**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](https://mathoverflow.net/questions/138896/convergence-and-closed-form-of-an-integral-involving-bell-numbers).** 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*


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



Integral Identity Involving Bell Numbers
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 $2.5755599998001798\ldots > 2.57$.


