Evaluate $\int_0^\infty\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^t}\,dt$

Question

I've noticed that the following improper integral converges: $$\int_0^\infty\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^t}\,dt$$ I can't seem to find Wolfram Alpha syntax that is able to parse this, and can't find any other way to approximate – let alone solve – the integral.

What is this integral's decimal expansion, and is it a known constant?

Answer 1

Since the first two terms of the inner sum cancel for all $t$: $$\int_0^\infty\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^t}\,dt=\int_0^\infty\sum_{n=2}^\infty\frac{(-1)^n}{(n!)^t}\,dt$$ Swap the integral and sum: $$=\sum_{n=2}^\infty\int_0^\infty\frac{(-1)^n}{(n!)^t}\,dt=\sum_{n=2}^\infty(-1)^n\int_0^\infty(n!)^{-t}\,dt$$ The inner integral is easily seen to be $\frac1{\ln n!}$: $$=\sum_{n=2}^\infty\frac{(-1)^n}{\ln n!}=\sum_{n=2}^\infty\frac{(-1)^n}{\log\Gamma(n+1)}$$ I am sure this last sum has neither a closed form nor a name. Evaluating this in mpmath gives $$\sum_{n=2}^\infty\frac{(-1)^n}{\log\Gamma(n+1)}=1.076901027858\dots$$

Answer 2

You want $\int_0^\infty{\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^t} dt} $

Let $f(x) =\int_0^\infty \dfrac{dt}{x^t} $. Since $\int \dfrac{dt}{x^t} =\dfrac{-1}{x^t\ln(x)} $, $f(x) =\dfrac1{\ln(x)} $.

Therefore, assuming all manipulations are legal,

$\begin{array}\\ \int_0^\infty{\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^t} dt} &=\sum_{n=0}^{\infty}(-1)^n\int_0^\infty\frac{1}{(n!)^t} dt\\ &=\sum_{n=0}^{\infty}(-1)^n\dfrac1{\ln(n!)}\\ \end{array} $

For this to be meaningful, we must have $n \ge 2$ since $\ln(1) = 0$. Therefore I will start the sum at $2$, and get

$\begin{array}\\ \int_0^\infty{\sum_{n=2}^{\infty}\frac{(-1)^n}{(n!)^t} dt} &=\sum_{n=2}^{\infty}(-1)^n\dfrac1{\ln(n!)}\\ \end{array} $

By the alternating series test, this sum converges.

Since $\ln(n!) < n \ln(n)$, the sum only converges conditionally.

The sum of the first 1000 terms according to Wolfy is 1.0769855503.

My usual technique of dealing with alternating series gives

$\begin{array}\\ \int_0^\infty{\sum_{n=2}^{\infty}\frac{(-1)^n}{(n!)^t} dt} &=\sum_{n=2}^{\infty}(-1)^n\dfrac1{\ln(n!)}\\ &=\sum_{n=1}^{\infty}(\dfrac1{\ln((2n)!)}-\dfrac1{\ln((2n+1)!)})\\ &=\sum_{n=1}^{\infty}(\dfrac{\ln((2n+1)!)-\ln((2n)!)}{\ln((2n)!)\ln((2n+1)!)})\\ &=\sum_{n=1}^{\infty}\dfrac{\ln(2n+1)}{\ln((2n)!)\ln((2n+1)!)}\\ \end{array} $

and this converges absolutely since the terms are, within a multiplicative constant, $\dfrac{\ln(n)}{n^2\ln^2(n)} =\dfrac{1}{n^2\ln(n)} $ and the sum of this converges, but not very rapidly.

The inverse symbolic calculator does not recognize this.