Integral Identity Involving Bell Numbers

Question

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

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

Answer 1

It seems that the integral on the left-hand side exceeds $2.57$, so it's not even close to the numerical value $2.5179\ldots$ of the integral on the right-hand side.

I told gp:

b(x) = suminf(n=1,n^x/n^n)
B(x) = suminf(n=1,n^x/n!)
r(x) = b(x)/B(x)
intnum(x=0,25,r(x))

and got $2.5793$+. Since the integrand $r(x)=b(x)/B(x)$ is positive but apparently decreasing, the Riemann sum $.01 \sum_{n=1}^{2500} r(n/100)$ should be a lower bound on $\int_0^{25} r(x)\,dx$, and thus on $\int_0^\infty r(x)\,dx$; and already that lower bound exceeds $2.57$: replacing the last gp command above by

sum(n=1,2500,.01*r(.01*n))

returns $2.5755599998001798\ldots > 2.57$.