Damped Oscillations on an Infinite Interval

Why Damping Makes an Oscillatory Integral Converge

An oscillating function can keep changing sign indefinitely, so its positive and negative contributions may appear to balance. But oscillation by itself does not guarantee that an improper integral converges. Here we study a sine wave whose amplitude is gradually reduced by an exponential factor. The damping makes the contributions small enough to obtain a convergent integral, and integration by parts gives its exact value.

Throughout, assume \(a>0\) and \(b\in\mathbb{R}\). The parameter \(a\) is the damping rate, while \(b\) determines the frequency and direction of the oscillation. The case \(b=0\) is included.

Definition: The damped sine integral is the improper integral \(\displaystyle \int_0^\infty e^{-ax}\sin(bx)\,dx =\lim_{R\to\infty}\int_0^R e^{-ax}\sin(bx)\,dx\), provided this limit exists as a finite real number.

The definition matters: first integrate over a finite interval, then take the limit. This gives us a precise way to keep track of the endpoint at infinity while using integration by parts.

Convergence Before Evaluation

Before calculating a value, we should establish that the integral exists. For every \(x\geq 0\), \(|\sin(bx)|\leq 1\), so

\left|e^{-ax}\sin(bx)\right| \leq e^{-ax}.

Since \(a>0\), the comparison integral is finite:

\int_0^\infty e^{-ax}\,dx =\lim_{R\to\infty}\left[-\frac{e^{-ax}}{a}\right]_0^R =\frac{1}{a}.

Thus the integral of the absolute value is bounded by \(1/a\), and the damped sine integral converges absolutely. The cosine integral that will appear in the calculation is also absolutely convergent, because \(|\cos(bx)|\leq 1\) gives the same comparison. In particular, we may take limits of both finite-interval integrals below.

Key condition: The assumption \(a>0\) ensures both convergence and the vanishing of the exponential boundary terms. The calculation does not assert convergence when \(a=0\).

Two Integrations by Parts

For a finite endpoint \(R>0\), define

I_R=\int_0^R e^{-ax}\sin(bx)\,dx, \qquad J_R=\int_0^R e^{-ax}\cos(bx)\,dx.

We derive relations between \(I_R\) and \(J_R\) rather than applying integration by parts directly on an infinite interval. For the first integral, take \(u=\sin(bx)\) and \(dv=e^{-ax}\,dx\). Then \(du=b\cos(bx)\,dx\) and \(v=-e^{-ax}/a\), so

I_R =\left[-\frac{e^{-ax}\sin(bx)}{a}\right]_0^R +\frac{b}{a}J_R =-\frac{e^{-aR}\sin(bR)}{a}+\frac{b}{a}J_R.

For \(J_R\), take \(u=\cos(bx)\) and again \(dv=e^{-ax}\,dx\). Here \(du=-b\sin(bx)\,dx\). Keeping the minus sign in the integration-by-parts formula gives

J_R =\left[-\frac{e^{-ax}\cos(bx)}{a}\right]_0^R -\frac{b}{a}I_R =\frac{1-e^{-aR}\cos(bR)}{a}-\frac{b}{a}I_R.

The lower endpoint contributes \(1/a\) in this second equation: at \(x=0\), \(e^0\cos(0)=1\). Substituting the expression for \(J_R\) into the equation for \(I_R\) gives

\begin{aligned} I_R &=-\frac{e^{-aR}\sin(bR)}{a} +\frac{b}{a}\left(\frac{1-e^{-aR}\cos(bR)}{a} -\frac{b}{a}I_R\right),\\ \left(1+\frac{b^2}{a^2}\right)I_R &=-\frac{e^{-aR}\sin(bR)}{a} +\frac{b\left(1-e^{-aR}\cos(bR)\right)}{a^2}. \end{aligned}

Multiplying the second line by \(a^2\) and solving for \(I_R\) yields an exact finite-interval identity:

I_R = \frac{b-e^{-aR}\bigl(a\sin(bR)+b\cos(bR)\bigr)} {a^2+b^2}.

This derivation also explains why the denominator is \(a^2+b^2\): each integration by parts introduces a factor involving \(b/a\), and collecting the terms in \(I_R\) produces \(1+b^2/a^2\).

Taking the Improper Limit

To pass from \(I_R\) to the desired integral, we must check the boundary terms rather than discard them merely because they oscillate. Since \(|\sin(bR)|\leq 1\) and \(|\cos(bR)|\leq 1\),

\left|e^{-aR}\bigl(a\sin(bR)+b\cos(bR)\bigr)\right| \leq (a+|b|)e^{-aR} \longrightarrow 0 \qquad (R\to\infty),

where \(a+|b|\) is constant and \(e^{-aR}\to 0\) because \(a>0\). Therefore the finite-interval identity has the limit

\boxed{\displaystyle \int_0^\infty e^{-ax}\sin(bx)\,dx =\frac{b}{a^2+b^2} } \qquad (a>0,\ b\in\mathbb{R}).

Theorem: For every positive damping rate \(a\) and every real frequency \(b\), the integral \(\int_0^\infty e^{-ax}\sin(bx)\,dx\) converges absolutely and equals \(b/(a^2+b^2)\).

The proof has two distinct parts. The comparison with \(e^{-ax}\) establishes convergence. Integration by parts on \([0,R]\), followed by a verified limit, establishes the value. Keeping those parts separate prevents an informal manipulation at infinity from hiding an unjustified boundary step.

Worked Examples

Worked Example: Damping Rate 2 and Frequency 3

Evaluate \(\displaystyle\int_0^\infty e^{-2x}\sin(3x)\,dx\). Here \(a=2\) and \(b=3\), so the formula gives

\int_0^\infty e^{-2x}\sin(3x)\,dx =\frac{3}{2^2+3^2} =\frac{3}{4+9} =\frac{3}{13}.

We can check the result using the antiderivative \(\displaystyle F(x)=\frac{e^{-2x}(-2\sin(3x)-3\cos(3x))}{13}\). Indeed, differentiating its numerator by the product rule gives \[ \frac{d}{dx}\left(e^{-2x}(-2\sin(3x)-3\cos(3x))\right) =e^{-2x}\bigl(4\sin(3x)+6\cos(3x)-6\cos(3x)+9\sin(3x)\bigr) =13e^{-2x}\sin(3x), \] so \(F'(x)=e^{-2x}\sin(3x)\). At \(x=0\), \(F(0)=-3/13\), and \(F(x)\to 0\) as \(x\to\infty\), since its trigonometric factor is bounded and \(e^{-2x}\to0\). Hence the improper integral is \(0-(-3/13)=3/13\), as required.

Worked Example: A Smaller Damping Rate

Evaluate \(\displaystyle\int_0^\infty e^{-x/2}\sin(2x)\,dx\). Substituting \(a=1/2\) and \(b=2\) gives

\int_0^\infty e^{-x/2}\sin(2x)\,dx =\frac{2}{(1/2)^2+2^2} =\frac{2}{1/4+4} =\frac{2}{17/4} =\frac{8}{17}.

For a direct check, an antiderivative is \(\displaystyle F(x)=\frac{e^{-x/2}(-2\sin(2x)-8\cos(2x))}{17}\). Differentiating the expression inside the exponential product gives \[ \frac{d}{dx}\left(e^{-x/2}(-2\sin(2x)-8\cos(2x))\right) =e^{-x/2}\bigl(\sin(2x)+4\cos(2x)+4\cos(2x)+16\sin(2x)\bigr) =17e^{-x/2}\sin(2x). \] Thus \(F'(x)=e^{-x/2}\sin(2x)\). At zero, \(F(0)=-8/17\); at infinity, \(F(x)\to0\), since the trigonometric terms are bounded and the exponential decays. The endpoint difference is \(0-(-8/17)=8/17\), confirming the answer.

Worked Example: Negative Frequency

Evaluate \(\displaystyle\int_0^\infty e^{-x}\sin(-2x)\,dx\). With \(a=1\) and \(b=-2\),

\int_0^\infty e^{-x}\sin(-2x)\,dx =\frac{-2}{1^2+(-2)^2} =-\frac{2}{5}.

The negative sign is consistent with \(\sin(-2x)=-\sin(2x)\). To check by differentiation, use \(\displaystyle F(x)=\frac{e^{-x}(\sin(2x)+2\cos(2x))}{5}\). The derivative of the expression after \(e^{-x}\) is \(2\cos(2x)-4\sin(2x)\). Applying the product rule gives \[ F'(x) =\frac{e^{-x}}{5}\bigl(-\sin(2x)-2\cos(2x)+2\cos(2x)-4\sin(2x)\bigr) =-e^{-x}\sin(2x) =e^{-x}\sin(-2x). \] Now \(F(0)=2/5\), and \(F(x)\to0\) at infinity because the trigonometric factors are bounded. The integral is \(0-2/5=-2/5\), as claimed.

If \(b=0\), the integrand is identically zero, and the formula gives \(0/(a^2+0)=0\). Thus the result includes the zero-frequency case without any separate limiting argument.

A Common Boundary-Term Mistake

A tempting but incorrect calculation is to integrate by parts twice and then say, “The boundary terms vanish, so the integral is zero.” This overlooks that not every boundary contribution vanishes. In the finite-interval equation for \(J_R\), the lower endpoint produces \(1/a\), and that term remains as \(R\to\infty\). For instance, with \(a=2\) and \(b=3\), discarding it would incorrectly predict zero, while the correct value is \(3/13\).

Correction: Evaluate the boundary expression at both endpoints of \([0,R]\). The terms containing \(e^{-aR}\) vanish as \(R\to\infty\), but terms evaluated at \(x=0\) need not vanish. Only after retaining those lower-end contributions should the limit be taken.

There is a related sign check in the cosine integration by parts. Because \(d(\cos(bx))/dx=-b\sin(bx)\), the resulting \(I_R\) term is \(-bI_R/a\), not \(+bI_R/a\). A sign error there changes the denominator and can invalidate the result. Writing \(du\), \(v\), and the full boundary evaluation explicitly helps prevent both mistakes.

What the Formula Tells Us

The answer has the same sign as \(b\): positive frequencies give a positive integral, negative frequencies give a negative one, and zero frequency gives zero. The sign also follows directly from the identity \(\sin(-bx)=-\sin(bx)\). The denominator \(a^2+b^2\) is positive for the stated parameters, so no sign reversal comes from it.

The exponential factor has a second important role beyond shrinking the oscillations: it ensures that the boundary terms are controlled uniformly by a constant times \(e^{-aR}\). This is why the argument can safely take the improper limit after working on a finite interval. The result is useful whenever an exponentially damped oscillation must be integrated over an unbounded time or distance range; it replaces an infinite-interval limit with a simple expression in the damping rate and frequency.

Check Your Understanding

Answer these questions using the convergence condition and finite-interval calculation.

Why is \(a>0\) needed to justify convergence and to make the exponential boundary terms vanish?
Evaluate \(\displaystyle\int_0^\infty e^{-3x}\sin(4x)\,dx\).
What is the value of the integral when \(b=0\), and why?
In the integration-by-parts equation for \(J_R\), which lower-endpoint term survives as \(R\to\infty\)?
Why is it not enough to say that the sine oscillates when justifying convergence?