First, some notation. I'll write $f(x)=o(g(x))$ if $\lim_{x\to\infty} \left|\frac{f(x)}{g(x)}\right|=0$. I'll also write $g(x)=\omega(f(x))$ if $f(x)=o(g(x))$, i.e. $\limsup_{x\to\infty} \left|\frac{g(x)}{f(x)}\right|=\infty$. I'll say $f(x)\sim g(x)$ as $x\to\infty$ if $f(x)=g(x)+o(g(x))$ for all large enough $x$, i.e. $f(x)/g(x)=1+o(1)$ for all large enough $x$.
I'm interested in whether, given a certain class of sequence $a(k)$ (see below), we have that
$$\sum_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}\sim\int_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt$$
as $x\to\infty$. The sequence $a(k)$ is non-negative, increasing, and approaches $+\infty$ as $k\to\infty$ fast enough that the series on the left hand side converges for every real number $x\in[0,\infty)$—it turns out that this is equivalent to $a(k)=\omega\left(\frac{k}{\log k}\right)$ by [1]. I know this relationships holds true when $a(k):=k$ (see [2], [3]), in which case the series is simply $e^x$, as well as when $a(k):=2k$, in which case the series is $\cosh(\sqrt{x})$.
It seems tempting to use Euler-Maclaurin summation, but the resulting difference between the sum and the integral already ends up being a divergent series for large enough $x$ in the case $a(k):=k$.
Edit: Given the answer below, I require that $a(t)\in C^\infty$ on its domain $[0,\infty)$ and the sequence $a(k)$ is defined to be the restriction $a(k):\mathbb{N}_0\to[0,\infty)=a(t)|_{t\in\mathbb{N}_0}$.
This question is cross-posted at MSE.
Best Answer
This is not true without further regularity assumptions on $a$.
Indeed, take any sequence $(a(k))_{k\ge0}$ as in your post and then extend it to the function $a$ on $[0,\infty)$ by the formula $a(t):=a(\lfloor t\rfloor)$. Then for $x>1$ (say) $$\int_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt =\frac{x-1}{\ln x}\,\sum_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))} \not\sim\sum_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}$$ as $x\to\infty$.