E8 - Reduction formulae

E8 - Reduction formulae

A family of difficult integrals can sometimes be reduced to one or two simple starting values. You will derive these links by integration by parts, retain the endpoint terms, and use the correct base case to calculate higher powers.

Derive recurrence via integration by parts

How can you evaluate ∫0π/2cos⁡6x dx\int_0^{\pi/2}\cos^6x\,\mathrm{d}x without expanding a sixth power? A rule that lowers the power by two would turn it into a fourth power, then a square, then a constant. First, see how integration by parts creates a link between neighbouring integrals.

Reduction formula

A reduction formula relates an indexed integral InI_n to integrals with smaller indices, together with any boundary terms. It becomes a calculation method when suitable starting values are known.

For a polynomial times an exponential, differentiate the polynomial factor: its degree falls by one. For a power of a trigonometric function, split off one factor to integrate, then use an identity to recover related powers.

Definite integration by parts

∫abudvdx dx=[uv]ab−∫abvdudx dx\int_a^b u\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x =[uv]_a^b-\int_a^b v\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x

Use differentiable factors with the displayed integrals defined. Integration by parts is in the formula booklet. A recurrence for a particular family must be derived or supplied in the question.

Worked example 1: lowering a polynomial degree

Question. For nonnegative integers nn, let Jn=∫01xnex dxJ_n=\int_0^1x^n\mathrm{e}^x\,\mathrm{d}x. Prove that Jn=e−nJn−1J_n=\mathrm{e}-nJ_{n-1} for integers n≥1n\ge1.

Plan. Use u=xnu=x^n and v′=exv'=\mathrm{e}^x, so differentiation lowers the index.

Jn=[xnex]01−n∫01xn−1ex dx=e−nJn−1n≥1 makes the lower product zero\begin{aligned} J_n&=[x^n\mathrm{e}^x]_0^1-n\int_0^1x^{n-1}\mathrm{e}^x\,\mathrm{d}x\\ &=\mathrm{e}-nJ_{n-1} && n\ge1\text{ makes the lower product zero} \end{aligned}

Answer. Jn=e−nJn−1J_n=\mathrm{e}-nJ_{n-1} for integers n≥1n\ge1, with J0=e−1J_0=\mathrm{e}-1 calculated separately.

Check. At n=1n=1 the recurrence gives J1=1J_1=1. Direct integration gives [(x−1)ex]01=1[(x-1)\mathrm{e}^x]_0^1=1, confirming the first reduction and the lower-end contribution.

Keep the interval and polynomial factor, but change the exponential rate. Its antiderivative introduces a factor of 1/21/2 in both places.

Worked example 2: changing the exponential rate

Question. Let Kn=∫01xne2x dxK_n=\int_0^1x^n\mathrm{e}^{2x}\,\mathrm{d}x for nonnegative integers nn. Derive a reduction formula for n≥1n\ge1.

Plan. Take u=xnu=x^n and v=e2x/2v=\mathrm{e}^{2x}/2.

Kn=[xne2x2]01−n2∫01xn−1e2x dx=e22−n2Kn−1\begin{aligned} K_n&=\left[\frac{x^n\mathrm{e}^{2x}}2\right]_0^1 -\frac n2\int_0^1x^{n-1}\mathrm{e}^{2x}\,\mathrm{d}x\\ &=\frac{\mathrm{e}^2}{2}-\frac n2 K_{n-1} \end{aligned}

Answer. Kn=e2/2−(n/2)Kn−1K_n=\mathrm{e}^2/2-(n/2)K_{n-1}, for integers n≥1n\ge1. The base is K0=(e2−1)/2K_0=(\mathrm{e}^2-1)/2.

Check. For n=1n=1, the formula gives (e2+1)/4(\mathrm{e}^2+1)/4. Differentiating e2x(x/2−1/4)\mathrm{e}^{2x}(x/2-1/4) gives xe2xx\mathrm{e}^{2x}; its endpoint difference is the same value.

The recurrence's index range matters. Define J0J_0 separately using the integrand ex\mathrm{e}^x: here the zeroth-power polynomial is the constant function 1. Its boundary product at zero is therefore 1; there is no need to evaluate 000^0. Do not extend a proof past the conditions used in its boundary calculation.

Watch out

Wrong: Jn=[xnex]01+nJn−1J_n=[x^n\mathrm{e}^x]_0^1+nJ_{n-1}. Integration by parts subtracts the new integral. Write the sign and the full boundary product before simplifying.

Derive the corresponding rule for a fresh exponential rate.

Base cases, parity and boundary terms

For powers of cosine, the reduction usually lowers the index by two. That preserves parity: even indices reach I0I_0, while odd indices reach I1I_1. You therefore need both base cases.

Let In(a)=∫0acos⁡nx dxI_n(a)=\int_0^a\cos^n x\,\mathrm{d}x, where 0<a≤π/20<a\le\pi/2 and nn is a nonnegative integer. For n≥2n\ge2, split cos⁡nx=cos⁡n−1xcos⁡x\cos^n x=\cos^{n-1}x\cos x. Parts, followed by sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x, gives

In(a)=[sin⁡xcos⁡n−1x]0a+(n−1)∫0asin⁡2xcos⁡n−2x dx=sin⁡acos⁡n−1a+(n−1)(In−2(a)−In(a))\begin{aligned} I_n(a)&=[\sin x\cos^{n-1}x]_0^a +(n-1)\int_0^a\sin^2x\cos^{n-2}x\,\mathrm{d}x\\ &=\sin a\cos^{n-1}a+(n-1)\bigl(I_{n-2}(a)-I_n(a)\bigr) \end{aligned}

The derivative of cos⁡n−1x\cos^{n-1}x is negative, so the subtraction in parts becomes an addition. Move the final −(n−1)In-(n-1)I_n to the left.

A cosine-power recurrence

For integers n≥2n\ge2 and fixed 0<a≤π/20<a\le\pi/2,

In(a)=sin⁡acos⁡n−1an+n−1nIn−2(a)I_n(a)=\frac{\sin a\cos^{n-1}a}{n}+\frac{n-1}{n}I_{n-2}(a)

with I0(a)=aI_0(a)=a and I1(a)=sin⁡aI_1(a)=\sin a. This family-specific recurrence is not in the formula booklet.

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