Regarding center of mass of an infinite area

[Anixx]

TL;DR

How does one find the projection of center of mass of an infinite area on the Y axis?

Regarding finding centers of mass of infinite figures, how one can show that

$$
\int_{-\infty}^\infty \left(\frac1{x^2}-\cos \frac1x\right)dx=\pi
$$

for instance, and other similar integrals, like the following?

$$
\int_0^\infty (x^2-\frac6{x^4})dx=0
$$

 

[Mark44]

Summary:: How does one find the projection of center of mass of an infinite area on the Y axis?

The projection onto the y-axis of the center of mass would simply be the y-coordinate of the C.M.

Regarding finding centers of mass of infinite figures, how one can show that

$$
\int_{-\infty}^\infty \left(\frac1{x^2}-\cos \frac1x\right)dx=\pi
$$

for instance, and other similar integrals, like the following?

$$
\int_0^\infty (x^2-\frac6{x^4})dx=0
$$

Are you asking how to evaluate these two integrals or similar one? What do they have to do with the center of mass?
I don't know how to evaluate the first integral, off the top of my head, but the second one should be relatively easy to do.

 

[mathman]

$\int_0^\infty(x^2-\frac{6}{x^4})dx$ is the sum of two divergent integrals - diverging at opposite ends of the domain. Makes no sense!

 

[Anixx]

Are you asking how to evaluate these two integrals or similar one? What do they have to do with the center of mass?

If the integral is zero, then the center of mass of the figure is on the X-axis.

 

[Anixx]

$\int_0^\infty(x^2-\frac{6}{x^4})dx$ is the sum of two divergent integrals - diverging at opposite ends of the domain.

That's why I am asking about infinite areas. This exact integral is zero though. At least the Laplace transform points in this direction.

 

[PeroK]

That's why I am asking about infinite areas. This exact integral is zero though. At least the Laplace transform points in this direction.

Try evaluating $$\int_{b/a}^a (x^2 - \frac{6}{x^4}) \ dx$$ where $b = \sqrt[3]{6}$