Evaluate the integral $\int_0^1 e^{-x^{4}}(1-x^{4}) dx $ How to find the definite integral of  $$\int_0^1 e^{-x^{4}}(1-x^{4})dx $$ I tried solving this by using integration by parts and then by substitution but want able to solve this by either of those methods .
 A: this integral can not expressed by the known elementary functions.
the result is given by $$\frac{4+3 e \Gamma \left(\frac{1}{4}\right)-3 e \Gamma \left(\frac{1}{4},1\right)}{16 e}
$$
A: By parts,
$$\int x^4e^{-x^4}dx=-\frac x4e^{-x^4}+\frac14\int e^{-x^4}dx$$ and on both terms you are left with
$$\int_0^1 e^{-x^4}dx$$ for which there is no analytical antiderivative.
It requires the incomplete Gamma function, or can be evaluated by the fast converging series
$$\sum_{k=0}^\infty\frac{(-1)^k}{(4k+1)k!}.$$
A: Let $y=x^{4}$
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{-3/4} dy
= \frac{1}{4} \gamma\left(\frac{1}{4},1 \right)
\end{equation}
Using the same substitution, we also have
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} x^{4} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{1/4} dy
= \frac{1}{4} \gamma\left(\frac{5}{4},1 \right)
\end{equation}
Thus we obtain
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx
= \frac{1}{4} \Big[ \gamma\left(\frac{1}{4},1 \right) - \gamma\left(\frac{5}{4},1 \right) \Big]
\approx 0.7256
\end{equation}
We have used the lower incomplete gamma function:
\begin{equation}
\gamma(s,z) = \int\limits_{0}^{z} \mathrm{e}^{-x} x^{s-1} dx
\end{equation}
