Skip to main content

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.
Labels:

Comments

Post a Comment

Popular posts from this blog

Factorials, and the Gamma Function

What is the Gamma function? What does it have to do with the factorial of a number? And why does such a formula even work? The Factorial In our high school combinatorics course, we were introduced to the factorial of a natural number. We defined the factorial of a number $n$ to be the product of all the natural numbers till $n$. Or symbolically: $$n! = 1 . 2 . 3 ... (n-2)  (n-1)  n$$ This is a very useful function in Counting problems. This function also pops up in the exponential function: $$e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$ Setting this aside, we think: can we 'extend' this function to all real number? Surely, even if we were to come up with such a function, what would it mean to take this new 'factorial' of some real number? Well, let's take a look at one important property of the factorial function: $$n! = n(n-1)!$$ and this is true for all natural numbers $n$. So, even if we came up with this homologous function, we...
Post a Comment
Read more

Why $y^2 = x^3 + 7$ has no integral solutions?

Consider the given equation... $$y^2-7=x^3=(x^{\frac{3}{2}})^2$$ $$y^2-(x^{\frac{3}{2}})^2=7$$ If $x^{\frac{3}{2}}$ is a decimal, then $y$ should also be some decimal to make the equation an integer, i.e., $7$. So $y$ will be a decimal in this case. But consider when $x^{\frac{3}{2}}$ is an integer. The only integers whose difference of squares is $7$, is $4$ and $3$ (why?). $$4^2 - 3^2 =7$$ But compare it to the equation $y^2-(x^{\frac{3}{2}})^2=7$. Then $$x^{\frac{3}{2}} = 3$$ $$x = 3^{\frac{2}{3}}$$ But here $x$ is a decimal! Therefore, at least one of the variables $x$ and $y$ is a decimal in the given equation, thus proving the fact that it has no integral solutions.
Post a Comment
Read more

The $\operatorname{erf}$ function

Can a function be 'defined' as the anti-derivative of another function? $\operatorname{erf}$ is one such function. Even though it has various proper infinite series expansions (as in Wolfram MathWorld - Erf ), it is defined by mathematicians as such: $$\operatorname{erf}(z) \equiv \frac{2}{\sqrt{\pi}}\int_0^ze^{-t^2}dt$$ This function has extensive implications in statistics, and can be used to express the integral of $e^{-x^2}$ and $e^{x^2}$. See the post that tries to find  integral of $e^{x^2}$  - we need not use integration by parts there, but rather can use the $\operatorname{erf}$ function. Source: Wolfram MathWorld
Post a Comment
Read more