# Welcome to $\mathsf{Pr} \infty \mathsf{fWiki}$

 ProofWiki is an online compendium of mathematical proofs! Our goal is the collection, collaboration and classification of mathematical proofs. If you are interested in helping create an online resource for math proofs feel free to register for an account. Thanks and enjoy! If you have any questions, comments, or suggestions please post on the discussion page, or contact one of the administrators. Also, feel free to take a look at the frequently asked questions because you may not be the first with your idea. To see what's currently happening in the community, visit the community portal.
19,699 Proofs 15,453 Definitions Help

# Featured Proof

## 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.

$\blacksquare$