I am trying to solve the equation of a forced harmonic oscillator using Fourier Transform. I know that if a function $f(t)$ is such that $\lim_{x->\pm \infty} f(t) = 0$, then $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{df}{dt}\exp(i\omega t)dt = -i\omega \tilde{f}(\omega)$$ Now if we apply this to the equation of motion of forced harmonic oscillator $$\ddot{x} + \omega_0^2x = F(x).$$ Then applying Fourier transform to the above equation and assuming that $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} x(t) \exp(i \omega_0 t) dt = g(\omega)$, then in Fourier transform of $\ddot{x}(t)$ I am stuck here $$\mathscr{F}[\ddot{x}(t)] = \frac{1}{\sqrt{2\pi}} \left[ (\dot{x}(t) \exp(i \omega t))\vert^{\infty}_{0}+ i \omega (x(t) \exp(i \omega t))\vert^{\infty}_{0} - \omega^2g(\omega)\right ]$$ Now we can say that $x(t = 0) = 0$ and $\dot{x}(t = 0) = 0$. But how can I argue the same at $t=\infty$?

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    $\begingroup$ This question has a close vote for violating the homework policy. While this is true at the surface, the question has a clearly recognizable conceptual core (and the concrete computation can be seen as just an illustration of this) – @ShubhamDas feel free to edit the question to highlight the conceptual problem to prevent the question being closed. $\endgroup$ Commented Sep 25, 2023 at 15:05
  • $\begingroup$ Furthermore, while the conceptual problem being a maths problem, I don't feel Math.SE would be a good home for this question as it is about the pragmatic "physicst's way" of doing Fourier transformation in cases where the integrals don't converge. $\endgroup$ Commented Sep 25, 2023 at 15:06

2 Answers 2


It is often helpful to add a regularization factor $e^{-\epsilon \left| t \right|}$ to the integrands in the Fourier transformation to ensure convergence. You then take the limit $\epsilon \to 0$ after computing the integral. This can be understood as approximating the problem by a similar one, that differs from the real one only far away from $t = 0$.

This procedure prevents problems with oscillating terms at infinity. Rigorously the integrals don't exist otherwise (but physicists typically don't care as the regularization procedure gives consistent results).

There are several ways you can understand this procedure in a rigorous way (and why our "arbitrary" regularization gives the correct result). If you have some background about distributions this Math.SE post offers an explanation.


Without any claiming to be a complete answer with all the details you need, you could start using the (unilateral) Laplace transform \begin{equation} F(s) := \int_{t=0}^{\infty} f(t) e^{-st} dt \ . \end{equation} If the system is stable, the solution of the homogeneous equation containing the effects of the initial condition is a decaying transient solution, so it goes to zero. Then, you can get the forced solution setting $s = j \omega$ in the particular solution of the forced system.

Laplace transform of the equation \begin{equation} m \ddot{x} + c \dot{x} + k x = f(t) \ , \end{equation} reads \begin{equation} [ s^2 m + s c + k ]X(s) = F(s) + s m x_0 + c x_0 + m v_0 \end{equation} with initial conditions $x(0) = x_0$, $\dot{x}(0) = v_0$, and the solution in the equation reads \begin{equation} X(s) = \dfrac{F(s)}{s^2 m + s c + k } + \dfrac{s m x_0 + c x_0 + m v_0}{s^2 m + s c + k } \ , \end{equation} being the first the solution of the forced equation, and the second one the (decaying) solution of the homogeneous equation with the prescribed initial conditions.

So, you get the harmonic solution after the transient is over as \begin{equation} X(j \omega) = \dfrac{1}{-\omega^2 m + j\omega c + k } F(j \omega) \ . \end{equation}

Some details about Fourier transform of non-decaying signals. You could either use a decaying regularization factor on the forcing, or cut the forcing after the time interval of your interest, to avoid troubles with the infinity. Namely, you can set your forcing as \begin{equation} \widetilde{f}(t) = \begin{cases} f(t) \quad & , \quad t < T \\ 0 \quad & , \quad t \ge T \end{cases} \end{equation}


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