Primitive of Pointwise Sum of Functions

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Theorem

Let $f_1, f_2, \ldots, f_n$ be real functions which are integrable.

Then:

$\displaystyle \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\displaystyle \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$


$\map P 1$ is true, as this just says:

$\displaystyle \int \map {f_1} x \rd x = \int \map {f_1} x \rd x$


Basis for the Induction

$\map P 2$ is the case:

\(\displaystyle \int \map {\paren {f_1 \pm f_2} } x \rd x\) \(=\) \(\displaystyle \int \paren {\map {f_1} x \pm \map {f_2} x} \rd x\) Definition of Pointwise Addition of Real-Valued Functions
\(\displaystyle \) \(=\) \(\displaystyle \int \map {f_1} x \rd x \pm \map {f_2} x \rd x\) Linear Combination of Integrals

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$\displaystyle \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x$


Then we need to show:

$\displaystyle \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_{k + 1} } x \rd x$


Induction Step

This is our induction step:


\(\displaystyle \) \(\) \(\displaystyle \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \paren {\map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x + \map {f_{k + 1} } x} \rd x\) Definition of Pointwise Addition of Real-Valued Functions
\(\displaystyle \) \(=\) \(\displaystyle \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x + \int \map {f_{k + 1} } x \rd x\) Basis for the Induction
\(\displaystyle \) \(=\) \(\displaystyle \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x + \int \map {f_{k + 1} } x \rd x\) Induction Hypothesis

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N: \displaystyle \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$

$\blacksquare$


Sources