Polynomial Factors with Exponential Decay

Why polynomial factors still give finite tails

A polynomial grows as \(x\) increases, while an exponential with a negative exponent decays. For a positive decay rate, the exponential eventually dominates any fixed nonnegative integer power of \(x\). This makes integrals of the form considered here converge, and integration by parts gives an exact way to evaluate them.

The calculation has a useful structure: one integration by parts reduces the power of \(x\) by one. Repeating the process ends after finitely many steps, at the integral of a pure exponential. The resulting answer is a finite sum rather than an infinite expansion.

Define the tail integral

Fix a real lower endpoint \(b\), a positive real number \(a\), and a nonnegative integer \(n\). Define the tail integral by taking a finite upper cutoff and then letting that cutoff tend to infinity:

I_n(b;a)=\int_b^\infty x^n e^{-ax}\,dx =\lim_{R\to\infty}\int_b^R x^n e^{-ax}\,dx, \qquad a>0,\quad n\in\{0,1,2,\ldots\}.

The positive condition on \(a\) is essential: it means \(e^{-ax}\) decays as \(x\) tends to infinity. The lower endpoint \(b\) may be negative, zero, or positive; changing a finite lower endpoint does not affect the behavior at infinity. In the recurrence below, \(I_0(b;a)\) is the starting case.

Definition: \(I_n(b;a)\) denotes the improper integral of \(x^n e^{-ax}\) from \(b\) to infinity, with \(b\) real, \(a>0\), and \(n\) a nonnegative integer.

Why the boundary term vanishes

Integration by parts on a finite interval produces a term involving \(R^n e^{-aR}\) at the upper cutoff \(R\). Before using the recurrence, we need to know that this term tends to zero. For \(R>0\), the exponential series gives \(e^{aR}\ge (aR)^{n+1}/(n+1)!\). Since both sides are positive, taking reciprocals and multiplying by \(R^n\) gives

0\le R^n e^{-aR} \le \frac{(n+1)!}{a^{n+1}R}.

Because \(a>0\) is fixed, the right-hand side tends to zero as \(R\to\infty\). Thus \(R^n e^{-aR}\to 0\). The same argument applies to every smaller nonnegative power of \(R\). It verifies the boundary limits needed as integration by parts is repeated.

Derive the recurrence

For \(n\ge 1\), first integrate over a finite interval \([b,R]\). Choose \(u=x^n\) and \(dv=e^{-ax}\,dx\), so \(du=nx^{n-1}\,dx\) and \(v=-e^{-ax}/a\). Integration by parts gives

\int_b^R x^n e^{-ax}\,dx = \left[-\frac{x^n e^{-ax}}{a}\right]_b^R +\frac{n}{a}\int_b^R x^{n-1}e^{-ax}\,dx.

Expanding the endpoint term yields \(-R^n e^{-aR}/a+b^n e^{-ab}/a\). Now let \(R\to\infty\). The first term tends to zero by the boundary estimate, and the two integrals tend to their respective tails. Therefore:

Theorem: For \(a>0\), \(b\in\mathbb R\), and integer \(n\ge1\), the tails satisfy \(I_n(b;a)=\frac{b^n e^{-ab}}{a}+\frac{n}{a}I_{n-1}(b;a)\). The base case is \(I_0(b;a)=\frac{e^{-ab}}{a}\).

To see the base case directly, integrate the exponential over \([b,R]\):

\int_b^R e^{-ax}\,dx =\left[-\frac{e^{-ax}}{a}\right]_b^R =\frac{e^{-ab}-e^{-aR}}{a} \longrightarrow \frac{e^{-ab}}{a}.

The recurrence is a finite procedure: each use lowers the index by one, so starting at \(n\) takes exactly \(n\) reductions to reach the base case. In particular, it establishes convergence for every integral in the definition.

Worked Example: A quadratic factor with rate three

Worked Example: Evaluate \(I_2(1;3)\)

Here \(n=2\), \(a=3\), and \(b=1\). Begin with the base case:

I_0(1;3)=\frac{e^{-3}}{3}.

Apply the recurrence for \(n=1\):

I_1(1;3) =\frac{1^1e^{-3}}{3}+\frac{1}{3}I_0(1;3) =\frac{e^{-3}}{3}+\frac{e^{-3}}{9} =\frac{4e^{-3}}{9}.

Apply it again for \(n=2\):

I_2(1;3) =\frac{1^2e^{-3}}{3}+\frac{2}{3}I_1(1;3) =\frac{e^{-3}}{3}+\frac{2}{3}\cdot\frac{4e^{-3}}{9} =e^{-3}\left(\frac{1}{3}+\frac{8}{27}\right) =\frac{17e^{-3}}{27}.

Check the result by differentiating an antiderivative. Let \(P(x)=x^2/3+2x/9+2/27\). Then \(3P(x)-P'(x)=x^2\), so \(\frac{d}{dx}[-e^{-3x}P(x)]=x^2e^{-3x}\). Also \(e^{-3R}P(R)\to0\) as \(R\to\infty\), by the polynomial-exponential boundary limit. Evaluating at the lower endpoint gives \(e^{-3}P(1)=e^{-3}(1/3+2/9+2/27)=17e^{-3}/27\), as found.

Work down to the base case

The recurrence can be applied one step at a time, or expanded into a finite sum. The first few steps display the pattern:

\begin{aligned} I_1(b;a)&=e^{-ab}\left(\frac{b}{a}+\frac{1}{a^2}\right),\\ I_2(b;a)&=e^{-ab}\left(\frac{b^2}{a}+\frac{2b}{a^2}+\frac{2}{a^3}\right),\\ I_3(b;a)&=e^{-ab}\left(\frac{b^3}{a}+\frac{3b^2}{a^2}+\frac{6b}{a^3}+\frac{6}{a^4}\right). \end{aligned}

For every nonnegative integer \(n\), the complete finite formula is

I_n(b;a) =e^{-ab}\sum_{k=0}^{n}\frac{n!}{(n-k)!}\frac{b^{\,n-k}}{a^{k+1}}.

Here \(n!=n(n-1)\cdots 1\) for \(n\ge1\), and \(0!=1\). The index \(k\) counts how many times the recurrence contributes a factor involving \(1/a\). The sum stops at \(k=n\), because after \(n\) reductions the base case has been reached. Equivalently, changing the summation index gives \(e^{-ab}n!\sum_{j=0}^n b^j/(j!a^{n-j+1})\). Both forms contain the same terms in reverse order.

Worked Example: A cubic factor starting at zero

Worked Example: Evaluate \(I_3(0;2)\)

Set \(b=0\), \(a=2\), and \(n=3\). The base case is

I_0(0;2)=\frac{e^0}{2}=\frac12.

For each positive index, the boundary contribution \(b^n e^{-ab}/a\) is zero. Thus:

\begin{aligned} I_1(0;2)&=0+\frac12 I_0(0;2)=\frac14,\\ I_2(0;2)&=0+\frac22 I_1(0;2)=\frac14,\\ I_3(0;2)&=0+\frac32 I_2(0;2)=\frac38. \end{aligned}

For a check, an antiderivative is \(-e^{-2x}P(x)\), where \(P(x)=x^3/2+3x^2/4+3x/4+3/8\). Direct differentiation gives \(2P(x)-P'(x)=x^3\), hence the derivative of \(-e^{-2x}P(x)\) is \(x^3e^{-2x}\). The upper endpoint contributes zero as \(x\to\infty\), and \(P(0)=3/8\); therefore the integral from zero to infinity is \(3/8\), agreeing with the recurrence.

Worked Example: A nonzero lower endpoint

Worked Example: Evaluate \(I_1(2;2)\)

Here \(n=1\), \(a=2\), and \(b=2\). First compute the base case:

I_0(2;2)=\frac{e^{-4}}{2}.

Then use the recurrence:

I_1(2;2) =\frac{2e^{-4}}{2}+\frac12\cdot\frac{e^{-4}}{2} =e^{-4}+\frac{e^{-4}}{4} =\frac{5e^{-4}}{4}.

For verification, differentiate \(-e^{-2x}(x/2+1/4)\). Its derivative is \(e^{-2x}(x+1/2-(1/2))=xe^{-2x}\). At infinity the antiderivative tends to zero; at \(x=2\) its value is \(-e^{-4}(1+1/4)\). Subtracting the lower value from the upper limit gives \(5e^{-4}/4\), as required.

A common recurrence error

A frequent mistake is to forget the factor \(n\) produced when differentiating \(x^n\). For example, an incorrect recurrence for \(I_2(1;3)\) might use \(I_2=e^{-3}/3+I_1/3\), omitting the factor \(2\). That calculation gives \(e^{-3}(1/3+4/27)=13e^{-3}/27\), which is not the correct value.

The correction follows directly from \(\frac{d}{dx}(x^2)=2x\): the correct step is \(I_2(1;3)=e^{-3}/3+(2/3)I_1(1;3)=17e^{-3}/27\). Keeping the index attached to its coefficient, \(\frac{n}{a}\), helps prevent this error.

Check the conditions: This recurrence requires \(a>0\) and an integer \(n\ge1\) for the reduction step. The base case handles \(n=0\). If \(a=0\), the exponential decay disappears, and this convergence argument and formula do not apply.

Why the finite formula is useful

These evaluations arise whenever a polynomial factor multiplies an exponentially decaying term. The recurrence gives an exact value for a tail without requiring a new integration-by-parts calculation from the beginning for each power. It also makes clear which factors depend on the lower endpoint and which come from the decay rate.

The assumptions matter to the method. The power must be a nonnegative integer so repeated reduction reaches zero in finitely many steps, and the rate must be positive so the endpoint terms vanish. This tutorial treats \(a\) as a fixed positive number. Dependence on a varying parameter is a separate question.

When using the result, preserve the endpoint contribution at every reduction. It is not generally correct to keep only the final base-case term: the terms involving \(b^n,b^{n-1},\ldots,b\) are part of the value of the tail. The finite-sum formula is a useful check that none have been lost.

Check Your Understanding

Answer each question by using the recurrence or the finite formula, and keep the assumptions on \(a\) and \(n\) in view.

State the recurrence for \(I_n(b;a)\) and its base case, including the conditions under which they apply.
Evaluate \(I_1(0;3)\) exactly.
Use the recurrence to evaluate \(I_2(0;1)\) exactly.
Why does the upper boundary term \(R^n e^{-aR}\) vanish when \(a>0\)?
Which factor is missing from the incorrect reduction \(I_n=b^ne^{-ab}/a+I_{n-1}/a\) when \(n>1\)?