📂Odinary Differential Equations

Laplace Transform of the Dirac Delta Function

Theorem¹

The Laplace Transform of the Dirac Delta Function is as follows.

$$ \mathcal{L} \left\{ \delta (t-t_{0}) \right\} = e^{-st_{0}} $$

Proof

Let’s define as shown in the picture above $d_\tau (t) = \dfrac{1}{2\tau}$ $-\tau \le t \le \tau$. Then, the limit below is the same as the Dirac Delta Function.

$$ \lim \limits_{\tau \to 0^+}d_\tau (t)=\delta (t) \\ \lim \limits_{\tau \to 0^+}d_\tau (t-t_{0})=\delta (t-t_{0}) $$

Thus $\mathcal{L} \left\{ \delta (t-t_{0}) \right\}=\mathcal{L} \left\{ \lim \limits_{ \tau \to 0^+}d_\tau (t-t_{0}) \right\}$. Therefore,

$$ \begin{align*} \int_{0}^\infty e^{-st}\delta (t-t_{0})dt &=\lim_{\tau \to 0^+} \int_{0} ^\infty e^{-st}d_\tau (t-t_{0})dt \\ &= \lim_{\tau \to 0^+} \int_{0} ^\infty e^{-st}d_\tau (t-t_{0})dt \\ &= \lim_{\tau \to 0^+} \int_{t_{0}-\tau}^{t_{0}+\tau}e^{-st}d_\tau (t-t_{0})dt \\ &= \lim_{\tau \to 0^+} \int_{t_{0}-\tau}^{t_{0}+\tau}e^{-st}\dfrac{1}{2\tau}dt \\ &= \lim_{\tau \to 0^+} \dfrac{1}{2\tau}\dfrac{-1}{s}\left[ e^{-st} \right]_{t_{0}-\tau}^{t_{0}+\tau} \\ &= \lim_{\tau \to 0^+} \dfrac{1}{2s\tau }\left( e^{-s(t_{0}-\tau)}-e^{-s(t_{0}+\tau)}\right) \\ &= \lim_{\tau \to 0^+} \dfrac{1}{2s\tau }e^{-st_{0}}\left( e^{s\tau}-e^{-s\tau}\right) \\ &= \lim_{\tau \to 0^+} e^{-st_{0}}\dfrac{1}{s\tau }\dfrac{e^{s\tau}-e^{-s\tau}}{2} \\ &= \lim_{\tau \to 0^+} e^{-st_{0}}\dfrac{1}{s\tau }\sinh (s\tau) \end{align*} $$

By the L’Hospital’s Rule,

$$ \lim \limits_{\tau \to 0^+}\dfrac{\sinh (s\tau) }{s\tau}=\lim \limits_{\tau \to 0^+} \dfrac { s\cosh (s\tau)}{s}=1 $$

Therefore,

$$ \int_{0}^\infty e^{-st}\delta (t-t_{0})dt =e^{-st_{0}} $$

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See Also

• Laplace Transform Table