# 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 .

• Substitute x^4=t – kayak Dec 4 '16 at 9:31
• @Heonji Ha I've tried that . It didn't help me at all – Tejus Dec 4 '16 at 9:32

As said, there is not elementary solution.

However, you can use series expansions to get, for example, $$e^{-x^{4}}(1-x^{4})=1-2 x^4+\frac{3 x^8}{2}-\frac{2 x^{12}}{3}+\frac{5 x^{16}}{24}-\frac{x^{20}}{20}+O\left(x^{22}\right)$$ and integrate each term.

Let us call $J_n$ the value of the integral up to $O\left(x^{4n+1}\right)$. In this case, you would obtain the following values $$\left( \begin{array}{ccc} n & J_n & \approx \\ 1 & \frac{3}{5} & 0.600000 \\ 2 & \frac{23}{30} & 0.766667 \\ 3 & \frac{93}{130} & 0.715385 \\ 4 & \frac{19297}{26520} & 0.727640 \\ 5 & \frac{44879}{61880} & 0.725259 \\ 6 & \frac{20206379}{27846000} & 0.725647 \\ 7 & \frac{195313597}{269178000} & 0.725593 \\ 8 & \frac{2455393823}{3383952000} & 0.725599 \end{array} \right)$$

Edit

What is good using Yves Daoust's answer is that we can a priori know how many terms have to be added in the expansion for a given number $p$ of accurate digits for the final result.

Since the series ia alternating, the remainder is smaller that the first negelected term. So, if we add $n$ terms the remainder if such that $$R_n=\frac 1 {(4n+5)(n+1)!}$$ what we want to be smaller than $10^{-p}$. This write $${(4n+5)(n+1)!}>10^{p}$$ and a rough approximation gives $$n=2.33676 \,p^{0.724341}-1.70173$$

• Hi Claude. Integrating by parts, you can get rid ot the $x^4$ term, which simplifies the process a little. – Yves Daoust Dec 4 '16 at 10:14
• @YvesDaoust. Perfectly correct (once more !). This starts to be tiring !! (joke). Cheers. – Claude Leibovici Dec 4 '16 at 10:52
• sorry, I cannot work without simplifying things first, so that I can understand them ;-) – Yves Daoust Dec 4 '16 at 10:55

this integral can not expressed by the known elementary functions. the result is given by $$\frac{4+3 e \Gamma \left(\frac{1}{4}\right)-3 e \Gamma \left(\frac{1}{4},1\right)}{16 e}$$

• I am feeling confused here . I can't understand any of this and is that result a. Definite value ? – Tejus Dec 4 '16 at 9:34
• $$\Gamma$$ is the so called gamma function – Dr. Sonnhard Graubner Dec 4 '16 at 9:35
• So there is no other way to calculate the integral ?? – Tejus Dec 4 '16 at 9:36
• Any other way either leads to the same answer or is wrong. – Ivan Neretin Dec 4 '16 at 9:57

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

$$\sum_{k=0}^\infty\frac{(-1)^k}{(4k+1)k!}.$$

Let $y=x^{4}$ $$\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{-3/4} dy = \frac{1}{4} \gamma\left(\frac{1}{4},1 \right)$$

Using the same substitution, we also have $$\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} x^{4} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{1/4} dy = \frac{1}{4} \gamma\left(\frac{5}{4},1 \right)$$

Thus we obtain $$\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx = \frac{1}{4} \Big[ \gamma\left(\frac{1}{4},1 \right) - \gamma\left(\frac{5}{4},1 \right) \Big] \approx 0.7256$$

We have used the lower incomplete gamma function: $$\gamma(s,z) = \int\limits_{0}^{z} \mathrm{e}^{-x} x^{s-1} dx$$