Evaluate $\int_0^\infty\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^t}\,dt$ 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?
 A: 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
$$\sum_{n=2}^\infty\frac{(-1)^n}{\log\Gamma(n+1)}=1.076901027858\dots$$
A: 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.
