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How do you evaluate the inverse transform below using convolution ?

$$ \mathcal{ L^{-1} } \left[ {\frac{s}{(s^2 + a^2)^2}} \right] $$

I tried

$$\begin{align} \mathcal{ L^{-1} } \left[ {\frac{s}{(s^2 + a^2)^2}} \right](t) &= \int\limits_0^t \sin t \cdot \cos(at - a\tau) \, d\tau = \sin t \cdot \int\limits_0^t \cos(at - a\tau) \, d\tau \\ &= \sin t \cdot \left[ -\frac{1}{a} \sin(at - a\tau) \right]_0^t = \sin t \cdot \left[ 0 - \left(-\frac{1}{a} \sin(at) \right) \right] \\ &= \frac{1}{a} \sin^2(at) \end{align} $$

What have I done wrong? The answer in my book is

$$ \frac{t}{2a} \sin(at) $$

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    $\begingroup$ In the convolution, it should be $\sin a \tau$ instead of $\sin t$... $\endgroup$
    – Fabian
    Commented Jan 5, 2013 at 13:41

2 Answers 2

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Assume someone tells you to derivative $ \frac{1}{(s^2+a^2)}$ respect to $s$ wherein $a$ is a constant. You certainly will reply $$\frac{-2s}{(s^2+a^2)^2}$$ $$\left(\frac{1}{(s^2+a^2)}\right)'=\frac{-2s}{(s^2+a^2)^2}$$ or $$\left(\frac{-1}{2(s^2+a^2)}\right)'=\frac{s}{(s^2+a^2)^2}$$ or $$\frac{1}{2}\times(-1)^1\left(\frac{1}{(s^2+a^2)}\right)'=\frac{s}{(s^2+a^2)^2}$$ But surely know that $$\mathcal{L}(t^1f(t))=(-1)^1F'(s)$$. Now can you find the proper $f(t)$ form tow last equalities when you know that $$\mathcal{L}\left(\frac{1}{a}\sin(at)\right)=\frac{1}{s^2+a^2}$$. This another approach besides to yours and @saz's.

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Instead of

$$\mathcal{L}^{-1} \left( \frac{s^2}{(s^2+a^2)^2} \right)(t) = \int_0^t \sin t \cdot \cos(a t-a \tau) \, d\tau$$

you should have

$$\mathcal{L}^{-1} \left( \frac{s^2}{(s^2+a^2)^2} \right)(t) = \frac{1}{a} \int_0^t \sin (a \tau) \cdot \cos(a t-a \tau) \, d\tau$$

because the convolution of functions $f,g$ is defined as $$(f \ast g)(t) = \int_{\mathbb{R}} g(\tau) \cdot f(t-\tau) \, d\tau$$ where $f(t) := \cos(a \cdot t) \cdot 1_{[0,\infty)}(t)$, $g(t) := \frac{1}{a} \cdot \sin (a \cdot t) \cdot 1_{[0,\infty)}(t)$.

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