What a Reduction Formula Does
A complicated integral can be easier to evaluate when it belongs to a family whose members are linked by a recurrence. Rather than compute each member from scratch, we relate one value of a parameter to a simpler value. Repeating that relation can reduce the original problem to one or more integrals that can be evaluated directly.
The essential bookkeeping is as important as the algebra: name the family, specify which parameter changes, state where the recurrence is valid, and identify the base integrals where the process stops. A recurrence without those details may not determine an answer.
For example, a recurrence might relate \(I_n\) to \(I_{n-1}\), or it might relate \(I_n\) to \(I_{n-2}\). The parameter being lowered is \(n\); it is not necessarily the variable of integration. A useful reduction formula also has a stopping point: a specified value of \(n\), or a small set of values, for which the corresponding integrals are known explicitly.
The Parts of a Useful Recurrence
Before using a reduction formula, identify four features:
- The family: What integral does \(I_n\) represent?
- The parameter: Which symbol changes from one member to another?
- The step and domain: Does the recurrence lower the parameter by one, two, or another amount, and for which parameter values is it valid?
- The base integrals: What explicit values will stop the repeated application?
For instance, a relation \(I_n=a_n I_{n-1}+b_n\) lowers an integer parameter by one, provided it is valid for the values being used. If instead the right side contains \(I_{n-2}\), the recurrence proceeds separately through even and odd indices. In that case, one base value may not be enough.
Here and in the examples, \(\mathbb{Z}_{\ge 0}\) denotes the nonnegative integers. The formulas may be valid only for a smaller range, such as \(n\ge 2\); that restriction is part of the formula, not an optional detail.
Worked Example: A Recurrence That Lowers the Index by One
Worked Example: A One-Step Recurrence
Define \(I_n=\int_0^1 \frac{x^n}{1+x}\,dx\) for \(n\in\mathbb{Z}_{\ge 0}\). Find a recurrence and use it to evaluate \(I_3\).
For \(n\ge 1\), rewrite the numerator as \(x^n=x^{n-1}(1+x)-x^{n-1}\). Dividing by \(1+x\) gives
Integrating both sides from \(0\) to \(1\) gives the reduction formula
The base integral is \(I_0=\int_0^1 \frac{1}{1+x}\,dx=\ln 2\). Apply the recurrence in descending order:
To check the result independently, polynomial division gives \(\frac{x^3}{1+x}=x^2-x+1-\frac{1}{1+x}\). Thus
which agrees with the recurrence calculation. The parameter decreases by one at each step, and the sequence terminates at the explicit base value \(I_0\).
When the Parameter Drops by More Than One
A reduction formula need not involve the immediately preceding index. If it lowers \(n\) by two, repeated use preserves whether \(n\) is even or odd. Consequently, the even-index calculations reach \(I_0\), while the odd-index calculations reach \(I_1\). Both base values are needed to evaluate every member of the family.
Worked Example: A Two-Step Recurrence and Two Base Integrals
Let \(J_n=\int_0^1 \frac{x^n}{1+x^2}\,dx\) for \(n\in\mathbb{Z}_{\ge 0}\). Derive a recurrence for \(n\ge 2\), then calculate \(J_4\).
Use \(x^n=x^{n-2}(1+x^2)-x^{n-2}\). Dividing by \(1+x^2\) yields
Therefore, for \(n\ge 2\),
The base integrals are found directly. First, \(J_0=\int_0^1 \frac{1}{1+x^2}\,dx=\arctan(1)-\arctan(0)=\frac{\pi}{4}\). For \(J_1\), the substitution \(u=1+x^2\), \(du=2x\,dx\), gives
Since the target index \(4\) is even, this calculation uses \(J_2\) and then \(J_0\):
Check by division: \(\frac{x^4}{1+x^2}=x^2-1+\frac{1}{1+x^2}\), so
The check confirms the value, while the index pattern explains why \(J_0\), rather than \(J_1\), is the relevant base integral for \(J_4\). The value \(J_1\) is still needed if the recurrence is to evaluate odd indices.
Why a Finite Recurrence Reaches an Explicit Answer
For a one-step recurrence of the form \(I_n=a_n I_{n-1}+b_n\), suppose it holds for every integer \(n\) from \(r+1\) through the target index \(N\), and suppose \(I_r\) is known. Repeated substitution gives a finite expression involving the base integral and the intervening terms:
An empty product is defined to be \(1\), so the term with \(k=N\) contains \(b_N\) alone. This formula shows precisely what “reducing” an integral means: after finitely many valid steps, the target is expressed using a base value and a finite number of known quantities.
Write the recurrence for \(I_N\), making sure its stated conditions include \(N\).
Replace \(I_{N-1}\) using the recurrence at \(N-1\), if that index is in its valid range.
Stop only when every remaining integral is an explicitly known base integral.
Combine the terms and verify the value when possible by simplifying the original integrand or using another direct calculation.
The finite formula follows by induction. For \(N=r\), the sum has no terms and the product is empty, so the right side is \(I_r\). Now suppose the formula holds at \(N-1\). The recurrence gives \(I_N=a_NI_{N-1}+b_N\). Multiplying the formula for \(I_{N-1}\) by \(a_N\) appends the factor \(a_N\) to each product. The additional term \(b_N\) is exactly the summand with \(k=N\), whose product is empty. This gives the stated expression at \(N\), completing the induction.
For a recurrence that lowers the index by two, the same reasoning applies along a chain of indices with the same parity. The chain must reach a base value, and all recurrence applications must lie in the valid range.
Worked Example: A Recurrence with a Different Multiplier
Worked Example: Tracking the Recurrence and Its Base Case
Define \(Q_n=\int_0^1 \frac{x^n}{2-x}\,dx\) for \(n\in\mathbb{Z}_{\ge 0}\). Derive a one-step recurrence and calculate \(Q_2\).
For \(n\ge 1\), the identity \(x^n=2x^{n-1}-x^{n-1}(2-x)\) implies
Integrating gives \(Q_n=2Q_{n-1}-\frac1n\), valid for \(n\ge 1\). The base value is
Now lower the index twice:
For a direct check, division gives \(\frac{x^2}{2-x}=-x-2+\frac{4}{2-x}\). Hence
as obtained from the recurrence. Here the recurrence multiplies the lower-index integral by \(2\); a reduction formula does not have to subtract the lower-index value.
Conditions, Termination, and Common Errors
A recurrence is useful only when its conditions are respected. Check that the integrals exist, that the parameter lies in the stated range, and that any coefficient or denominator in the recurrence is defined. For a discrete parameter, also check that repeated steps actually reach a base index. If the index can vary continuously, lowering it may not produce a finite chain to a known value.
A second common error is to confuse the amount by which the parameter drops. For the family \(J_n\), an incorrect step would be
This is wrong: the integrand identity for \(J_n\) contains \(x^{n-2}/(1+x^2)\), so its remaining integral is \(J_{n-2}\). The correct calculation is \(J_4=\frac13-J_2\), followed by \(J_2=1-J_0\). Tracking the exponent in the identity prevents the index error.
It is also incomplete to stop at a recurrence and present it as though it were already a value. For example, \(Q_2=2Q_1-\frac12\) does not evaluate \(Q_2\) until \(Q_1\) is evaluated or given as a base value. A complete reduction specifies the starting integral and follows the chain until no unknown family members remain.
These habits matter because reduction formulas organize repeated integration into a controlled process. They expose the structure of a family, make calculations reusable for different parameter values, and show exactly which simple integrals underpin the results.
Check Your Understanding
For each question, identify both the parameter step and the base information needed.
- In the family \(I_n=\int_0^1 \frac{x^n}{1+x}\,dx\), what index does the recurrence use to evaluate \(I_5\)?
- Why are both \(J_0\) and \(J_1\) useful base values for the recurrence \(J_n=\frac{1}{n-1}-J_{n-2}\)?
- State two conditions that should be checked before applying a reduction formula repeatedly.
- For \(Q_n=2Q_{n-1}-\frac1n\), what is the explicit base integral needed to begin evaluating \(Q_3\)?
- Explain why a recurrence relation alone may not determine a numerical or closed-form value.
