The function $f(x) = \int_0^\infty \frac{x^t}{\Gamma(t+1)} \, dt$ Does anyone know if this function has a name? I came up with it by looking at the power series for $e^z$, changing the summation to an integral, and substituting the gamma function for the factorial function.
 A: Laplace transform
We consider the function for $x\ge 0$.
As found by @Mathlover, it is not hard to get the Laplace transform of $f(x)$. 
Inverting, we find
$$f(x) = \frac{1}{2\pi i} \int_\gamma ds\, e^{s x}\frac{1}{s\log s},$$
where $\gamma$ is the Bromwich contour.
There is a singularity at $s=1$ and a cut from $s=0\to -\infty$.
We deform the contour in the usual way and simplify, with the result
$$\begin{eqnarray*}
f(x) &=& e^x - \int_0^\infty d\rho\,
\frac{e^{-\rho x}}{\rho(\log^2\rho + \pi^2)} \\
&=& e^x - \int_{-\infty}^\infty d\sigma\,
\frac{e^{-x e^\sigma}}{\sigma^2 + \pi^2}.
\end{eqnarray*}$$
It is possible to go further with the last integral, but we'll stop here. 
Addendum: Series
Let's examine the series for $f(x)$. 
Define
$$\begin{eqnarray*}
\Delta(x) &=& \int_0^\infty d\sigma\,
\frac{e^{-x e^\sigma} + e^{-x e^{-\sigma}}}{\sigma^2 + \pi^2}
\end{eqnarray*}$$
Note that $f(x) = e^x - \Delta(x)$.
We have $e^{-x e^\sigma} + e^{-x e^{-\sigma}} \le e^{-x}+1$, 
and $\int_0^\infty d\sigma\, \frac{1}{\sigma^2+\pi^2}$ converges,
so $\Delta(x)$ converges uniformly.
Clearly the derivative of $\Delta(x)$ doesn't exist at $x=0$
(the integral 
$\int_0^\infty d\sigma\,\frac{\cosh \sigma}{\sigma^2+\pi^2}$ diverges)
so there is no Maclaurin series for $f(x)$. 
We can do a Taylor series about any $x>0$. 
Of course, we must evaluate integrals of the form 
$$\left.\frac{d^k}{d x^k} \Delta(x)\right|_{x=x_0} = (-1)^k \int_0^\infty d\sigma\,
\frac{e^{k\sigma -x_0 e^\sigma} + e^{-k\sigma -x_0 e^{-\sigma}}}{\sigma^2 + \pi^2}.$$
These integrals are uniformly convergent for $x_0 >0$. 
In this regard it is useful to notice that
$e^{-k\sigma -x_0 e^{-\sigma}} \le 1$ and 
$$e^{k\sigma -x_0 e^\sigma} \le
\begin{cases}
e^{-x_0}, & x_0 \ge k \\
\left(\frac{k}{x_0 e}\right)^k, & \mathrm{else}.
\end{cases}$$
For large $x_0$ the dominant terms in the expansion come from the series for $e^x$, as expected. 
Trapezoidal rule
Approximate the integral by a sum.
Let the step size be $1/m$, where $m=1,2,\ldots$, and let 
$g_k = \frac{x^{k/m}}{\Gamma(k/m+1)}$.
Then,
$$\begin{eqnarray*}
\int_0^\infty dt\, \frac{x^t}{\Gamma(t+1)} &\simeq&
\lim_{n\to\infty}
\left(
\frac{1}{m}\sum_{k=0}^n g_k -\frac{1}{2m}(g_0+g_n) \right)\\
&=& \frac{1}{m}E_{1/m}(x^{1/m}) -\frac{1}{2m} \\
&=& \frac{1}{m} e^x\left(1+\sum_{k=1}^{m-1}\frac{\gamma(1-k/m,x)}{\Gamma(1-k/m)}\right) - \frac{1}{2m},
\end{eqnarray*}$$
where $E_\alpha(z)$ is the Mittag-Leffler function and $\gamma$ is the lower incomplete gamma function.
(The relation for $E_{1/m}(x^{1/m})$ used above can be found in this paper.)
In the limit $m\to\infty$ the sum is equal to the original integral, so
$$\begin{eqnarray*}
f(x) &=& e^x \lim_{m\to\infty} \frac{1}{m} \sum_{k=1}^{m-1}\frac{\gamma(1-k/m,x)}{\Gamma(1-k/m)} \\
&=& e^x \int_0^1 ds\,\frac{\gamma(s,x)}{\Gamma(s)}
\end{eqnarray*}$$
This is the integral formula found by @sos440 using another method. 
Here are the sums for $m=1$ and $2$. 
$$\begin{array}{ll}
m & f(x) \\ \hline
1 & e^x - 1/2 \\
2 & \frac{1}{2}e^x \left(\mathrm{erf}\left(\sqrt{x}\right) + 1\right) -\frac{1}{4} \end{array}$$
Since $f(x) \simeq e^x-1/2$, $f^{-1}(x)\simeq \log(x+1/2)$. 
Below we plot $f(x)$ (solid), $e^x$ (dashed, black), and the approximations for $m=1$ (dashed, red), and $m=2$ (dashed, blue). 


A: This is a bit too long for a comment.
I don't think the function has any specific name. One plausible reason why this function doesn't have its own name is probably because it's growth is very closely related to $\exp(x)$ as expected. A very rough analysis as shown below gives some growth of this function with respect to $x$. The argument can be improved to get better growth rates of the function.
If we let $n = \lfloor t \rfloor$, then
$$(n+1)! = \Gamma(n+2) \geq \Gamma(t+1) \geq \Gamma(n+1) = n!$$
Hence for $t \in [n,n+1)$, assuming $x>0$, $$\dfrac{x^t}{(n+1)!} \leq \dfrac{x^{t}}{\Gamma(t+1)} \leq \dfrac{x^t}{n!}$$
Now $$\int_0^{\infty} \dfrac{x^t}{\Gamma(t+1)} dt = \sum_{n=0}^{\infty} \int_n^{n+1} \dfrac{x^t}{\Gamma(t+1)} dt$$
Hence, $$\sum_{n=0}^{\infty} \int_n^{n+1} \dfrac{x^t}{(n+1)!} dt \leq \int_0^{\infty} \dfrac{x^t}{\Gamma(t+1)} dt \leq \sum_{n=0}^{\infty} \int_n^{n+1} \dfrac{x^t}{n!} dt$$
$$\sum_{n=0}^{\infty} \dfrac{x^{n+1}-x^n}{(n+1)! \log(x)} \leq \int_0^{\infty} \dfrac{x^t}{\Gamma(t+1)} dt \leq \sum_{n=0}^{\infty} \dfrac{x^{n+1}-x^n}{n! \log(x)}$$
$$\dfrac{(x-1)}{\log(x)} \left(\sum_{n=0}^{\infty} \dfrac{x^n}{(n+1)!} \right) \leq \int_0^{\infty} \dfrac{x^t}{\Gamma(t+1)} dt \leq \dfrac{(x-1)}{\log(x)} \left(\sum_{n=0}^{\infty} \dfrac{x^n}{n!} \right)$$
$$\dfrac{(x-1)}{\log(x)} \dfrac{\exp(x)-1}{x} \leq \int_0^{\infty} \dfrac{x^t}{\Gamma(t+1)} dt \leq \dfrac{(x-1)}{\log(x)} \exp(x)$$
A: $$f(x) = \int_0^\infty \frac{x^t}{\Gamma(t+1)} \, dt$$
I have noticed that we can find Laplace transform of this function in closed form. You can define the function as inverse Laplace Transform in closed form.
$$ {\mathcal L}\left\{ {x^t} \right\}(s)=\int_0^\infty x^t e^{-sx} \, dx=\frac{\Gamma(t+1)}{s^{t+1}}$$
$$ {\mathcal L}\left\{ {f(x)} \right\}(s)=\int_0^\infty \frac{1}{\Gamma(t+1)}\int_0^\infty x^t e^{-sx} \, dx \, dt=\int_0^\infty \frac{1}{s^{t+1}} \, dt=\frac{1}{s}\int_0^\infty s^{-t} \, dt=\frac{1}{s\ln{s}}$$
UPDATE:
I am trying to find a method to define the function as series as well . I am going to post if I find a way in series methods.
My starting point for series methods:.
$$ {\mathcal L}\left\{ {f(x)e^{-x}} \right\}(s)=\frac{1}{(s+1)\ln{(s+1)}}$$
A: Almost tautological, but we can also find that
$$ f(x) = e^{x} \left( 1 - \int_{0}^{1} \frac{\Gamma(s, x)}{\Gamma(s)} \; ds \right),$$
where $\Gamma(s, x) = \int_{x}^{\infty} t^{s-1}e^{-t} \; dt$ is the incomplete gamma function. Indeed, 
$$ f'(x) = \int_{0}^{\infty} \frac{s x^{s-1}}{\Gamma(s+1)} \; ds = \int_{0}^{1} \frac{x^{s-1}}{\Gamma(s)} \; ds + f(x)$$
and thus
$$ (f(x)e^{-x})' = \int_{0}^{1} \frac{x^{s-1}e^{-x}}{\Gamma(s)} \; ds.$$
Then integrating both sides with respect to $x$ gives the desired relation.
Now we try to find its asymptotic formula. Since $\Gamma(s,x) / \Gamma(s) \searrow 0$ as $x \to \infty$, we find
$$f(x) = (1 - o(1))e^{x}$$
asymptotically. To obtain a better approximation, note that $1/\Gamma(s) \asymp s$ near $s = 0$. Thus
$$ \begin{align*}
\int_{0}^{1} \frac{\Gamma(s, x)}{\Gamma(s)} \; ds
&\asymp \int_{0}^{1} s\Gamma(s, x) \; ds \\
&= \int_{0}^{1} (\Gamma(s+1,x) - x^{s}e^{-x}) \; ds \\
&= \int_{0}^{1} \int_{x}^{\infty} t^{s}e^{-t} \; dt ds - \frac{x-1}{\log x}e^{-x} \\
&= \int_{x}^{\infty} \left( \frac{t-1}{\log t}\right) e^{-t} \; dt - \frac{x-1}{\log x}e^{-x} \\
&= \int_{x}^{\infty} \left( \frac{t-1}{\log t}\right)' e^{-t} \; dt \\
&\asymp \int_{x}^{\infty} \frac{e^{-t}}{\log t} \; dt \\
&\asymp \frac{e^{-x}}{\log x}.
\end{align*}$$
In particular, we have
$$f(x) = e^{x} - O\left( \frac{1}{\log x} \right).$$
