Deriving the Polynomial-Exponential Recurrence

Why a Recurrence Helps

When an integrand contains a polynomial multiplied by an exponential, integration by parts can transform it into an integral of the same kind but with a lower polynomial degree. Repeating that step eventually reaches a simple starting case. A formula connecting successive integrals in this way is called a recurrence relation, or a reduction formula.

Here the goal is to derive that relation carefully. The choice of parts matters: differentiating the polynomial lowers its degree, while the exponential has an elementary antiderivative as long as its coefficient $a$ is nonzero.

Define the Integral Family

Definition: Let $a$ be a fixed nonzero real constant, and let $n$ be an integer with $n\ge 0$. Define $I_n$ to denote an antiderivative of $x^n e^{ax}$: $$ I_n=\int x^n e^{ax}\,dx. $$

Because this is an indefinite integral, $I_n$ represents a family of antiderivatives that differ by an additive constant. A recurrence written with indefinite integrals is understood with that freedom in mind: its right-hand side gives an antiderivative, and any other antiderivative differs from it by a constant.

The recurrence will apply when $n\ge 1$, so that $x^n$ can be differentiated to a polynomial of lower degree. The case $n=0$ is the base case. The condition $a\ne 0$ is needed because the antiderivative of $e^{ax}$ used in integration by parts is $e^{ax}/a$.

Deriving the Recurrence

For $n\ge 1$, choose the parts so that the polynomial is differentiated and the exponential is integrated. This is the useful choice because differentiating $x^n$ produces $nx^{n-1}$, reducing the exponent by one.

Choose $u$ and $dv$.
Set $u=x^n$ and $dv=e^{ax}\,dx$. Then $du=nx^{n-1}\,dx$ and $v=e^{ax}/a$, since $a\ne 0$.

Apply integration by parts.
Use $\int u\,dv=uv-\int v\,du$, retaining the minus sign in the second term.

Recognize the lower-degree integral.
The remaining integrand contains $x^{n-1}e^{ax}$, which is the integrand defining $I_{n-1}$.

Writing out the calculation gives

\begin{aligned} I_n &=\int x^n e^{ax}\,dx\\ &=x^n\frac{e^{ax}}{a} -\int \frac{e^{ax}}{a}\,n x^{n-1}\,dx\\ &=\frac{x^n e^{ax}}{a} -\frac{n}{a}\int x^{n-1}e^{ax}\,dx. \end{aligned}

Recurrence: For a fixed real constant $a\ne 0$ and integer $n\ge 1$, $$ I_n=\frac{x^n e^{ax}}{a}-\frac{n}{a}I_{n-1}, $$ with equality understood up to an additive constant for indefinite integrals.

The recurrence reduces the index from $n$ to $n-1$. It does not claim that $I_n$ is a constant multiple of $I_{n-1}$: the explicit term $x^n e^{ax}/a$ is also part of the relation.

Why the Recurrence Works

The integration-by-parts calculation establishes the formula. It is also useful to verify directly that the right-hand side has the required derivative. Suppose $F_{n-1}$ is any antiderivative of $x^{n-1}e^{ax}$, so $F_{n-1}'=x^{n-1}e^{ax}$. Define

F_n(x)=\frac{x^n e^{ax}}{a}-\frac{n}{a}F_{n-1}(x).

Differentiate using the product rule:

\begin{aligned} F_n'(x) &=\frac{1}{a}\left(nx^{n-1}e^{ax}+a x^n e^{ax}\right) -\frac{n}{a}F_{n-1}'(x)\\ &=\frac{n}{a}x^{n-1}e^{ax}+x^n e^{ax} -\frac{n}{a}x^{n-1}e^{ax}\\ &=x^n e^{ax}. \end{aligned}

The two terms involving $x^{n-1}e^{ax}$ cancel. Thus $F_n$ is indeed an antiderivative of the target integrand. This check explains the structure of the formula: the added explicit term has a derivative containing a lower-degree contribution, and the recurrence subtracts exactly that contribution.

The Base Case

The recurrence needs a starting value at $n=0$. In that case the polynomial factor is $x^0=1$, so the integral is simply an exponential antiderivative.

Base case: For $a\ne 0$, $$ I_0=\int e^{ax}\,dx=\frac{e^{ax}}{a}+C. $$

Indeed, differentiating $e^{ax}/a$ gives $(a e^{ax})/a=e^{ax}$. The recurrence together with this base case specifies a step-by-step method for reaching any nonnegative integer index. The base case is not obtained by substituting $n=0$ into the recurrence: the recurrence was derived for $n\ge 1$, and at $n=0$ its lower-index term would not be part of the defined family.

Worked Examples

Worked Example: Verify the Base Case for a Specific Parameter

Take $a=3$. Since $n=0$, use the base case:

I_0=\int e^{3x}\,dx=\frac{e^{3x}}{3}+C.

To check, differentiate the nonconstant part:

\frac{d}{dx}\left(\frac{e^{3x}}{3}\right) =\frac{3e^{3x}}{3} =e^{3x}.

This recovers the integrand. The factor $1/3$ is necessary because differentiating $e^{3x}$ introduces a factor of $3$.

Worked Example: Apply the Recurrence at $n=1$

Let $a=2$ and $n=1$. The recurrence gives

\begin{aligned} I_1 &=\frac{x e^{2x}}{2}-\frac{1}{2}I_0\\ &=\frac{x e^{2x}}{2}-\frac{1}{2}\left(\frac{e^{2x}}{2}\right)+C\\ &=\frac{x e^{2x}}{2}-\frac{e^{2x}}{4}+C. \end{aligned}

Differentiate the result term by term:

\begin{aligned} \frac{d}{dx}\left(\frac{x e^{2x}}{2}-\frac{e^{2x}}{4}\right) &=\frac{1}{2}\left(e^{2x}+2x e^{2x}\right)-\frac{2e^{2x}}{4}\\ &=\frac{e^{2x}}{2}+x e^{2x}-\frac{e^{2x}}{2}\\ &=x e^{2x}. \end{aligned}

The derivative matches the integrand, confirming the recurrence application and the base-case substitution.

Worked Example: Reduce $I_3$ When $a=-2$

For $a=-2$, $n=3$, the recurrence gives

I_3=\frac{x^3e^{-2x}}{-2}-\frac{3}{-2}I_2 =-\frac{x^3e^{-2x}}{2}+\frac{3}{2}I_2.

This relation reduces the integral with $x^3$ to one with $x^2$. To verify it without needing to calculate $I_2$ explicitly, let $F_2$ be any antiderivative satisfying $F_2'=x^2e^{-2x}$, and define

F_3(x)=-\frac{x^3e^{-2x}}{2}+\frac{3}{2}F_2(x).

Differentiate each term, using the product rule on the first:

\begin{aligned} F_3'(x) &=-\frac{1}{2}\left(3x^2e^{-2x}-2x^3e^{-2x}\right) +\frac{3}{2}x^2e^{-2x}\\ &=-\frac{3}{2}x^2e^{-2x}+x^3e^{-2x} +\frac{3}{2}x^2e^{-2x}\\ &=x^3e^{-2x}. \end{aligned}

The lower-degree terms cancel, so the proposed expression is an antiderivative of the required integrand.

Conditions and Interpretation

Each condition in the recurrence has a role:

$a\ne 0$: the integration step $v=e^{ax}/a$ requires division by $a$.
$n\ge 1$: the recurrence refers to $I_{n-1}$, with $n-1$ still a nonnegative integer, and differentiating $x^n$ lowers its degree.
$a$ is fixed: the same parameter is used in $I_n$ and $I_{n-1}$ throughout the recurrence.
Indefinite integrals: constants of integration can be chosen independently; when using the recurrence, any difference between antiderivatives is absorbed into an additive constant.

If $a=0$, the integrand becomes $x^n$, and the displayed recurrence is not available because it divides by $a$. That case must be treated separately rather than by substituting zero into the formula.

A Common Sign Error

A frequent mistake is to replace the minus sign in $\int u\,dv=uv-\int v\,du$ with a plus sign. For example, at $a=2$ and $n=1$, the incorrect sign would give

I_1\stackrel{\text{wrong}}{=}\frac{x e^{2x}}{2}+\frac{1}{2}I_0 =\frac{x e^{2x}}{2}+\frac{e^{2x}}{4}+C.

Differentiate this incorrect candidate:

\frac{d}{dx}\left(\frac{x e^{2x}}{2}+\frac{e^{2x}}{4}\right) =\frac{e^{2x}}{2}+x e^{2x}+\frac{e^{2x}}{2} =x e^{2x}+e^{2x}.

This is not the required integrand $x e^{2x}$; an extra $e^{2x}$ remains. The correction is to keep the minus sign from integration by parts. With $I_0=e^{2x}/2+C$, the recurrence subtracts $e^{2x}/4$, and differentiation then gives $x e^{2x}$.

What the Recurrence Accomplishes

The main benefit is reduction: each application replaces an index $n$ by $n-1$. Repeated use therefore leads through lower polynomial degrees until the base case $I_0$ is reached. The relation also makes clear why the polynomial-exponential product is well suited to integration by parts: the polynomial gets simpler at each step, while integrating the exponential preserves its form up to a constant factor.

For this tutorial, the essential result is the derivation and correct interpretation of the recurrence, not a general expanded expression for every $I_n$. When applying it, keep the parameter $a$ fixed, use the formula only for $n\ge1$, and check signs and constants by differentiation.

Check Your Understanding

Answer these questions using the recurrence and its conditions.

For which values of $n$ and $a$ is the recurrence $I_n=\frac{x^n e^{ax}}{a}-\frac{n}{a}I_{n-1}$ applicable?
What choice of $u$ and $dv$ reduces the polynomial degree, and what are the resulting $du$ and $v$?
Why does the recurrence contain a minus sign before the $I_{n-1}$ term?
State the base case $I_0$, and verify it by differentiation for a general nonzero $a$.
Why is substituting $a=0$ into the recurrence invalid?

Deriving the Polynomial-Exponential Recurrence

What You'll Learn

Why a Recurrence Helps

Define the Integral Family

Deriving the Recurrence

Why the Recurrence Works

The Base Case

Worked Examples

Worked Example: Verify the Base Case for a Specific Parameter

Worked Example: Apply the Recurrence at \(n=1\)

Worked Example: Reduce \(I_3\) When \(a=-2\)

Conditions and Interpretation

A Common Sign Error

What the Recurrence Accomplishes

Check Your Understanding

Test yourself: Calculus II final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Deriving the Polynomial-Exponential Recurrence

What You'll Learn

Why a Recurrence Helps

Define the Integral Family

Deriving the Recurrence

Why the Recurrence Works

The Base Case

Worked Examples

Worked Example: Verify the Base Case for a Specific Parameter

Worked Example: Apply the Recurrence at \(n=1\)

Worked Example: Reduce \(I_3\) When \(a=-2\)

Conditions and Interpretation

A Common Sign Error

What the Recurrence Accomplishes

Check Your Understanding

Test yourself: Calculus II final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Previous

Next