Deriving Recurrences for Exponential-Trigonometric Moments

Why Couple the Sine and Cosine Moments?

Integrals involving a polynomial, an exponential, and a trigonometric factor combine two useful patterns: differentiating a power of \(x\) lowers its degree, while integrating an exponential times a sine or cosine preserves the same exponential-trigonometric form. Integration by parts can therefore reduce a moment of degree \(n\) to moments of degree \(n-1\).

The sine and cosine integrals do not generally reduce independently. Integrating either exponential-trigonometric factor produces a combination of both sine and cosine. The resulting formulas are a coupled pair: to compute one moment at degree \(n\), we need both moments at the lower degree.

Define the Moments and Their Conditions

Let \(a,b\in\mathbb R\), let \(n\) be a nonnegative integer, and let \(x\in\mathbb R\). We use moments anchored at zero; when \(x<0\), the integral is understood as an oriented definite integral.

The integration variable is \(t\), while \(x\) is the upper endpoint. This distinction helps keep the boundary terms in the recurrence clear. The formulas below use the positive quantity \(D=a^2+b^2\), so they apply when \(a\) and \(b\) are not both zero. The case \(a=b=0\) is stated separately.

The Coupled Recurrence

Each formula contains a boundary term at \(x\) and a linear combination of moments of degree \(n-1\). There is no additional boundary term at zero because \(n\ge1\) makes \(0^n=0\).

To start the recurrence, evaluate the degree-zero moments directly:

These base formulas are valid under the same condition \(a^2+b^2>0\). In the exceptional case \(a=b=0\), the cosine integrand is \(t^n\) and the sine integrand is zero, so \(C_n(x)=x^{n+1}/(n+1)\) and \(S_n(x)=0\).

Why the Recurrence Works

For \(D=a^2+b^2>0\), define the two exponential-trigonometric expressions

Differentiating gives \(V_C'(t)=e^{at}\cos(bt)\) and \(V_S'(t)=e^{at}\sin(bt)\). For instance, in the first derivative the sine terms cancel, while the cosine terms combine to \(a^2+b^2\). Thus these expressions are antiderivatives of the corresponding exponential-trigonometric factors.

This derivation explains why the degree drops by one and why the moments are coupled. The derivative of the polynomial factor supplies the factor \(n\), and the antiderivative of each exponential-trigonometric factor mixes sine and cosine.

Worked Example: Degree One When \(a=b=1\)

Worked Example: Reducing Degree Two for Sine and Cosine

Worked Example: Degree One When \(a=1\) and \(b=2\)

A Common Sign Error

The sine recurrence contains \(aS_{n-1}-bC_{n-1}\), not \(aS_{n-1}+bC_{n-1}\). The minus sign comes from the cosine term \(-b\cos(bt)\) in the antiderivative of \(e^{at}\sin(bt)\).

For a concrete check, use \(a=b=1\) and \(n=1\). Replacing the minus sign by a plus would incorrectly give \(S_1(x)=x e^x(\sin x-\cos x)/2-(S_0+C_0)/2=e^x\bigl(x(\sin x-\cos x)-\sin x\bigr)/2\). Differentiating this wrong expression gives \(e^x(x\sin x-\cos x)\), not \(x e^x\sin x\). The corrected recurrence uses \(S_0-C_0\) and gives the verified sine moment from the first example.

Using the Recurrence Effectively

For a degree \(n\) moment, the practical sequence is to compute the degree-zero pair, then use the coupled formulas to obtain degrees one through \(n\). At every stage, keep both the sine and cosine moments: discarding one generally leaves the recurrence incomplete. These relations are useful when the degree is larger than one because they organize repeated integration by parts into a predictable reduction rather than requiring a new calculation from the beginning.

The formulas are for moments anchored at zero. An indefinite antiderivative can be obtained from a moment expression by adding an arbitrary constant. If a different lower endpoint is used, its boundary contribution must also be included; it does not generally vanish. Keeping the endpoint convention explicit prevents a correct recurrence from being mistaken for an equality between unspecified antiderivatives.