Frequency Differentiation in Damped Trigonometric Integrals

Damping Makes Frequency Differentiation Work

Oscillatory factors such as \(\sin(wx)\) and \(\cos(wx)\) vary with the frequency \(w\). Differentiating with respect to \(w\) introduces a factor of \(x\), which can turn a basic damped integral into a weighted one. The key analytic point is that, when the damping parameter is positive, the resulting factor \(x e^{-ax}\) is integrable. This supplies the domination needed to differentiate under the integral sign.

Throughout, the measure space is \(([0,\infty),\mathcal{L},m)\), with Lebesgue measure, which is sigma-finite and complete but not finite. The functions below are real-valued and Lebesgue measurable. All integrals are Lebesgue integrals; the exponential damping ensures absolute integrability for the integrands we use. The parameters satisfy \(a>0\) and \(w\in\mathbb{R}\).

Differentiate with Respect to Frequency

We use the Dominated Convergence Theorem, in its differentiation-under-the-integral application established earlier in the course. In this application the difference quotients are measurable, converge pointwise, and are bounded in absolute value by an integrable function. The bound must include the factor introduced by differentiating the trigonometric function.

Proof. Fix \(w\in\mathbb{R}\). For \(h\ne0\), the difference quotient for the cosine integral has integrand \(e^{-ax}[\cos((w+h)x)-\cos(wx)]/h\). For each \(x\ge0\), this tends as \(h\to0\) to \(-x e^{-ax}\sin(wx)\). By the mean value theorem applied to \(t\mapsto\cos(tx)\), its absolute value is at most \(x e^{-ax}\), since the absolute value of the derivative \(-x\sin(tx)\) is at most \(x\). This bound is integrable: integration by parts gives \(\int_0^\infty x e^{-ax}\,dm(x)=1/a^2\). The boundary term \(x e^{-ax}\) tends to zero at infinity and is zero at \(x=0\). For any sequence of nonzero \(h\)'s tending to zero, the Dominated Convergence Theorem therefore gives convergence of the integrals of the difference quotients to the integral of their pointwise limit. The sequential characterization of a real limit then gives the asserted derivative of \(C_a\). For \(S_a\), the difference quotient uses \(\sin((w+h)x)-\sin(wx)\); the mean value theorem for \(t\mapsto\sin(tx)\) gives the same bound \(x e^{-ax}\), and its pointwise limit is \(x e^{-ax}\cos(wx)\). Applying the same theorem proves the asserted derivative of \(S_a\). \(\square\)

For completeness, the integration-by-parts value used in the proof follows first on \([0,R]\): \(\int_0^R x e^{-ax}\,dx=-R e^{-aR}/a+(1-e^{-aR})/a^2\). As \(R\to\infty\), \(R e^{-aR}\to0\) and \(e^{-aR}\to0\), giving \(1/a^2\). Thus the dominating function is integrable on this infinite-measure space.

Evaluate the Basic Sine and Cosine Integrals

The derivatives above are useful once the basic integrals \(C_a(w)\) and \(S_a(w)\) are known. We obtain them directly by integrating derivatives with respect to \(x\). The boundary terms at infinity vanish because \(a>0\); the remaining integrals converge absolutely, since their integrands are bounded by \(e^{-ax}\).

Proof. On \([0,R]\), differentiate \(e^{-ax}\sin(wx)\) with respect to \(x\) and integrate. Its derivative is \(e^{-ax}(w\cos(wx)-a\sin(wx))\). The boundary values are \(e^{-aR}\sin(wR)\) and \(0\), which tend to zero as \(R\to\infty\). Taking this limit gives \(wC_a(w)-aS_a(w)=0\). Similarly, the derivative of \(e^{-ax}\cos(wx)\) is \(-e^{-ax}(a\cos(wx)+w\sin(wx))\); its value at \(R\) tends to zero and its value at \(0\) is \(1\). Hence \(aC_a(w)+wS_a(w)=1\). These two equations have determinant \(a^2+w^2>0\). From the first, \(S_a(w)=wC_a(w)/a\). Substitution into the second gives \((a^2+w^2)C_a(w)/a=1\), so \(C_a(w)=a/(a^2+w^2)\), and then \(S_a(w)=w/(a^2+w^2)\). This calculation also covers \(w=0\): it gives \(C_a(0)=1/a\) and \(S_a(0)=0\). \(\square\)

Differentiating these explicit formulas and using the theorem gives the weighted integrals. The quotient rule yields \(C_a'(w)=-2aw/(a^2+w^2)^2\) and \(S_a'(w)=(a^2-w^2)/(a^2+w^2)^2\). Therefore \(\int_0^\infty x e^{-ax}\sin(wx)\,dm(x)=2aw/(a^2+w^2)^2\), while \(\int_0^\infty x e^{-ax}\cos(wx)\,dm(x)=(a^2-w^2)/(a^2+w^2)^2\). In particular, the sine-weighted integral changes sign with \(w\), and the cosine-weighted one can be negative. Neither integral is asserted to be nonnegative.

Domination and a Common Pitfall

The proof depends on the correct bound for the frequency difference quotient. A common mistake is to omit the factor \(x\) and claim that the derivative of the cosine integrand is bounded by \(e^{-ax}\). In fact, that derivative is \(-x e^{-ax}\sin(wx)\). For \(x=2\) and \(w=\pi/4\), its absolute value is \(2e^{-2a}|\sin(\pi/2)|=2e^{-2a}\), which is strictly greater than the proposed bound \(e^{-2a}\). The valid bound is \(x e^{-ax}\), whose integral is \(1/a^2<\infty\). The same bound works for the sine derivative and for the corresponding difference quotients.

Positive damping is also essential to this argument. If \(a=0\) and \(w=0\), the cosine integrand is the constant function \(1\) on \([0,\infty)\), whose Lebesgue integral is infinite. There is no integrable bound of the form used above, and the basic integral formula is not asserted for \(a=0\). In particular, one must not substitute \(a=0\) into a formula proved under \(a>0\) and treat the resulting expression as an evaluation of a Lebesgue integral.