# Integrals/Non-Elementary Antiderivatives

• leonardthecow
In summary, I am not sure how to solve the following definite integral, \int^\infty_{-\infty}{x^4e^{-x^2}dx}.I understand how to evaluate the definite "Gaussian" integral $$\int^\infty_{-\infty}{e^{-x^2}}=\sqrt \pi$$ using a switch to polar coordinates and a u sub, but not sure if/how that applies here.Extending the question, I'm also wondering about an integral like $$\int^\infty_{-\infty}{e^{ikx}e^{-x^2}}$$.
leonardthecow
Hey everyone,

I'm wondering how to solve the following definite integral,

$\int^\infty_{-\infty}{x^4e^{-x^2}dx}$.

I know the answer is ##\frac{3 \sqrt{\pi}}{4}##, but I'm not positive how to get there.

I understand how to evaluate the definite "Gaussian" integral $$\int^\infty_{-\infty}{e^{-x^2}}=\sqrt \pi$$ using a switch to polar coordinates and a u sub, but not sure if/how that applies here.

Extending the question, I'm also wondering about an integral like $$\int^\infty_{-\infty}{e^{ikx}e^{-x^2}}$$.

Thanks!

Try integration by parts for the first one. Try completing the square for the second one.

So I tried parts for the first one, and assuming I didn't mess up somewhere, if I choose u to be the exponential term then I end up with $$e^{-x^2}\frac{x^5}{5} - \int^{\infty}_{-\infty}\frac{x^5}{5}(-2xe^{-x^2})$$, with the first term being evaluated from minus infinity to infinity as well. Wouldn't I just have to keep doing partial fractions, this way, and it would just keep going indefinitely?

Conversely, if I chose u to be the x term, then I think I'd get $$x^4 \sqrt{\pi} - \int^{\infty}_{-\infty}\sqrt{\pi}4x^3$$, with the first term being evaluated from x is minus infinity to positive infinity and the ##\sqrt{\pi}## coming from the exponential term being integrated over all space. Doing this, then wouldn't you just get 0?

What happens if you use integration by parts on

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

I do not like the idea of integration by parts here. Step by step

1. Instead solve a slightly more difficult but related problem definite integral (x squared exp - a times x squared).
2. Call this definite integral I(a).
3. Take the derivative with respect to a (not x) of the integrand inside the integral. Here you may note you end up multiplying the integrand by a times x squared.
4. Equate this integral with dI(a) / da.
5. do step 3 to the previous integral expression to get a x fourth power in the integrand.
6 equate this integral to d2I / da2.
7. Plug in a = 1.
8 This will give you the answer.

Correction: Step 3 you end up multiplying the integrand by -x squared not a times x squared, but you get the idea.

Step 1 can be obtained with the change of variables x = sqrt(a) * y. then when you get the result later substitute in x for y to preserve the notation in terms of x not y. (this is optional)

Sorry step 1 is still wrong. Start with definite int( exp-x squared ) = sqrt(pi). Change variables y = sqrt(a) times x to eventually get I(a) = definte int( -x squared times exp (- x sqared)).

Step 4 should be calculate dI(a)/da and equate coefficient (take the derivative)
Step 6 is similar.

In the long run you change variables and take two derivatives. You do not even have to evaluate an integral (i.e. you already have all you need as long as you do not consider changing variables in step 1 to be evaluating an integral)

Sorry step 1 is still wrong. Start with definite int( exp-x squared ) = sqrt(pi). Change variables y = sqrt(a) times x to eventually get I(a) = definte int( -x squared times exp (- a times x sqared)).

Integration by parts also gives the answer.

Notice that the integral can be written as

$$\int_{-\infty}^{\infty} x^3(xe^{-x^2})\,dx$$
Can you see how to use by parts here? :)

To: Pranav-Arora

Very good. It still takes several steps but it can be integrated by parts without the dodge of derivating inside the integral with respect to a and introducing the a in the exponent.

Better idea is to write your integral: Def integral x (x cubed times x squared exp(- x squared) ). Then substitute y in for x squared.

Internet kicked me off.

Correction:

Better idea is to write your integral: Def integral x times (x cubed times exp(- x squared) ). Then substitute y in for x squared.

## 1. What is an integral and why is it important?

An integral is a mathematical concept that represents the area under a curve on a graph. It is important because it allows us to find the total value or the accumulated value of a certain quantity, such as distance or volume, over a specified interval.

## 2. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, which determine the starting and ending points for finding the area under the curve. An indefinite integral does not have limits of integration and represents a family of curves, giving us the general form of the antiderivative.

## 3. What is a non-elementary antiderivative?

A non-elementary antiderivative is an integral that cannot be expressed in terms of elementary functions such as polynomials, exponential functions, and trigonometric functions. These integrals often require more advanced techniques, such as substitution or integration by parts, to solve.

## 4. How can I determine if a function has a non-elementary antiderivative?

There is no simple method to determine if a function has a non-elementary antiderivative. However, we can use the Risch algorithm, which is a set of rules and algorithms for determining if a function has a closed-form solution in terms of elementary functions. If the Risch algorithm fails to find a solution, then the function likely has a non-elementary antiderivative.

## 5. Can non-elementary antiderivatives be evaluated numerically?

Yes, non-elementary antiderivatives can be evaluated numerically using approximation methods such as the trapezoidal rule, Simpson's rule, or the Monte Carlo method. These methods involve breaking the integral into smaller parts and using numerical techniques to approximate the area under the curve.

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