What is the series representation of $\int_{0}^{t} \exp(-x^n)\, dx$ where $n$ is a positive integer? I have searched in the web to get series representation of $$\int_{0}^{t} e^{-x^n}\,dx$$  where $n$ is a positive integer, but I haven’t succeeded; really I want its representation as hypergeometric series.
 A: Integrating the Power Series
We have the power series
$$
e^{-x^n}=\sum_{k=0}^\infty\frac{(-1)^kx^{kn}}{k!}\tag1
$$
which leads to the alternating power series
$$
\begin{align}
\int_0^t e^{-x^n}\,\mathrm{d}x
&=\sum_{k=0}^\infty\frac{(-1)^kt^{kn+1}}{(kn+1)k!}\\
&=\bbox[5px,border:2px solid #C0A000]{t\,{}_1F_1\!\left(\tfrac1n;1+\tfrac1n;-t^n\right)}\tag2
\end{align}
$$
Mathematica verification:
Simplify[D[t Hypergeometric1F1[1/n,1+1/n,-t^n],t]]

Unilateral Power Series
The problem with $(2)$ is that for large $t$, the terms get huge and oscillate. This means that a lot of significance can get lost in cancellation. It is better to have a unilateral series where nothing is lost to cancellation.
Setting
$$
f(t)=e^{t^n}\int_0^te^{-x^n}\,\mathrm{d}x\tag3
$$
we have
$$
f'(t)=nt^{n-1}f(t)+1\tag4
$$
which, since $f(0)=0$, says not only that
$$
f(t)=t+O\!\left(t^{n+1}\right)\tag5
$$
but also gives the coefficient recurrence
$$
a_{k+n}=\frac{n}{k+n}\,a_k\tag6
$$
From $(5)$ and $(6)$, we get
$$
a_{kn+1}=\frac{\Gamma\!\left(1+\frac1n\right)}{\Gamma\!\left(k+1+\frac1n\right)}\tag7
$$
and $a_k=0$ for $k\not\equiv1\pmod n$. This gives the unilateral series:
$$
\begin{align}
\int_0^te^{-x^n}\,\mathrm{d}x
&=e^{-t^n}\sum_{k=0}^\infty\frac{\Gamma\!\left(1+\frac1n\right)}{\Gamma\!\left(k+1+\frac1n\right)}\,t^{kn+1}\\
&=\bbox[5px,border:2px solid #C0A000]{t\,e^{-t^n}{}_1F_1\!\left(1;1+\tfrac1n;t^n\right)}\tag8
\end{align}
$$
Note that the terms in the sum in $(8)$ peak when $k\sim t^n$.
Mathematica verification:
Simplify[D[t Exp[-t^n]Hypergeometric1F1[1,1+1/n,t^n],t]]

Asymptotic Expansion
$$
\begin{align}
\int_t^\infty e^{-x^n}\,\mathrm{d}x
&=\frac1n\int_{t^n}^\infty e^{-x}x^{\frac1n-1}\,\mathrm{d}x\tag9\\
&=\frac{e^{-t^n}}n\int_0^\infty e^{-x}\left(t^n+x\right)^{\frac1n-1}\mathrm{d}x\tag{10}\\
&=\frac{e^{-t^n}t^{1-n}}n\int_0^\infty e^{-x}\left(1+\frac{x}{t^n}\right)^{\frac1n-1}\mathrm{d}x\tag{11}\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\int_0^{t^n/2}e^{-x}\left(1+\frac{x}{t^n}\right)^{\frac1n-1}\mathrm{d}x+O\!\left(e^{-t^n/2}\right)\right]\tag{12}\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\int_0^{t^n/2} e^{-x}\frac{\Gamma\left(k+1-\frac1n\right)}{k!\,\Gamma\left(1-\frac1n\right)}\frac{x^k}{t^{nk}}\mathrm{d}x+O\!\left(\frac1{t^{nm}}\right)\tag{13}\right]\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\frac{\Gamma\left(k+1-\frac1n\right)}{\Gamma\left(1-\frac1n\right)}\frac1{t^{nk}}+O\!\left(\frac1{t^{nm}}\right)\right]\tag{14}
\end{align}
$$
Explanation:
$\phantom{0}(9)$: substitute $x\mapsto x^{1/n}$
$(10)$: substitute $x\mapsto t^n+x$
$(11)$: substitute $x\mapsto x/t^n$
$(12)$: $\int_{t^n/2}^\infty e^{-x}\,\mathrm{d}x=e^{-t^n/2}$
$(13)$: Binomial Theorem
$(14)$: $\int_{t^n/2}^\infty e^{-x}x^k\,\mathrm{d}t=O\!\left(e^{-t^n/3}\right)$
Therefore,
$$
\begin{align}
\int_0^te^{-x^n}\,\mathrm{d}x
&=\Gamma\left(1+\frac1n\right)-\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\frac{\Gamma\left(k+1-\frac1n\right)}{\Gamma\left(1-\frac1n\right)}\frac1{t^{nk}}+O\!\left(\frac1{t^{nm}}\right)\right]\\
&=\bbox[5px,border:2px solid #C0A000]{\Gamma\left(1+\frac1n\right)-\frac{e^{-t^n}t^{1-n}}n\,{}_2F_0\!\left(1,1-\tfrac1n;;-t^{-n}\right)}\tag{15}
\end{align}
$$
where ${}_2F_0\!\left(1,1-\tfrac1n;;-t^{-n}\right)$ is not convergent, but is to be taken in the asymptotic sense.
A: I am unfamiliar with hypergeometric series (such as this?), but since in the comments you indicated that a Riemann sum could suffice, here is a brute-force approach:

Consider that by definition $$\int_a^b f(X) \, dX = \lim_{N\to\infty} \sum_{i=1}^N f(x_i) \, \Delta x$$ where $x_i$ could be the righthand points of $N$ subintervals $[a,b]$ each of width $\Delta x$.
For your example, $a=0$ and $b=t$. Therefore the length of the domain of integration is merely $b-a=t$, and when this is partitioned into $N$ subintervals, $\Delta x = t/N$.
The righthand $x_i$ of each Riemannian rectangle is given by $a+i\,\Delta x$, in this case $x_i = it/N$.
Hence $$\int_0^t \exp\left( -x^n\right) \, dx = \lim_{N\to\infty}\sum_{i=1}^N \exp\left[ -\left(\frac{it}{N}\right)^n\right] \frac tN$$

If you are interested in pursuing this approach further I can try to continue the evaluation and/or construct a visual if necessary.
A: The exponential power series is
$$ e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} $$
Plugging $y = -x^n$
$$ e^{-x^n} = \sum_{k=0}^{\infty} \frac{(-x^n)^k}{k!} = \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{k!} x^{k n} \right) $$
Then integrating in $x$ from $0$ to $t$ we obtain
$$\begin{align}
\int_0^t e^{-x^n} \,dx &= \int_0^t \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{k!} \,x^{k n} \right) \,dx \\
&= \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{k!} \int_0^t x^{k n} \,dx \right) \\
&= \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{k!} \frac{t^{k n + 1}}{(k n + 1)} \right) \\
&= \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{k! (k n + 1)} \,t^{k n + 1} \right)
\end{align}$$
A: Let 
$$x^n=t \implies x=y^{\frac{1}{n}}\implies dx=\frac{y^{\frac{1}{n}-1}}{n}$$ making
$$\int \exp(-x^n)\,dx=\frac 1 n \int e^{-y} y^{\frac{1}{n}-1}\,dy=-\frac 1 n \Gamma \left(\frac{1}{n},y\right)$$ and, then for the definite integral
$$\int_0^t \exp(-x^n)\,dx=-\frac{t\, \Gamma \left(\frac{1}{n},t^n\right)}{n}=-\frac{t\,E_{\frac{n-1}{n}}\left(t^n\right)}{n}$$ where appear either the incomplete gamma function or the exponential integral function.
