Why Termination Matters
A reduction formula replaces an integral by one or more simpler integrals. But writing down a recurrence is only part of solving the problem: you must also know that repeated use of the recurrence reaches integrals you can evaluate directly. An incorrect index shift can leave the same integral in place, while a missing base case can leave the calculation unfinished even when the indices have decreased correctly.
The key diagnostic is simple: follow the index at every step. It should move toward a finite set of explicitly evaluated cases. This tutorial develops that test through logarithmic, exponential, and trigonometric examples.
For indefinite integrals, recurrence identities are understood up to an additive constant. During repeated substitution, the constants combine into one arbitrary constant at the end.
A Test for Termination
Suppose a family of integrals is denoted by \(R_n\), and the recurrence replaces \(R_n\) with an expression involving \(R_{n-d}\), where \(d\) is a fixed positive integer. If the recurrence is valid whenever \(n\) is at least some threshold, then each application lowers the index. Termination still requires evaluated base cases for every index that the process can reach.
For a step of size \(d\), the starting index determines which sequence of indices appears. For example, a step of two preserves parity: an even index eventually reaches zero, and an odd index eventually reaches one. Thus a recurrence with a step of two generally needs both \(R_0\) and \(R_1\) as base cases.
A recurrence that leaves the same index on the right-hand side does not make progress by iteration. A recurrence that lowers the index but stops at an unevaluated integral has made progress, but has not yet produced a complete answer.
Worked Example: Reducing Powers of a Logarithm
Worked Example: Reducing Powers of a Logarithm
For \(x>0\), let \(J_n=\int (\ln x)^n\,dx\), where \(n\) is a nonnegative integer. Find \(J_2\), and diagnose a tempting incorrect index shift.
For \(n\geq 1\), integration by parts with \(u=(\ln x)^n\) and \(dv=dx\) gives \(du=n(\ln x)^{n-1}x^{-1}\,dx\) and \(v=x\). Therefore,
The index drops by one. The required base case is \(J_0\), which is direct:
For \(n=1\), the recurrence gives \(J_1=x\ln x-J_0=x\ln x-x\). For \(n=2\),
To check, differentiate the result:
A mistaken shift would write \(J_n=x(\ln x)^n-nJ_n\). This does not reduce the index: it leaves the same unknown integral on the right. Worse, at \(n=1\) it implies \(2J_1=x\ln x\), so that \(J_1=\tfrac12x\ln x\). Its derivative is \(\tfrac12\ln x+\tfrac12\), not \(\ln x\). The correct differentiation in integration by parts produces \((\ln x)^{n-1}\), so the remaining integral is \(J_{n-1}\), not \(J_n\).
Exponential Recurrence: Decrease by One
For a nonnegative integer \(n\), define \(I_n=\int x^n e^x\,dx\). Integration by parts with \(u=x^n\) and \(dv=e^x\,dx\) gives, for \(n\geq1\),
Here each step reduces the index by one, so the single base case \(I_0=\int e^x\,dx=e^x\) is enough for every nonnegative starting index. If the recurrence is written down but \(I_0\) is never evaluated, the procedure has not been carried through to an antiderivative.
Worked Example: Reducing an Exponential Integral to Its Base Case
Evaluate \(I_3=\int x^3e^x\,dx\) by repeatedly applying the recurrence.
The index path is \(3\to2\to1\to0\), and \(I_0\) is evaluated explicitly. To check the answer, set \(P(x)=x^3-3x^2+6x-6\). Then \(P'(x)=3x^2-6x+6\), so
A useful error check is to substitute \(n=1\) into the recurrence: it must give \(I_1=xe^x-I_0=xe^x-e^x+C\). Any proposed shift that leaves \(I_1\) on both sides has not performed a reduction and should not be iterated as though it had.
Trigonometric Recurrence: Decrease by Two
Let \(S_n=\int \sin^n x\,dx\), for nonnegative integers \(n\). The sine-power reduction formula, valid for \(n\geq2\), is
Because the index drops by two, an even starting index eventually reaches zero and an odd starting index eventually reaches one. Both base cases are needed: \(S_0=x\) and \(S_1=-\cos x\).
Worked Example: Following Both Trigonometric Base Cases
Use the recurrence to find \(S_4\), and then see what happens if the base case for odd indices is omitted.
First reduce the even index \(4\) to \(2\):
Then reduce \(S_2\) to its even base case \(S_0\):
Substitution gives the antiderivative
For a direct check, differentiating the recurrence expression for \(S_n\) gives
Now consider the odd index \(S_3\). The same recurrence gives \(S_3=-\tfrac13\sin^2x\cos x+\tfrac23S_1\). If \(S_1\) is missing from the stated base cases, the reduction stops at an unevaluated integral; the formula is not a complete evaluation procedure for odd powers. Supplying \(S_1=-\cos x\) finishes it. A recurrence that lowers indices by two therefore needs both parity branches represented among its base cases.
Common Failure Modes
Several checks catch most termination errors before they spread through a calculation:
- Read the subscript, not just the formula’s appearance. A term \(R_n\) on the right has not been reduced. A term \(R_{n-1}\) or \(R_{n-2}\) has a lower index, but you must still confirm that the full chain reaches a stated base case.
- Check the step size against the bases. A step of two preserves parity, so one base case cannot cover both even and odd starting indices.
- Check the smallest index where the recurrence is used. A recurrence stated only for \(n\geq2\) cannot be applied at \(n=1\). At that point you need a base case or a separately valid rule.
- Evaluate the terminal integral. Writing “continue until \(I_0\)” is a useful plan, but it is not the final antiderivative until \(I_0\) has been computed.
These checks distinguish two different problems: a recurrence can be algebraically valid but incompletely supplied with base cases, or its index shift can be wrong and prevent the intended reduction. A valid recurrence, a strictly decreasing index, and explicit base cases together provide a finite route to the answer.
Check Your Understanding
For each question, trace the index path before deciding whether the recurrence gives a complete evaluation method.
- A recurrence replaces \(R_n\) by an expression involving \(R_{n-1}\). What base case is sufficient for all nonnegative starting indices, provided the recurrence applies for \(n\geq1\)?
- A recurrence replaces \(T_n\) by an expression involving \(T_{n-2}\). Which base cases are needed to cover every nonnegative starting index?
- Why does a formula involving \(J_n\) on both sides fail to reduce the index by iteration?
- For \(S_5\), which base case does the sine-power recurrence reach?
- What is missing from a reduction procedure that ends with an unevaluated base integral?
