Find $L^{-1}[\frac{1}{(p^2+a^2)^2}]$ by convolution.

I know that $L^{-1}[\frac{1}{(p^2+a^2)^2}]=\frac{1}{2a^2}(\frac{\sin ax}{a}-x\cos ax)$ but I don't know how to do this using convolutions. Some help would really be appreciated.


2 Answers 2


You must evaluate $\mathcal{L}^{-1}\left\{\dfrac{1}{(s^2+a^2)^2}\right\}$

Recall that: $$\mathcal{L}^{-1}\{F(s)\cdot G(s)\}=f(t)\ast g(t)=\int_0^t f(t-\tau)g(\tau)~d\tau \tag{1}$$ Where $F(s)=\mathcal{L}\left\{f(t)\right\}$ and $G(s)=\mathcal{L}\left\{g(t)\right\}$.

The most logical and obvious choice in your case is to select $F(s)=G(s)=\dfrac{1}{s^2+a^2}$.

The inverse laplace transform of $\dfrac{1}{s^2+a^2}$ is obviously $f(t)=g(t)=\dfrac{\sin(at)}{a}$. Hence, from $(1)$ we must evaluate the following integral:

$$\begin{align} \mathcal{L}^{-1}\left\{\frac{1}{(s^2+a^2)^2}\right\}&=\int_0^t \frac{\sin(a(t-\tau))}{a}\cdot \frac{\sin(a\tau)}{a}~d\tau\\&=\frac{1}{a^2}\int_0^t \sin(a(t-\tau))\sin(a\tau)~d\tau \tag{2} \end{align}$$ Evaluating this will give you the same answer you have: $$\mathcal{L}^{-1}\left\{\frac{1}{(s^2+a^2)^2}\right\}=\frac{\sin(at)-at\cos(at)}{2a^3}$$

  • $\begingroup$ All is clear. The only thing that I don't understand is why is the inverse Laplace transform of $\frac{1}{s^2+a^2}$ $\frac{\sin(at)}{a}$ instead of $\sin (at)$? $\endgroup$
    – thisisme
    Jul 11, 2017 at 1:55
  • $\begingroup$ It is well known that $\mathcal{L}\{\sin(at)\}=\frac{\color{red}{a}}{s^2+a^2}$ (You can find this on almost every table). I proved it on this answer. Also recall the identity $\mathcal{L}\{bf(t)\}=b\mathcal{L}\{f(t)\}$. Dividing both sides by $a$: $$\frac{1}{a}\mathcal{L}\{\sin(at)\}=\frac{1}{s^2+a^2}$$ Using the identity: $$\mathcal{L}\left\{\frac{\sin(at)}{a}\right\}=\frac{1}{s^2+a^2}$$ Now take the inverse laplace transform on both sides and conclude. $\endgroup$ Jul 11, 2017 at 8:00

While projectilemotion's answer is sound, it's worth emphasizing that the convolution theorem should hold for any factorization of the transform into well-behaved functions. For instance, we could factorize the transform of interest as \begin{align} F(p) &=\frac{1}{(p^2+a^2)^2}\\ &=\frac{1}{(p-ia)^2}\frac{1}{(p+ia)^2}\\ &=\frac{1}{p-ia}\frac{1}{p-ia}\frac{1}{p+ia}\frac{1}{p+ia}=G_+(p)^2 G_-(p)^2 \end{align} and then the convolution theorem yields

$$f(x)=L^{-1}[F(p)]=((g_+\ast g_+)\ast(g_- \ast g_-))(x)$$ where $g_\pm (x)$ are the inverse Laplace transforms of $G_\pm (p)$. But $L^{-1}[p^{-1}]=1$, so the frequency-shift formula implies $g_{\pm}(x)=L^{-1}[(p\pm i a)^{-1}]=e^{\mp i a x}$. Hence

$$(g_{\pm}\ast g_{\pm})(x) =\int_0^x g_{\pm}(y)g_{\pm}(x-y)\,dy =\int_0^x e^{\mp i a y}e^{\mp i a (x-y)}\,dx=xe^{\mp i a x}$$ and therefore

\begin{align} f(x) &=\int_0^x (g_{\pm}\ast g_{\pm})(y)(g_{\pm}\ast g_{\pm})(x-y)\,dy\\ &=\int_0^x ye^{-i a y}(x-y)e^{i a (x-y)}\,dy\\ &=e^{i a x}\int_0^x y(x-y)e^{-2i a y}\,dy. \end{align} This last integral is not simple to carry out by hand, but if we use Mathematica we correctly obtain $f(x)=\frac{1}{2a^3}\left[\sin(ax)-ax \cos(ax)\right]$. So both factorizations are equally valid as far as the convolution theorem goes (though the one used in the other answer is a good deal more efficient!)


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