Definite integral into indefinitie series

Question

Convert $\displaystyle \int_0^1 e^{x^2}\, dx$ to an infinite series.

Answer 1

$$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$

$$e^{x^2}=\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$$

$$\int_0^1 e^{x^2} dx=\int_0^1 \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}dx=\sum_{n=0}^{\infty} \frac{1}{n!} \left [\frac{x^{2n+1}}{2n+1} \right ]_0^1=\sum_{n=0}^{\infty} \frac{1}{n!} \cdot \frac{1}{2n+1}=\sum_{n=0}^{\infty} \frac{1}{n!(2n+1)}$$

Answer 2

As $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$ $$\int_0^1e^{x^2}dx\\\eqalign{ &=\int_0^1\left(1+x^2+\frac{(x^2)^2}{2!}+\frac{(x^2)^3}{3!}+\cdots\right) dx\\ &=\int_0^1\left(1+x^2+\frac{(x^4)}{2!}+\frac{(x^6)}{3!}+\cdots\right)dx\\ &=\left(x+\frac{x^3}{3.1!}\frac{x^5}{(5.2)}+\frac{x^7}{(7.3!)}+\cdots\right)_0^1\\ &=\frac1{1.0!}+\frac1{3.1!}+\frac1{(5.2!)}+\frac1{(7.3!)}+...\\ &=\sum_{k=0}^{\infty}\frac1{k!(2k+1)}}$$