Polynomial-Weighted Cyclic Integrals on Finite Intervals

Why the Finite-Interval Form Matters

An integral such as \(\int_L^U x e^{ax}\sin(bx)\,dx\) combines three features: a polynomial factor, exponential growth or decay, and oscillation. The linear factor calls for integration by parts, while the sine and cosine terms are linked when differentiated or integrated. Treating those trigonometric integrals as a pair produces a compact method that works for any finite endpoints.

For a definite integral, the endpoints are part of the calculation, not an afterthought. In particular, the factor \(x\) means the upper and lower boundary terms need not disappear even when one endpoint is zero. We will keep those terms explicit and use the resulting antiderivatives to evaluate both the sine-weighted and cosine-weighted cases.

The Paired Antiderivatives

Write \(D=a^2+b^2\). When \(D>0\), equivalently when \(a\) and \(b\) are not both zero, define

A product-rule differentiation gives \(S_0'(x)=e^{ax}\sin(bx)\) and \(C_0'(x)=e^{ax}\cos(bx)\). These are the paired antiderivatives for the exponential-sine and exponential-cosine terms. Integrating those antiderivatives once more gives the useful cyclic identities

For example, the derivative of \((aS_0-bC_0)/D\) is \[ \frac{a e^{ax}\sin(bx)-b e^{ax}\cos(bx)}{D}=S_0(x). \] Similarly, the derivative of \((aC_0+bS_0)/D\) is \(C_0(x)\). The two antiderivatives are coupled: integrating the sine expression involves both \(S_0\) and \(C_0\), and integrating the cosine expression does too.

Deriving the Linear-Weight Formulas

Apply integration by parts to \(x e^{ax}\sin(bx)\), taking \(u=x\) and \(dv=e^{ax}\sin(bx)\,dx\). Then \(du=dx\) and \(v=S_0(x)\). Thus an antiderivative is \(xS_0-\int S_0\,dx\). The cyclic identity above evaluates the remaining integral. The cosine case follows by taking \(v=C_0(x)\) instead.

The derivative check is also the reason the method works. For the sine case, differentiating \(xS_0\) gives \(S_0+x e^{ax}\sin(bx)\); differentiating the subtracted cyclic term gives \(S_0\). They cancel, leaving the desired integrand. For the cosine case the same cancellation leaves \(x e^{ax}\cos(bx)\). The Fundamental Theorem of Calculus then gives the endpoint differences.

If desired, the two antiderivatives can be written directly without \(S_0\) and \(C_0\):

These expanded forms follow by substituting \(S_0\) and \(C_0\) into the theorem. They are convenient for endpoint evaluation, but the paired form often makes the integration-by-parts logic easier to see.

Worked Examples

Conditions, Endpoint Choices, and a Sign Pitfall

The formulas with \(D\) in the denominator require \(a^2+b^2>0\). If \(a=b=0\), the integrands reduce directly to \(x\sin 0=0\) and \(x\cos 0=x\). Therefore

This handles the only degenerate parameter pair. There is no need for \(a\) or \(b\) individually to be nonzero: for example, the formulas apply when \(a=0,b\ne0\), or \(b=0,a\ne0\). The interval endpoints can be any finite real numbers; if the order is reversed, the usual signed-integral rule gives \(\int_U^L=-\int_L^U\).

A frequent error is to lose a minus sign in the cyclic sine step. For \(a=b=1\), the correct identity is \(\int S_0\,dx=(S_0-C_0)/2\). If one incorrectly uses \((S_0+C_0)/2\), its derivative is \((e^x\sin x+e^x\cos x)/2\), which is not \(S_0=e^x(\sin x-\cos x)/2\). The incorrect expression therefore cannot be used as an antiderivative of \(S_0\). Keeping the paired identities visible until after integration by parts helps prevent this sign error.

Another endpoint mistake is to keep only the term containing \(x\) in \(F_s\) or \(F_c\). The remaining cyclic term is part of the antiderivative and can be nonzero even when \(x=0\), as the examples show. The safe procedure is to evaluate the complete \(F(U)\), evaluate the complete \(F(L)\), and subtract.

Why This Method Is Useful

The polynomial factor is reduced by one degree through integration by parts, while the exponential-trigonometric pair returns to the same two-function system. For a linear factor, the reduction ends after one integration-by-parts step. The cyclic antiderivatives make that final remainder explicit, so there is no unresolved integral left.

The same structure also explains why endpoint simplification can be substantial. When \(bU\) or \(bL\) is a familiar angle, sine and cosine may take simple exact values. That simplifies evaluation, but it does not remove the need to include all terms in the endpoint antiderivative. For general finite endpoints, the formula remains valid without any period assumption.