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I am currently working on a paper on this topic, but I can give you a link to my Wiki page where I summed up the main ideas and results.

In short, any divergent integral or series can be interpreted as an extended (infinite) number, in which the regularized value is the regular part in a similar way as in complex numbers the real part is the part of the complex number.

In other words the expression $(\infty)!$ means something like a mean value of the factorial (from zero to infinity), but since it is infinite, it is not a normal real number, but an extended number, albeit with real("finite") part $\sqrt{2\pi}$. It is some kind of $\lambda+\sqrt{2\pi}$ where $\lambda$ is some purely irregular number.

As such, the answer to your third question is, you can divide the both sides by a real number and the regularized value will remain correct. Notice though that taking only odd or even numbers in the left part is not equal to dividing it by 2.

On the other hand, you cannot invert (1/x) or exponentiate the both sides without taking into accout the irregular part.

So if you want the regularization to remain valid after you do any operations on the left part, you have to find a closed-form expression for the irregular part ($\lambda$), and depending on its properties apply the required operation to the sum $\lambda+\sqrt{2\pi}$. Then you would have to find the regular part of the result.

I am currently working on a paper on this topic, but I can give you a link to my Wiki page where I summed up the main ideas and results.

In short, any divergent integral or series can be interpreted as an extended (infinite) number, in which the regularized value is the regular part in a similar way as in complex numbers the real part is the part of the complex number.

In other words the expression $(\infty)!$ means something like a mean value of the factorial (from zero to infinity), but since it is infinite, it is not a normal real number, but an extended number, albeit with real("finite") part $\sqrt{2\pi}$. It is some kind of $\lambda+\sqrt{2\pi}$ where $\lambda$ is some purely irregular number.

Speaking with more precise language, all the regularization techniques are represented by linear operators.

As such, the answer to your third question is, you can divide the both sides by a real number and the regularized value will remain correct. Notice though that taking only odd or even numbers in the left part is not equal to dividing it by 2.

On the other hand, you cannot invert (1/x) or exponentiate the both sides without taking into accout the irregular part.

So if you want the regularization to remain valid after you do any operations on the left part, you have to find a closed-form expression for the irregular part ($\lambda$), and depending on its properties apply the required operation to the sum $\lambda+\sqrt{2\pi}$. Then you would have to find the regular part of the result.

I am currently working on a paper on this topic, but I can give you a link to my Wiki page where I summed up the main ideas and results.

In short, any divergent integral or series can be interpreted as an extended (infinite) number, in which the regularized value is the regular part in a similar way as in complex numbers the real part is the part of the complex number.

In other words the expression $(\infty)!$ means something like a mean value of the factorial (from zero to infinity), but since it is infinite, it is not a normal real number, but an extended number, albeit with real("finite") part $\sqrt{2\pi}$. It is some kind of $\lambda+\sqrt{2\pi}$ where $\lambda$ is some purely irregular number.

As such, the answer to your third question is yes, you can divide the both sides by a real number and the regularized value will remain correct.

On the other hand, you cannot invert (1/x) or exponentiate the both sides without taking into accout the irregular part.

So if you want the regularization to remain valid after you do any operations on the left part, you have to find a closed-form expression for the irregular part ($\lambda$), and depending on its properties apply the required operation to the sum $\lambda+\sqrt{2\pi}$. Then you would have to find the regular part of the result.

Notice though that taking only odd or even numbers in the left part is not equal to dividing it by 2.

I am currently working on a paper on this topic, but I can give you a link to my Wiki page where I summed up the main ideas and results.

In short, any divergent integral or series can be interpreted as an extended (infinite) number, in which the regularized value is the regular part in a similar way as in complex numbers the real part is the part of the complex number.

In other words the expression $(\infty)!$ means something like a mean value of the factorial (from zero to infinity), but since it is infinite, it is not a normal real number, but an extended number, albeit with real("finite") part $\sqrt{2\pi}$. It is some kind of $\lambda+\sqrt{2\pi}$ where $\lambda$ is some purely irregular number.

As such, the answer to your third question is, you can divide the both sides by a real number and the regularized value will remain correct. Notice though that taking only odd or even numbers in the left part is not equal to dividing it by 2.

On the other hand, you cannot invert (1/x) or exponentiate the both sides without taking into accout the irregular part.

So if you want the regularization to remain valid after you do any operations on the left part, you have to find a closed-form expression for the irregular part ($\lambda$), and depending on its properties apply the required operation to the sum $\lambda+\sqrt{2\pi}$. Then you would have to find the regular part of the result.

I am currently working on a paper on this topic, but I can give you a link to my Wiki page where I summed up the main ideas and results.

In short, any divergent integral or series can be interpreted as an extended (infinite) number, in which the regularized value is the regular part in a similar way as in complex numbers the real part is the part of the complex number.

In other words the expression $(\infty)!$ means something like a mean value of the factorial (from zero to infinity), but since it is infinite, it is not a normal real number, but an extended number, albeit with real("finite") part $\sqrt{2\pi}$. It is some kind of $\lambda+\sqrt{2\pi}$ where $\lambda$ is some purely irregular number.

As such, the answer to your third question is yes, you can divide the both sides by a real number and the regularized value will remain correct.

On the other hand, you cannot invert (1/x) or exponentiate the both sides without taking into accout the irregular part.

So if you want the regularizetion to remain valid after you do any operations on the left part, you have to find a closed-form expression for the irregular part ($\lambda$), and depending on its properties apply the required operation to the sum $\lambda+\sqrt{2\pi}$. Then you would have to find the regular part of the result.

I am currently working on a paper on this topic, but I can give you a link to my Wiki page where I summed up the main ideas and results.

In short, any divergent integral or series can be interpreted as an extended (infinite) number, in which the regularized value is the regular part in a similar way as in complex numbers the real part is the part of the complex number.

In other words the expression $(\infty)!$ means something like a mean value of the factorial (from zero to infinity), but since it is infinite, it is not a normal real number, but an extended number, albeit with real("finite") part $\sqrt{2\pi}$. It is some kind of $\lambda+\sqrt{2\pi}$ where $\lambda$ is some purely irregular number.

As such, the answer to your third question is yes, you can divide the both sides by a real number and the regularized value will remain correct.

On the other hand, you cannot invert (1/x) or exponentiate the both sides without taking into accout the irregular part.

So if you want the regularization to remain valid after you do any operations on the left part, you have to find a closed-form expression for the irregular part ($\lambda$), and depending on its properties apply the required operation to the sum $\lambda+\sqrt{2\pi}$. Then you would have to find the regular part of the result.

Notice though that taking only odd or even numbers in the left part is not equal to dividing it by 2.