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December 5th, 2016, 09:02 PM  #1 
Newbie Joined: Dec 2016 From: Birmingham Posts: 2 Thanks: 0  What can be said about an integral that is only continuous from one side
I need help figuring out what is happening when it is only continuous from one side or not continuous.

December 6th, 2016, 03:48 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,849 Thanks: 742 
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.

December 6th, 2016, 10:05 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,103 Thanks: 2321 
Really? What examples did you have in mind?

December 7th, 2016, 03:55 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 
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 (antiderivative) 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. 
December 7th, 2016, 08:11 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,103 Thanks: 2321 
The integration is with respect to g (it's$\displaystyle \int_0^1\! fdg$), so the integration limits are values of g.

December 7th, 2016, 08:54 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 
Oh yes. I remember reading that a couple of days ago.

December 7th, 2016, 02:42 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,849 Thanks: 742 
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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