# Question about the symmetry of integrals

• LateToTheParty
In summary, the question is about the behavior of an integral when a change is made to the function being integrated over all space. The first case, with x in front, results in a separable integral that evaluates to zero, while the second case, with z in front, also evaluates to zero but without being separable. The question is why this arbitrary change affects the behavior of the integral.
LateToTheParty
O.K. , this question is inspired by a physics class I'm taking where we're working out the expectation values of wave functions, but I think the question really belongs in the math section. Thank you in advance for any help. Here goes nothing...
We have a function ψ(x,y,z) = x e$\sqrt{}x2 + y2+ z2$
Now I want to integrate this over all space. So I switch to spherical polar include the Jacobian and I end up with a separable integral of sinθ cubed form 0 to pi which is zero. No problem. But here's where the question pops up. I go to calculate the same function except this time with a z in front of the exponential, and I get a different answer, because I no longer get the sin^3 when I switch to spherical. It just seems really odd to me. I apologize if my formatting leaves something to be desired I'm new to the site, and any help would be greatly appreciated. Any insight on why a arbitrary change would affect the behavior of the integral?

In the first case(x in front),the integral is:
$\int_0^\infty \int_0^{2\pi} \int_0^\pi r\sin{\theta}\cos{\varphi} e^r r^2\sin{\theta}d\theta d\varphi dr=\int_0^\infty r^3 e^r dr \int_0^{2\pi} \cos\varphi d\varphi \int_0^\pi \sin^2{\theta} d\theta$
The $\varphi$ integral is equal to zero so the answer is zero.
In the second case(z in front),the integral is:
$\int_0^\infty \int_0^{2\pi} \int_0^\pi r\cos{\theta} e^r r^2\sin{\theta}d\theta d\varphi dr= \int_0^\infty r^3 e^r dr \int_0^{2\pi} d\varphi \int_0^\pi \cos{\theta}\sin{\theta}d\theta$
Now the $\theta$ integral is zero and so again the answer is zero.

## 1. What is the significance of symmetry in integrals?

Symmetry in integrals refers to the property of a function or equation remaining unchanged when certain transformations, such as reflection or rotation, are applied to it. In the context of integrals, symmetry allows for simplification of the integration process and can also help in solving more complex integrals.

## 2. How do you determine the symmetry of an integral?

The symmetry of an integral can be determined by checking if the function or equation is unchanged when certain transformations are applied. For example, if the integrand is an even function, meaning it is symmetric about the y-axis, the integral will be symmetric as well. Similarly, if the integrand is an odd function, meaning it is symmetric about the origin, the integral will also be symmetric.

## 3. Can an integral have both even and odd symmetry?

No, an integral cannot have both even and odd symmetry. If an integral has even symmetry, it means the function being integrated is even and if it has odd symmetry, the function is odd. Therefore, an integral can only have one type of symmetry at a time.

## 4. How does symmetry affect the limits of integration in an integral?

Symmetry can affect the limits of integration in an integral by allowing for simplification of the limits. For example, if the integrand is an even function, the limits of integration can be changed from -a and a to 0 and a, since the area under the curve will be the same for both intervals due to symmetry.

## 5. Are there any real-life applications of symmetry in integrals?

Yes, symmetry in integrals has many real-life applications. One common application is in physics, where symmetry allows for simplification of mathematical models and equations for physical systems. Symmetry is also used in engineering, economics, and other fields where integrals are used to solve problems.

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