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-   -   What can be said about an integral that is only continuous from one side (http://mymathforum.com/real-analysis/337891-what-can-said-about-integral-only-continuous-one-side.html)

 jiggs December 5th, 2016 08:02 PM

What can be said about an integral that is only continuous from one side

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I need help figuring out what is happening when it is only continuous from one side or not continuous.

 mathman December 6th, 2016 02:48 PM

From the description it looks like f(1/2) is the only value that matters. If it is continuous at that point, then the integral = f(1/2). If f is discontinuous then the integral is the average of the left and right values.

 skipjack December 6th, 2016 09:05 PM

Really? What examples did you have in mind?

 v8archie December 7th, 2016 02:55 AM

I would say that $$\int_0^1 f(x) \, \mathrm dx = \lim_{t \to \frac12^-} \int_0^t f(x) \, \mathrm dx + \lim_{t \to \frac12^+} \int_t^1 f(x) \, \mathrm dx$$
as long as both limits on the right hand side exist and are finite.
In this case we get
$$\int_0^1 f(x) \, \mathrm dx = \cancelto{0}{ \lim_{t \to \frac12^-} \int_0^t 0 \, \mathrm dx } + \lim_{t \to \frac12^+} \int_t^1 g(x) \, \mathrm dx = \lim_{t \to \frac12^+} \int_t^1 g(x) \, \mathrm dx$$
I don't know what you mean by the integral being continuous from one side. But if the integrand $g(x)$ is continuous on $\frac12 \le x \le 1$, the limit is just the obvious integral
$$\int_0^1 f(x) \, \mathrm dx = \int_{\frac12}^1 g(x) \, \mathrm dx = G(1) - G(\tfrac12)$$
where $G(x)$ is a primitive (anti-derivative) of $g(x)$.

If $g(x)$ is continuous everywhere (in the interval of integration) except at $x=\frac12$, then we get
$$\int_0^1 f(x) \, \mathrm dx = \lim_{t \to \frac12^+} \int_{t}^1 g(x) \, \mathrm dx = \lim_{t \to \frac12^+} \big( G(1) - G(t) \big) = G(1) - \lim_{t \to \frac12^+} G(t)$$
as long as the limit exists and is finite.

 skipjack December 7th, 2016 07:11 AM

The integration is with respect to g (it's$\displaystyle \int_0^1\! fdg$), so the integration limits are values of g.

 v8archie December 7th, 2016 07:54 AM

Oh yes. I remember reading that a couple of days ago.

 mathman December 7th, 2016 01:42 PM

Using g as defined it looks like you have a Stieltjes integral, where physicists would have an integral with a delta function at 1/2.

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