# Solving Integrals for e^-ax^2: (i), (ii) & (iii)

• sanitykey
In summary, To solve the given integrals, the first step is to use differentiation with respect to the parameter 'a' for (i) and (iii), and integration by parts for (ii). Then, using the given expression, we can solve for the integrals in terms of the parameter 'a'. The final answers are \frac{\sqrt{\pi}}{4(\sqrt{a})^3} for (i), \frac{1}{2a} for (ii), and \frac{3\sqrt{\pi}}{16(\sqrt{a})^5} for (iii).
sanitykey
Hi, i have a problem which is confusing me

Question:

Given that

$$\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}$$

What is

(i) $$\int_{0}^{\infty}e^{-ax^2} x^2 dx$$
(ii) $$\int_{0}^{\infty}e^{-ax^2} x^3 dx$$
(iii) $$\int_{0}^{\infty}e^{-ax^2} x^4 dx$$

It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)

I tried (ii):

$$u=x^3$$

$$\frac{du}{dx} = 3x^2$$

$$\frac{dv}{dx} = e^{-ax^2}$$

$$v=\frac{\sqrt{\pi}}{2\sqrt{a}}$$

$$\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \int_{0}^{\infty}\frac{3x^2 \sqrt{\pi}}{2\sqrt{a}}$$

$$= \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} = 0$$

I know limits should be in that last bit and the bit before but i thought it wouldn't matter as both parts are the same, but then i thought i can't say $v=\frac{\sqrt{\pi}}{2\sqrt{a}}$ anyway because the given expression wasn't general it was between limits, or does it not matter because I'm using the same limits?

I don't know what the question means when it tells me to "use differentiation for (i) and (iii) with respect to the a paramter" how will that help me get to a solution?

Last edited:
What happened to the exponential after you integrated by parts? And have you tried their suggestion for i and iii? You can move the d/da inside the integral (justifying this is a little tricky, but I'm sure you don't need to worry about that).

You can't say $v=\frac{\sqrt{\pi}}{2\sqrt{a}}$ for the part inside the integral

sanitykey said:
Hi, i have a problem which is confusing me

Question:

Given that

$$\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}$$

What is

(i) $$\int_{0}^{\infty}e^{-ax^2} x^2 dx$$
(ii) $$\int_{0}^{\infty}e^{-ax^2} x^3 dx$$
(iii) $$\int_{0}^{\infty}e^{-ax^2} x^4 dx$$

It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)

I tried (ii):

$$u=x^3$$

$$\frac{du}{dx} = 3x^2$$

$$\frac{dv}{dx} = e^{-ax^2}$$

$$v=\frac{\sqrt{\pi}}{2\sqrt{a}}$$
This is the definite integral from $-\infty$ to $\infty$, not the anti-derivative! $e^{-ax^2}$ has no simple anti-derivative. Try $u= x^2$, $dv= xe^{-ax^2}dx$ instead.

$$\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \int_{0}^{\infty}\frac{3x^2 \sqrt{\pi}}{2\sqrt{a}}$$

$$= \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} = 0$$

I know limits should be in that last bit and the bit before but i thought it wouldn't matter as both parts are the same, but then i thought i can't say $v=\frac{\sqrt{\pi}}{2\sqrt{a}}$ anyway because the given expression wasn't general it was between limits, or does it not matter because I'm using the same limits?

I don't know what the question means when it tells me to "use differentiation for (i) and (iii) with respect to the a paramter" how will that help me get to a solution?

sanitykey said:
Hi, i have a problem which is confusing me

Question:

Given that

$$\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}$$

What is

(i) $$\int_{0}^{\infty}e^{-ax^2} x^2 dx$$
(ii) $$\int_{0}^{\infty}e^{-ax^2} x^3 dx$$
(iii) $$\int_{0}^{\infty}e^{-ax^2} x^4 dx$$

It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)

For (ii) compute first

$$\int_{0}^{\infty} e^{-ax^2} \ x \ dx$$

And when you get the result, you can differentiate wrt "a" to get the needed integral.

HINT:Make the sub x^{2}=t and of course assume a>0.

Daniel.
(

Thanks for all these replies! :D

I would of responded sooner but i thought i'd best come back with something to show I've been quite busy.

I'll just show you what my answers are not sure if they're right but if they are credit to you all for helping me.

$$\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}$$

$$\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2}\right) dx = \frac{d}{da}\left(\frac{\sqrt{\pi}}{2\sqrt{a}}\right)$$

$$\int_{0}^{\infty}x^2 e^{-ax^2} dx = \frac{\sqrt{\pi}}{4(\sqrt{a})^3}$$

$$\int_{0}^{\infty}e^{-ax^2} x dx = \frac{1}{2a}$$

$$\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2} x \right) dx = \frac{d}{da}\left(\frac{1}{2a}\right)$$

$$\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{1}{2a^2}$$

$$\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2} x^2 \right) dx = \frac{d}{da}\left(\frac{\sqrt{\pi}}{4(\sqrt{a})^3}\right)$$

$$\int_{0}^{\infty}e^{-ax^2} x^4 dx = \frac{3\sqrt{\pi}}{8(\sqrt{a})^5}$$

Last edited:
You're missing a half in the last line, otherwise that looks fine.

Ok edited to correct that thank you :)

## What is the basic concept behind solving integrals for e^-ax^2?

The basic concept behind solving integrals for e^-ax^2 is to use a technique called integration by substitution. This involves replacing the variable in the integral with a new variable, and then solving the integral in terms of the new variable.

## What is the general formula for solving integrals for e^-ax^2?

The general formula for solving integrals for e^-ax^2 is ∫ e^-ax^2 dx = √π/a, where a is a constant. This formula can be derived using the integration by substitution technique.

## What are the steps to solve integrals for e^-ax^2?

The steps to solve integrals for e^-ax^2 are as follows:

1. Replace the variable in the integral with a new variable, typically u.
2. Use the chain rule to find the derivative of u with respect to x.
3. Substitute the value of u and du/dx into the integral, and simplify as much as possible.
4. Integrate the resulting expression in terms of u.
5. Substitute the original variable back into the solution to get the final answer.

## Can integrals for e^-ax^2 be solved using other techniques?

Yes, integrals for e^-ax^2 can also be solved using other techniques such as integration by parts or completing the square. However, the integration by substitution technique is generally the most efficient and straightforward method for these types of integrals.

## What are the real-life applications of solving integrals for e^-ax^2?

Solving integrals for e^-ax^2 has many real-life applications, particularly in physics and engineering. It can be used to calculate the probability of a particle's position in quantum mechanics, or to model the decay of radioactive materials. It is also used in the study of heat transfer and diffusion processes in materials.

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