Fundamental Theorem of Analysis for discontinous functions

Last week I discovered a very neat property of improper Integrals. There are some functions where it is possible to calculate the area under the curve with the Fundamental Therorem of Analysis even if the function is discontinuous .

Let's say you have a function $f\left( x \right)$ and you want to calculate $\int _{ a }^{ b }{ f\left( x \right) dx }$. But unfortunately there is a pole point at ${ x }_{ p }$ with $b>{ x }_{ p }>a$.Now I will prove that: If ${ x }_{ p }$ is the only point on $\left[ a;b \right]$ where $f$ is discontinuos and there exists a point $P\left( { x }_{ p };{ y }_{ p } \right)$ to which $f$ is point symmetric, then it is possible to calculate the integral with the Fundamental Therorem of Analysis.

Because it is an improper integral, you devide the ingral into two: $\int _{ a }^{ b }{ f\left( x \right) dx } =\lim _{ z\rightarrow { 0 }^{ + } }{ \left( \int _{ a }^{ { x }_{ p }-z }{ f\left( x \right) dx } +\int _{ { x }_{ p }+z }^{ b }{ f\left( x \right) dx } \right) }$ Because ${ x }_{ p }$ is the only point of discontinuity of $f$, $f$ is continous on $\left[ a;{ x }_{ p }-z \right]$ and on $\left[ { x }_{ p }+z;b \right]$ and therefore you can calculate this with the Fundamental Therorem of Analysis, if $F$ is the antiderivative of $f$: \begin{aligned} \int _{ a }^{ b }{ f\left( x \right) dx } &= \lim _{ z\rightarrow { 0 }^{ + } }{ \left( F\left( { x }_{ p }-z \right) -F\left( a \right) +F\left( b \right) -F\left( { x }_{ p }+z \right) \right) } \\ \int _{ a }^{ b }{ f\left( x \right) dx } &= F\left( b \right) -F\left( a \right) +\lim _{ z\rightarrow { 0 }^{ + } }{ \left( F\left( { x }_{ p }-z \right) -F\left( { x }_{ p }+z \right) \right) } \end{aligned} If ${ x }_{ p }$ is a pole point of $f$ then it is a pole point of $F$ as well. $f$ is point symmetric to $P\left( { x }_{ p };{ y }_{ p } \right)$ and because $f$ is a measure of the slope of $F$ (because $\frac { d }{ dx } F\left( x \right) =f\left( x \right)$), $F$ must be axially symmetrical to $x={ x }_{ p }$.

an example point symmetric function $f$ in green and a possible axial symmetrical antiderivative of $f$ in blue

And Therefore: $F\left( { x }_{ p }-z \right) =F\left( { x }_{ p }+z \right) \\ F\left( { x }_{ p }-z \right) -F\left( { x }_{ p }+z \right) =0$

$\int _{ a }^{ b }{ f\left( x \right) dx } =F\left( b \right) -F\left( a \right)$

The geometric interpretation of this relationship is, that the infinite areas on the left and right side of the pole cancel out each other because of the opposite sign.

It is important to mention, that the antiderivative must be defined for every $x\neq { x }_{ p }$ in $\left[ a;b \right]$, because otherwise it will not work.

Note by CodeCrafter 1
2 months, 3 weeks ago

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- 2 days, 18 hours ago

Splitting the interval of integration into multiple sub-intervals has the advantage that the function has discontinuity Assignment Writing Service only at the end points and these are removable. So the function can be modified accordingly at end points to make it continuous and the integral evaluated as usual.

- 1 month, 1 week ago

So does this mean, that it would work even if the pole point doesn't have any symmetry? I tested it with $\frac{1}{x}+x^2$ and it seems so (I know an example is not a proof).

- 1 month, 1 week ago

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- 2 months ago

Edit:

In this proof, I used the argument, that the antiderivative of $f$ must be axial symmetrical to $x={x}_{p}$. But this is not true for all functions.

For example if $f \left( x \right) = x^3 + 1$ then the antiderivative is $F \left( x \right) = \frac{1}{4} x^4 + x$. $f$ is point symmtric to $P \left( 0;1 \right)$ but $F$ is not axial symmetrical to $x=0$. But as we go closer and closer to the pole point, then we approach a "axial symmetry". Therefore near the pole point $F$ is axially symmetrical and hence the important limit in the proof above still converges to 0.

But unfortunately I cannot find a general argumentation, which guarantees that there are no counterexamples.

- 2 months, 1 week ago