Is the Improper Integral \int_{-1}^1 \frac{dx}{x} Convergent?

[JG89]

\int_{-1}^1 \frac{dx}{x} = \lim_{a \rightarrow 0} \int_{-1}^a \frac{dx}{x} + \int_a^1 \frac{dx}{x} = 0.

Since 2a also goes to 0 for 'a' going to 0, then we have \int_{-1}^1 \frac{dx}{x} = \lim_{a \rightarrow 0} \int_{-1}^{2a} \frac{dx}{x} + \int_a^1 \frac{dx}{x} = ln2.

It seems like the increase in area beneath the curve is because 2a goes to 0 slower than 'a' does. So the area from -1 to 2a is less than the area from 'a' to 1 as 'a' goes to 0, and so when you add the two areas there is a positive value left over, equaling ln2.

My question is though, do we say that the improper integral \int_{-1}^1 \frac{dx}{x} is convergent? I thought that we say it converges if \lim_{\epsilon \rightarrow \infty} \int_{-1}^{x_{\epsilon}} \frac{dx}{x} + \int_{y_{\epsilon}}^1 \frac{dx}{x} where x_{\epsilon}; y_{\epsilon} are both sequences tending to 0, always yields the same value regardless of how they go to 0.

 

[Mute]

The first method of computing the integral, which you described, is called the "Cauchy principal value" of the integral. When computing such an integral you symmetrically cut off the integration around a discontinuity and take the limit as a -> 0 afterwards.

Such integrals without the cutoff are not convergent, but the Cauchy principal value integration tends to show up from time to time in actual calculations, often when doing contour integrals.

 

[JG89]

I still don't understand though, how come in the second method of calculation I get a different value? I know that it's probably because 2a goes to 0 slower than 'a' does, but, when thinking of the area under the curve 1/x, intuitively I should get only one value for the area.

*Note* I've only taken first year analysis, so please don't use any explanations from complex analysis or other sophisticated things.

 

[JG89]

Okay, I think I might understand:

In my textbook it says that if a function f is continuous on (a,b), except for an interior point x = c, then \int_a^b f(x) dx exists if and only if both \int_a^c f(x) dx and \int_c^b f(x) dx exist and has the value \int_a^c f(x) dx + \int_c^b f(x) dx (the original integral must have this value)

In my example, \int_0^1 \frac{dx}{x} doesn't exist, so then \int_{-1}^1 \frac{dx}{x} doesn't exist. That's why we get different values when we do the two computations differently. But sometimes we might want to give some value to the integral of 1/x from -1 to 1, and so we define the Cauchy Principal Value of the integral as Mute did previously: "you symmetrically cut off the integration around the discontinuity and take the limit as a -> 0 afterwards".

Is this correct?

 

[Elucidus]

The problem is that the integral

\int_{-1}^{1} \frac{dx}{x}

and the expression

\lim_{a \rightarrow 0} \left[ \int_{-1}^{a} \frac {dx}{x} + \int_{a}^{1} \frac {dx}{x} \right]

are not equal. The original improper integral equals:

\lim_{a \rightarrow 0^-} \int_{-1}^{a} \frac {dx}{x} + \lim_{b \rightarrow 0^+} \int_{b}^{1} \frac {dx}{x}.

The issue is that the boundary values (a and b) must be able to vary and converge to 0 independently of each other. Making one a and the other 2a makes them dependent and you lose the equality.

--Elucidus

 

[JG89]

The original improper integral equals:

\lim_{a \rightarrow 0^-} \int_{-1}^{a} \frac {dx}{x} + \lim_{b \rightarrow 0^+} \int_{b}^{1} \frac {dx}{x}.

The issue is that the boundary values (a and b) must be able to vary and converge to 0 independently of each other. Making one a and the other 2a makes them dependent and you lose the equality.

--Elucidus

Thanks. This way of taking the limits separately and independently makes much more sense to me. I now see that the CPV is just a way to give some divergent integrals a value.