# Integrating exp(x^2): Solution & Explanation

• Wishe Deom
In summary: So, the last equation says that the integral is equal to the area of a circle with radius r centered at the origin and angle \theta.
Wishe Deom
Hello everyone,

This is my first post on the forum; I'm pretty sure my question fits in this section.

I am having a lot of difficulty finding the definite integral
$$\int^{+\infty}_{-\infty}e^{-2ax^{2}}$$ dx
where a is positive and real.
I know the answer is $$\\sqrt{\frac{\pi}{2a}}, but I have no idea at all how to get there. hint: write another integral of that form, but with y as the variable. multiply them together. convert to polar coordinates, and integrate. then take the square root. I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products. Is this what you are suggesting I try? [tex]\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy [tex] \int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy$$

(most of the time)

and yes, I'm suggesting you look at:

$$\sqrt{\int \int e^{-\alpha x^2} e^{-\alpha y^2} dxdy$$

As a bonus, once you convince yourself that
$$\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{{\frac{\pi}{\alpha}}$$,
you can easily calculate integrals of the form
$$\int_{-\infty}^{\infty}x^{2n} e^{-\alpha x^2} dx$$
by differentiating the first expression (under the integral sign, on the left) with respect to alpha n times. This is a very useful fact.

Wishe Deom said:
I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

To justify the interchange you can use "[URL Theorem[/URL].

Last edited by a moderator:
Thank you for all the help. I'm well on my way, but I am having trouble with the limits of integration. Since x and y go from -inf to inf, what are the new limits?

Since r = sqrt(x^2+y^2), it seems that r goes from inf to inf, which would result in 0. I know, logically, r should be 0 to infinity, but how do I calculate that?

Wishe Deom said:
I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

Is this what you are suggesting I try?

$$\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy fluxions said: [tex] \int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy$$

(most of the time)
This is "Fubini's Theorem"

1) $$\int_{-\infty}^\infty e^{-2ax^2}dx= 2\int_0^\infty e^{-2ax^2}dx [/itex] by symmetry. 2) If you let [tex]I= \int_{-\infty}^\infty} e^{-2ax^2}dx$$
then $$\frac{I}{2}= \int_0^\infty e^{-2ax^2}dx$$
and it is also true that
$$\frac{I}{2}= \int_0^\infty e^{-2ay^2}dy$$
since that is the same integral with a different "dummy" variable.

so
3) $$I^2/4= \left(\int_0^\infty e^{-2ax^2}dx\right)\left(\int_0^\infty e^{-2ay^2}dy\right)$$
$$= \int_0^\infty \int_0^\infty e^{-2a(x^2+ y^2)} dy dx$$
by Fubini's theorem.

You can interpret that last as a double integral over the first quadrant and change to polar coordinates. Remember that the "differential of area" in polar coordinates is $r drd\theta$ and that $\theta$ goes from 0 to $\pi/2$ in the first quadrant.

## 1. What is the formula for integrating exp(x^2)?

The formula for integrating exp(x^2) is ∫exp(x^2)dx = √π/2 * erf(x).

## 2. How do you solve an integral with exp(x^2)?

To solve an integral with exp(x^2), you can use the formula ∫exp(x^2)dx = √π/2 * erf(x) or use the substitution method by letting u = x^2. You can also use special techniques such as the Gaussian integral or the Laplace transform.

## 3. What is the significance of the solution to integrating exp(x^2)?

The solution to integrating exp(x^2) is significant because it is used to solve many important mathematical problems in fields such as statistics, physics, and engineering. It is also a special case of the error function, which has many practical applications in probability and statistics.

## 4. Can you provide an example of integrating exp(x^2)?

One example of integrating exp(x^2) is solving the integral ∫exp(x^2)dx from 0 to 1. Using the formula, we get ∫exp(x^2)dx = √π/2 * [erf(1) - erf(0)] = √π/2 * [0.8427 - 0] = √π/2 * 0.8427 ≈ 0.8862.

## 5. Are there any special properties of integrating exp(x^2)?

Yes, there are several special properties of integrating exp(x^2). One is that the integral of exp(x^2) is an entire function, meaning it is analytic and has no singularities. Another property is that the integral is not expressible in terms of elementary functions, making it a special function. Additionally, the integral evaluates to the square root of π over 2, which is a constant value.

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