Reduction Formulae

A reduction formula is a recurrence relation that expresses an integral I_n (involving a parameter n) in terms of I_(n-1), I_(n-2), or similar. This powerful technique allows you to evaluate families of integrals systematically without integrating from scratch each time.

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1 What Is a Reduction Formula?

A reduction formula takes the form:

I_n = (some expression involving n) + (some coefficient) * I_(n-k)

where I_n = integral of some expression involving a parameter n, and k is usually 1 or 2.

By applying the formula repeatedly, you reduce the index n down to a base case (I_0 or I_1) that is easy to evaluate.

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2 Deriving a Reduction Formula

The standard approach is:

1. Write down I_n = integral of the given expression.
2. Apply integration by parts (or another technique) to express I_n in terms of I_(n-1) or I_(n-2).
3. State the reduction formula clearly.
4. Find the base case(s).
5. Use the formula to compute specific values.

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3 Worked Example 1 — I_n = integral from 0 to 1 of x^n e^x dx

Step 1: Set up the integral.

I_n = integral from 0 to 1 of x^n e^x dx

Step 2: Integration by parts. Let u = x^n, dv = e^x dx. Then du = n x^(n-1) dx, v = e^x.

I_n = [x^n e^x] from 0 to 1 - integral from 0 to 1 of n x^(n-1) e^x dx

= (1^n * e^1 - 0) - n * integral from 0 to 1 of x^(n-1) e^x dx

= e - n * I_(n-1)

Step 3: The reduction formula.

I_n = e - n * I_(n-1)

Step 4: Base case.

I_0 = integral from 0 to 1 of e^x dx = [e^x] from 0 to 1 = e - 1

Step 5: Compute specific values.

I_1 = e - 1 * I_0 = e - (e - 1) = 1

I_2 = e - 2 * I_1 = e - 2

I_3 = e - 3 * I_2 = e - 3(e - 2) = e - 3e + 6 = -2e + 6 = 6 - 2e

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4 Worked Example 2 — I_n = integral from 0 to pi/2 of sin^n x dx

This is one of the most important reduction formulae in Further Mathematics.

Step 1: I_n = integral from 0 to pi/2 of sin^n x dx

Step 2: Write sin^n x = sin^(n-1) x * sin x and integrate by parts.

Let u = sin^(n-1) x, dv = sin x dx. Then du = (n-1) sin^(n-2) x cos x dx, v = -cos x.

I_n = [-sin^(n-1) x cos x] from 0 to pi/2 + (n-1) integral from 0 to pi/2 of sin^(n-2) x cos^2 x dx

The boundary term: at x = pi/2, cos(pi/2) = 0; at x = 0, sin(0) = 0. So the boundary term = 0.

Replace cos^2 x = 1 - sin^2 x:

I_n = (n-1) integral from 0 to pi/2 of sin^(n-2) x (1 - sin^2 x) dx

= (n-1) I_(n-2) - (n-1) I_n

Rearranging:

I_n + (n-1) I_n = (n-1) I_(n-2)

n * I_n = (n-1) I_(n-2)

I_n = ((n-1)/n) * I_(n-2)

Step 3: Base cases.

I_0 = integral from 0 to pi/2 of 1 dx = pi/2

I_1 = integral from 0 to pi/2 of sin x dx = [-cos x] from 0 to pi/2 = 1

Step 4: Compute specific values.

I_2 = (1/2) * I_0 = (1/2)(pi/2) = pi/4

I_3 = (2/3) * I_1 = 2/3

I_4 = (3/4) * I_2 = (3/4)(pi/4) = 3pi/16

I_5 = (4/5) * I_3 = (4/5)(2/3) = 8/15

I_6 = (5/6) * I_4 = (5/6)(3pi/16) = 5pi/32

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5 Wallis's Formulae

The results from Example 2 can be summarised as:

For even n = 2m: I_(2m) = ((2m-1)!! / (2m)!!) * (pi/2)

For odd n = 2m+1: I_(2m+1) = (2m)!! / (2m+1)!!

where n!! denotes the double factorial (product of all positive integers up to n with the same parity).

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6 Worked Example 3 — I_n = integral from 0 to pi/2 of cos^n x dx