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Is the following identity true ?

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


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

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

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

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At first glance this seems too good to be true. (But this also applies to the valid formula:'s_dream ) Have you tried to disprove it by (estimating and) calculating both sides by computer? (The left hand side is the tricky part, the right hand side is around: 2.51792 ) – Daniel Soltész Oct 19 '13 at 19:44
Of course. No luck whatsoever in calculating the left hand side either numerically or symbolically. I don't even know its first digit ! All I know is that it converges, and that a man who later deleted his own comments wrote that its numerical value is about 2.5. That's ALL I was able to find out... In almost an entire year ! :-( – Lucian Oct 19 '13 at 20:05
You should also mention your other question about the convergence of the left hand side:… Alexander Shamov's answer is clearly relevant. – Daniel Soltész Oct 19 '13 at 20:12
The value $2.5...$ can be obtained by the mathematica command: NIntegrate[Gamma[x + 1]/x^x, {x, 0, Infinity}] – Suvrit Oct 19 '13 at 21:43

1 Answer 1

up vote 8 down vote accepted

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)

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


returns $2.5755599998001798\ldots > 2.57$.

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What is ‘GP’ ? Mathematica, for instance, can't even handle the integral, numerically, at all, that's why I'm asking. What mathematical software would you recommend ? – Lucian Oct 20 '13 at 10:56
My version (5.2) has significant problems calculating even as much as a single value of the function in question, let alone 2500 of them... Not sure why, but that's it. – Lucian Oct 20 '13 at 13:41
@Lucian: Sorry, I think I spoke too soon; seems indeed not easy for mathematica to crunch this out of the box---perhaps a more clever reformulation is needed! – Suvrit Oct 20 '13 at 14:12
What version of Mathematica are you using ? Because I thought of installing a newer one, but I'm afraid they might not be any better either. At least the old one is small, and I'm generally pleased with it, a few small bugs notwithstanding. – Lucian Oct 20 '13 at 14:25
@Lucian: I'm using v8.0; also, I think we should move our "how to do this on mathematica" question to --- I downloaded and tried pari/gp, and it works like a charm! (sorry Noam, for the comment noise!) – Suvrit Oct 20 '13 at 15:26

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