Consider the following function:
$$ I(x,y)=\sum_{n=1}^{\infty}\bigg|\int_0^1 \frac{1}{t~(\log t)^y}\exp\bigg(\frac{n^x}{\log t}\bigg) ~dt~\bigg|$$
For $x=0$ and letting $y$ vary we get the Gamma function, $\Gamma(y-1).$ For $y=2$ letting $x$ vary, we get the Riemann zeta function $0!\zeta(x).$ For $y=3$ letting $x$ vary we get $1!\zeta(2x)$ and for $y=4$ letting $x$ vary we get $2!\zeta(3x)$ etc.
Adding up these values reminds me of a similar looking infinite series: $$\varphi(x)=\sum_{n=1}^\infty (e^{-n^{-x}}-1)=\sum_{n=1}^\infty \frac{(-1)^n}{n!}\zeta(nx)$$
Although the factorial is in the denominator and the terms alternate…
$$ \sum_{k=1}^\infty I(x,k)=\sum_{k=1}^\infty \sum_{n=1}^\infty \bigg|\int_0^1\frac{1}{t~(\log t)^k}\exp\bigg(\frac{n^x}{\log t}\bigg) ~dt~\bigg|=\zeta(x)+\zeta(2x)+2\zeta(3x)\cdot\cdot~\cdot$$
Is there an explicit closed form for $I(x,y)?$
Is $\sum_{k=1}^\infty I(x,k)$ related to $\varphi(x)?$
What does a plot of $I(x,y)$ look like?
Context:
This integral initially came from $$\Gamma(n-1)=\bigg|\int_0^1 \frac{1}{t(\log t)^n}\exp\bigg(\frac{1}{\log t}\bigg)~dt~\bigg|=\int_0^\infty t^{n-2}e^{-t}~dt $$
Best Answer
This is just to provide an answer to this question, which was answered in the comments.
The closed form of the integral is:
$$I(x,y)=\Gamma(y-1)\zeta((y-1)x).$$
This can be seen by looking at the integral:
$$\int_{0}^{1}\frac{1}{t(\log t)^y} \exp\left(\frac{n^x}{\log t}\right)\, dt=(-1)^{-y} \Gamma(y-1) (n^{-x})^{y-1}.$$