**Is the following identity true ?**

$$\int_0^\infty \frac{b(x)}{B(x)} dx \quad \overset{?}{=} \quad \int_0^\infty \frac{x!}{x^x} dx$$

*where*

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad \text{and} \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$

**NOTE:** *A short sketch of the demonstration proving the convergence of the integral on the left can be found here. Also, the numerical value of the integral on the right is about 2.5179^{+}. Furthermore, if the position of $x$ and $n$ in the numerator of each sum were reversed, and both sums were to start at n = 0, we would have the following identity:*

$$\int_0^\infty \frac{E(x)}{e^x} dx \quad = \quad \sum_{n=0}^\infty \frac{n!}{n^n}$$

*where* $\lim_{n \to 0} n^n = 1,$ *and*

$$E(x) = \sum_{n=0}^\infty \frac{x^n}{n^n} \qquad \text{and} \qquad e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$