Primitive of Power of x by Exponential of a x

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Theorem

Let $n$ be a positive integer.

Let $a$ be a non-zero real number.

Then:

\(\ds \int x^n e^{a x} \rd x\) \(=\) \(\ds \frac {e^{a x} } a \paren {x^n - \dfrac {n x^{n - 1} } a + \dfrac {n \paren {n - 1} x^{n - 2} } {a^2} - \dfrac {n \paren {n - 1} \paren {n - 2} x^{n - 3} } {a^3} + \cdots + \dfrac {\paren {-1}^n n!} {a^n} } + C\)
\(\ds \) \(=\) \(\ds \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C\)

where $n^{\underline k}$ denotes the $k$th falling factorial power of $n$.


Proof

Proof by induction:

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

$\ds \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$


$\map P 0$ is true, as from Primitive of $e^{a x}$:

$\ds \int e^{a x} \rd x = \frac {e^{a x} } a$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \int x e^{a x} \rd x\) \(=\) \(\ds \frac {e^{a x} } a \paren {x - \frac 1 a} + C\) Primitive of $x e^{a x}$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^1 \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} }\)

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.


So this is our induction hypothesis:

$\ds \int x^r e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } + C$


Then we need to show:

$\ds \int x^{r + 1} e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r + 1 - k} } {a^k} } + C$


Induction Step

This is our induction step:

\(\ds \int x^{r + 1} e^{a x} \rd x\) \(=\) \(\ds \frac {x^{r + 1} e^{a x} } a - \frac {r + 1} a \int x^r e^{a x} \rd x + C\) Primitive of $x^n e^{a x}$: Lemma
\(\ds \) \(=\) \(\ds \frac {x^{r + 1} e^{a x} } a - \frac {r + 1} a \paren {\frac {e^{a x} } a \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } } + C\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \frac {e^{a x} } a \paren {x^{r + 1} - \frac {r + 1} a \paren {\sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } } } + C\) factorising $\dfrac {e^{a x} } a$
\(\ds \) \(=\) \(\ds \frac {e^{a x} } a \paren {x^{r + 1} - \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {\paren {r + 1} r^{\underline k} x^{r - k} } {a^{k + 1} } } } + C\) moving $\dfrac {r + 1} a$ within the summation
\(\ds \) \(=\) \(\ds \frac {e^{a x} } a \paren {x^{r + 1} - \sum_{k \mathop = 1}^{r + 1} \paren {\paren {-1}^{k - 1} \frac {\paren {r + 1} r^{\underline {k - 1} } x^{r - k + 1} } {a^k} } } + C\) changing limits of summation
\(\ds \) \(=\) \(\ds \frac {e^{a x} } a \paren {x^{r + 1} + \sum_{k \mathop = 1}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r - k + 1} } {a^k} } } + C\) Definition of Falling Factorial, multplying summation by $-1$
\(\ds \) \(=\) \(\ds \frac {e^{a x} } a \sum_{k \mathop = 0}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r + 1 - k} } {a^k} } + C\) moving initial term into summation


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


Therefore:

$\ds \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$

$\blacksquare$


Examples

Primitive of $x^3 e^{-x}$

$\ds \int x^3 e^{-x} \rd x = -e^{-x} \paren {x^3 + 3 x^2 + 6 x + 6} + C$


Also see


Sources

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