe^{x} + x^{2}e^{x} + 2xe^{x} = 1 + 4x + \frac{9x^{2}}{2!} + \frac{16x^{3}}{3!} + ...$

Evaluating this at x = 1, we get:

$e + e + e + 2e = 1 + 4x + \frac{9x^{2}}{2!} + \frac{16x^{3}}{3!} + ...$

$5e = 1 + S$

$S = 5e - 1 = 12.59$

Which gets the answer of 13 .