Is there a deeper connection among $\int_0^1 x^{e-2} dx$, $\int_0^{\infty}\{x\}e^{-x}\,dx$, and $\sum_{n=1}^{\infty} e^{-n}$?

Question

Consider the following integrals $I_1,I_2$: $$ I_1=\int _1^{\infty} \lfloor x\rfloor e^{-x}\,dx = \sum_{n=1}^{\infty}n \int_{n}^{n+1}e^{-x}\,dx = (e-1)\sum_{n=1}^{\infty} n e^{-n-1} = \frac{e-1}{(e-1)^2}= \frac{1}{e-1} $$ $$ I_2 = \int_0^1 \sqrt{x\sqrt[3]{x\sqrt[4]x{\sqrt[5]{x\cdots }}}}\,dx = \int_0^1 x^{e-2}\,dx = \frac{1}{e-1}\left. x^{e-1} \right|_0^1= \frac{1}{e-1} $$Note that $I_1$ is an infinite series of definite integrals and $I_2$ is a definite integral of an infinite series, and they both evaluate to the same number. Millions of definite integrals evaluate to the same value but these are striking enough that I wonder if there is a deeper connection between them, or if one could possibly be transformed into the other. This might be a wild-goose chase but I'm curious if there is a transformation that takes one to the other, possibly using the closed-form of the geometric series $\sum_{n=1}^{\infty} e^{-n} = \frac{1}{e-1}$ as an intermediate step.

Answer 1

Here's a fun connection that takes $I_2$ to something like $I_1$ without passing through the exact result, but uses advanced tools.

Consider $\int_{0}^1 x^a dx$ and Taylor expand the integrand around $x=1$. The following string of equalities ensues

$$\int_0^1 x^a dx=\sum_{n=0}^{\infty}\frac{(a)_n}{n!}\int_0^1 (x-1)^n dx=\sum_{n=0}^{\infty}\frac{(-1)^n (a)_n}{(n+1)!}$$

We want to expand the falling factorials $(a)^n=a(a-1)...(a-n+1)$ into powers of $n$ (they are polynomials in it after all) and for that we use their connection to the Stirling numbers of 1st kind $(a)_n=\sum_{k=0}^ns(n,k)a^k$. After reversing the order of summation we obtain

$$\int_0^1 x^a dx=\sum_{k=0}^{\infty}a^k\sum_{n=k}^{\infty}\frac{(-1)^n}{(n+1)!}s(n,k)$$

Finally by manipulating the generating function we obtain the relation

$$\sum_{n=k}^{\infty}\frac{(z)^n}{(n+1)!}s(n,k)=\frac{1}{zk!}\int_{0}^zdt \log^k(1+t)$$

After putting $z=-1$ and performing the integral we finally obtain that

$$\sum_{n=k}^{\infty}\frac{(-1)^n}{(n+1)!}s(n,k)=(-1)^k$$

and hence

$$\int_0^1 x^{e-2}dx=\sum_{n=0}^{\infty}(2-e)^n$$

which is almost a surprise (at least from this point of view). I tried to connect $I_1$ and $I_2$ directly using this method and various cheap manipulations, and while it is possible to get a geometric series in $1/e$ (alas not quite the right one above) out of an appropriate expansion after a substition, I found the manipulation unenlightening. Hope this is interesting!