# Why the Laplace transform of the integral is 1/s?

I got this interpretation:

If we have $y'(t)=u(t)$ , it's like have $y(t)=\int{u(t)dt}$

If we solve this simple equation, we obtain:

$$sY(s)=U(s)$$ $$Y(s)=\frac{U(s)}{s}$$

So, we have the $U(s)$ multiplied for $\frac{1}{s}$

Is this conjunction correct? It doesn't seem quite complete. What do you think?

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If you take Laplace transform of the ode with the initial condition $y(0)=0$, then your derivation is correct. – Mhenni Benghorbal Sep 28 '13 at 14:14

Assuming the initial condition to be $y(0)=0$, your formulation is correct. My only suggestion would be that you should adjust your terminology a bit. The Laplace transform of the integral isn't $\frac{1}{s}$. It'd be more accurate to say

The Laplace transform of an integral is equal to the Laplace transform of the integrand multiplied by $\frac{1}{s}$.

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Laplace transform of f(t) is defined as $\text{F(s)=}\int _0^{+\infty }f(t)e^{-\text{st}}dt$

Thus, the Laplace transform of $\frac{df}{dt}$ can be derived as below.

$$\int _0^{+\infty }\left(\frac{d}{dt}f(t)\right)e^{-\text{st}}dt=\int_0^{+\infty } e^{-\text{st}} \, df=f(t)e^{-\text{st}}|_0^{+\infty }+\int _0^{+\infty }f(t)\text{se}^{-\text{st}}dt$$ Note that $\text{$\exists $F(s)=} \int _0^{+\infty }f(t)e^{-\text{st}}dt\Rightarrow f(t)e^{-\text{st}}|_{t=+\infty }=0$, we have $$\int _0^{+\infty }\left(\frac{d}{dt}f(t)\right)e^{-\text{st}}dt=\text{sF}(s)-f(0)$$

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