integral: i = ∫₀¹ e^(x²) dx

Explanation:

Step1: Recall error - function definition

The antiderivative of $e^{x^{2}}$ is not an elementary function. We use the error - function $\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}dt$. But for $\int e^{x^{2}}dx$, we can use power - series expansion. The power series of $e^{u}=\sum_{n = 0}^{\infty}\frac{u^{n}}{n!}$. Let $u = x^{2}$, then $e^{x^{2}}=\sum_{n=0}^{\infty}\frac{(x^{2})^{n}}{n!}=\sum_{n = 0}^{\infty}\frac{x^{2n}}{n!}$.

Step2: Integrate the power - series term - by - term

$\int_{0}^{1}e^{x^{2}}dx=\int_{0}^{1}\sum_{n = 0}^{\infty}\frac{x^{2n}}{n!}dx$. By the uniform convergence of power - series on a closed interval $[0,1]$, we can interchange the integral and the sum: $\sum_{n = 0}^{\infty}\frac{1}{n!}\int_{0}^{1}x^{2n}dx$.

Step3: Calculate the integral of $x^{2n}$

We know that $\int_{0}^{1}x^{2n}dx=\left[\frac{x^{2n + 1}}{2n+1}\right]_{0}^{1}=\frac{1}{2n + 1}$.

Step4: Find the sum

So $\int_{0}^{1}e^{x^{2}}dx=\sum_{n=0}^{\infty}\frac{1}{n!(2n + 1)}=1+\frac{1}{3}+\frac{1}{10}+\frac{1}{42}+\cdots$. Evaluating the sum: [ \begin{align*} \sum_{n = 0}^{\infty}\frac{1}{n!(2n+1)}&= 1+\frac{1}{3\times1!}+\frac{1}{5\times2!}+\frac{1}{7\times3!}+\cdots\ &\approx1.46265 \end{align*} ]

Answer:

$\sum_{n=0}^{\infty}\frac{1}{n!(2n + 1)}\approx1.46265$