Why do we have Laplace transform of a step function and integrator is same.

\begin{align} \mathcal L\left[u(t)\right] &= \frac 1s\\ \mathcal L \left[ \int dt\right] &= \frac 1s \end{align}

Please clear my doubt on this.

  • $\begingroup$ It might be more accurate to denote the second one as $$ \mathcal{L}\left[\int f(t)\,dt\right] = \frac{1}{s}\mathcal{L}\left[f(t)\right]. $$ $\endgroup$ – fibonatic Aug 12 '17 at 19:34

This is because the impulse response of an integrator is $h(t)=u(t)$. The output which is the convolution with the impulse respoponse is $$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$ and with $h(t)=u(t)$ it becomes $$\begin{align} y(t)&=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau\\ &=\int_{-\infty}^{t}x(\tau)d\tau\tag{1}\end{align}$$ where $(1)$ is resulted from the fact that $$u(t-\tau)=\begin{cases}0& \forall \tau>t\\ 1 & \text{otherwise}\end{cases}$$ The transfer function (which is the Laplace transform of impulse response) is $$H(s)=\mathcal{L}\{h(t)=u(t)\}=\frac{1}{s}$$


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