Indefinite integral with discontinuous integrand

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  • #1
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Suppose ##f## is defined as follows:
##f(x) = 1## for all ##x ≠ 1##, ##f(1) = 10##.
Is the indefinite integral (or the most general antiderivative) of ##f## defined at ##x = 1##?
I'm asking this question because I already know how to deal with, say, ##\int_0^2 f##; ##f## has only one removable discontinuity in ##[a,b]##, so it's integrable. When I do the following:
$$\int_0^2 f(x) dx = \int_0^1 dx + \int_1^2 dx$$
and then apply the second fundamental theorem of calculus on each of the integrals, I am assuming that the antiderivative of ##f## exists at ##x = 1## and is equal to ##1##.
Is this assumption wrong?
 

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  • #2
RUber
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That is one way to say it. However, if F is the antiderivative of f, the F' = f. Perhaps, you are defining a continuous function g such that g = f almost everywhere and finding its integral.
 
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  • #3
HallsofIvy
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Integration is a "smoothing" operator. If there is a singularity at a single point the integral will simply ignore it. The integral of "f(x)= 1 if x is not 1, f(1)= 10" is exactly the same as the integral of f(x)= 1.
 
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  • #4
hunt_mat
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Let [itex]\varepsilon >0[/itex] and split up the integrand in the following way:
[tex]\int_{0}^{2}f(x)dx=\int_{0}^{1-\varepsilon}f(x)dx+\int_{1-\varepsilon}^{1+\varepsilon}f(x)dx+\int_{1+\varepsilon}^{2}f(x)dx[/tex]
Now think about the consequences as [itex]\varepsilon\rightarrow 0[/itex]
 
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  • #5
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Integration is a "smoothing" operator. If there is a singularity at a single point the integral will simply ignore it. The integral of "f(x)= 1 if x is not 1, f(1)= 10" is exactly the same as the integral of f(x)= 1.
But if we apply the first fundamental theorem to the antiderivative ##F(x) = x + C## (i.e. ##F' = f##) we get ##1 = f(x)##, which is not true for ##x = 1##.
 
  • #6
Svein
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But if we apply the first fundamental theorem to the antiderivative ##F(x) = x + C## (i.e. ##F' = f##) we get ##1 = f(x)##, which is not true for ##x = 1##.
Well, in this case you must apply the extended fundamental theorem which says [itex] F'=f [/itex]almost everywhere (which means "except possibly on a set with measure 0).
 
  • #7
RUber
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The antiderivative F is defined such that F' = [f] where [f] = f almost everywhere. There is not going to be an antiderivative that captures point discontinuities, since the goal of the antiderivative is to have the property that ##F(b)-F(a) = \int_a^b f(x) dx ##. And the point discontinuities do not contribute to the integral.
 
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