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19,980 Proofs 15,711 Definitions Help

Featured Proof

Generalized Integration by Parts


Theorem

Let $\map f x, \map g x$ be real functions which are integrable and at least $n$ times differentiable.

Then:

\(\displaystyle \int f^{\paren n} g \rd x\) \(=\) \(\displaystyle \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^{\paren j} + \paren {-1}^n \int f g^{\paren n} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle f^{\paren {n - 1} } g - f^{\paren {n - 2} } g' + f^{\paren {n - 3} } g'' - \cdots + \paren {-1}^n \int f g^{\paren n} \rd x\)

where $f^{\paren n}$ denotes the $n$th derivative of $f$.


Proof

Proof by induction:

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

$\displaystyle \int f^{\paren n} g \rd x = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^\paren j + \paren {-1}^n \int f g^{\paren n} \rd x$


Basis for the Induction

$\map P 1$ is the case:

$\displaystyle \int f' g \rd x = f g - \int f g' \rd x$

which is proved in Integration by Parts.

This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis:

$\displaystyle \int f^{\paren k} g \rd x = \sum_{j \mathop = 0}^{k - 1} \paren {-1}^j f^{\paren {k - j - 1} } g^{\paren j} + \paren {-1}^k \int f g^{\paren k} \rd x$


Then we need to show:

$\displaystyle \int f^{\paren {k + 1} } g \rd x = \sum_{j \mathop = 0}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x$


Induction Step

This is our induction step:


\(\displaystyle \int f^{\paren {k + 1} } g \rd x\) \(=\) \(\displaystyle \int \paren {f^{\paren k} }' g \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle f^{\paren k } g - \int f^{\paren k} g' \rd x\) Basis for the Induction
\(\displaystyle \) \(=\) \(\displaystyle f^{\paren k} g - \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^j f^{\paren {k - j - 1} } g^\paren {j + 1} + \paren {-1}^k \int f g^{\paren {k + 1} } \rd x}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle f^{\paren k} g + \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^{j + 1} f^{\paren {k - \paren {j + 1} } } g^{\paren {j + 1} } + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x}\) moving $-1$ into the parenthesis
\(\displaystyle \) \(=\) \(\displaystyle f^{\paren k} g + \paren {\sum_{j \mathop = 1}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x}\) substituting $j$ for $j + 1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 0}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x\) as $f^{\paren k} g$ is the $0$th element of the summation

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


Therefore:

$\displaystyle \forall n \in \N_{>0}: \int f^{\paren n} g \rd x = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^{\paren j} + \paren {-1}^n \int f g^{\paren n} \rd x$

assuming that $f$ and $g$ are sufficiently differentiable.

$\blacksquare$


Sources