Calculating the Area Under an Inexisting Curve?

[alingy1]

Hi,
I always wondered,
what is the area under an inexisting curve.

That arctan(1/sqroot(x^2-1)) for example. Its domain does not include from -1 to 1.

If I take its integral from 0 to 10, what answer should I get?

 

[Mark44]

Hi,
I always wondered,
what is the area under an inexisting curve.

If the curve isn't defined, there's no region. Hence, area is meaningless.

That arctan(1/sqroot(x^2-1)) for example. Its domain does not include from -1 to 1.

If I take its integral from 0 to 10, what answer should I get?

Whatever answer you get will almost certainly be wrong. For an integral to exist, the function in the integral must be continuous on the interval over which you're integrating. Your arctan function isn't defined on part of the interval [0, 10]. If you were to ignore these requirements for integration, whatever antiderivative you get won't be defined at 0.

 

[haruspex]

You could interpret it by allowing y to be complex. Whether that leads to a meaningful integral I'm not sure. I tried it for x²+y²=1, x from 1 to 2. I got terms like $i \ln(2+\sqrt 3)$.

 

[skiller]

For an integral to exist, the function in the integral must be continuous on the interval over which you're integrating.

Hmmm... you sure about that?

 

[haruspex]

For an integral to exist, the function in the integral must be continuous on the interval over which you're integrating.

Hmmm... you sure about that?

Good catch. Even functions that would ordinarily be considered fairly pathological can be integrable. E.g. f(x) = 1 if x rational, 0 otherwise.
Depends partly on what definition of integration is being used, as in Riemann-Stieltjes v. Lebesgue.

 

[skiller]

Good catch.

No, I didn't think it was a particularly great catch. As I'm sure you know, there are extremely simple functions that have discontinuities and yet are integrable. A simple step function, for one.

 

[haruspex]

No, I didn't think it was a particularly great catch.

I just excusing myself for not having read Mark44's post fully enough to have caught it.