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Integral of $e^{x^2}$

Using the Taylor series is a straight forward method, but let us try to approach the problem in a different manner.

Integration by Parts

NOTE: This method requires working knowledge of integration by parts.

When I tried to solve this problem, I first started to expand it using the integration by parts method. The pattern I observed was the thing which helped me put myself on the right track to solve this problem. Here is the pattern:
$$\int e^{x^2}dx = xe^{x^2} - 2\int x^2e^{x^2}dx$$
$$3\int x^2e^{x^2}dx = x^3e^{x^2} - 2\int x^4e^{x^2}dx$$
$$5\int x^4e^{x^2}dx = x^5e^{x^2} - 2\int x^6e^{x^2}dx$$
To make sure this pattern sustains, I decided to find the integral of $x^{2n}e^{x^2}$.
$$Y = \int x^{2n}e^{x^2}dx$$
$$Y = x^{2n+1}e^{x^2} - \int x[2nx^{2n-1}e^{x^2}+2x^{2n+1}e^{x^2}]dx$$
$$(2n+1)Y = x^{2n+1}e^{x^2}-2\int x^{2n+2}e^{x^2}dx$$
$$\therefore Y = \frac{x^{2n+1}e^{x^2}}{2n+1}-\frac{2}{2n+1}\int x^{2n+2}e^{x^2}dx$$
Note that the $\int x^{2n+2}e^{x^2}dx$ is similar to Y, with $n$ replaced by $n+1$. Expanding according to the above,
$$Y = \frac{x^{2n+1}e^{x^2}}{(2n+1)}-\frac{2^1x^{2n+3}e^{x^2}}{(2n+1)(2n+3)}+\frac{2^2x^{2n+5}e^{x^2}}{(2n+1)(2n+3)(2n+5)}-...$$
Placing $n=0$ since we want $\int e^{x^2}$,
$$Y_{n=0} = \frac{2^0x^1e^{x^2}}{1} -\frac{2^1x^3e^{x^2}}{1.3} +\frac{2^2x^5e^{x^2}}{1.3.5}-\frac{2^3x^7e^{x^2}}{1.3.5.7} +...$$
$$Y_{n=0}=\sum_{i=1}^{\infty} (-1)^{i+1}\frac{2^{i-1}x^{2i-1}e^{x^2}}{1.3.5...(2i-1)}$$

Integration using the $\operatorname{erf}$ function

As stated in The $\operatorname{erf}$ function, one can use it to find the integral of $e^{x^2}$, as well as $e^{-x^2}$.
Since $\int_0^x{\mathrm e}^{-t^2}{\mathrm d}t = \sqrt{\frac{\pi}2}\operatorname{erf}(x)$,
$$\int_0^x{\mathrm e}^{t^2}{\mathrm d}t = -i\sqrt{\frac{\pi}2}\operatorname{erf}(ix)$$
You can proceed to use the various series expansion of $\operatorname{erf}(x)$ to complete this problem.

Footnote: If you have an another method to solve this particular problem through which we can obtain the integral in a more creative and easier manner, please contact any of the writers in this blog.
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