# A question arising from some misunderstanding involving the Laplace Transform of the Heaviside function

I am trying to compute the Laplace Transform of the Heaviside Step-function. I define the Heaviside step-function with the half-maximum convention:

$H(t-t_0) = 0$ for $t < t_0$ ; $H(t-t_0) = 1/2$ for $t=t_0$ ; $H(t-t_0) = 1$ for $t > t_0$.

Performing the integrations necessary to find the Laplace transform of the Heaviside step-function, I found that: (I use the symbol $L$ to denote the laplace transform.)

$L[ 0 ] = 0$ ; $L[1/2] = \frac{1}{2s}$ ; $L[1] = \frac{1}{s}$ .

I thought that this is the correct Laplace Transform for the different intervals on which the Heaviside Step-function is defined. However, my book mentions that: $$L [ H(t - t_0) f(t - t_0) ] = e^{-t_0 s } F(s) ,$$ where $F(s) = L [f(t) ]$. So if we look at the interval in which $t > t_0$, we find that $$L [ H(t - t_0) f(t - t_0) ] = \frac{ e^{-t_0 s } }{s} .$$

This doesn't agree with my calculations. Can you please point who goes wrong, and where and how?

• I'd recommend \mathscr L ($\mathscr L$) or \mathcal L ($\mathcal L$) for the Laplace transform. – kahen Jun 5 '13 at 13:25
• @kahen ah these symbols look much prettier indeed. – Max Muller Jun 5 '13 at 13:39

The unilateral Laplace transform $F(s)$ of a function $f(t)$ is defined by

$$F(s)=\int_{0}^{\infty}f(t)e^{-st} \,dt$$

For the shifted step function $H(t-t_0)$ this gives

$$F(s)=\int_{0}^{\infty}H(t-t_0)e^{-st} \,dt=\int_{t_0}^{\infty}e^{-st} \,dt=\frac{e^{-st_0}}{s}$$

So I guess your book is correct.

Your assumption about the value o the Laplace transform is wrong. Because the step function is not nonZero until $t=t_0$, the integral defining the LT does not have its lower limit at $0$, but rather $t_0$.

I do not think your question has been answered since it is merely provided to you the definition of the Laplace transform as you already know it. Even though this is quite an old question, other users may find it useful. Your way of thinking is half correct. As you know, the step function can be indeed defined as $\frac{1}{2}$ for $t=0$. Now, the Laplace transform takes into account whatever happens in $0$. A comprehensive definition of Laplace transform, as you can read also in Wikipedia, is $$\mathcal L \{ f\} (s) = \int_{0^{-}} ^{\infty} e^{-st}dt$$

As you can see, the integral starts with $0^{-}$. So, if you now take your definition of the step function, you can go ahead in this way

$$\mathcal L \{ f\} (s) = \int_{0^{-}} ^{0^{+}} \frac{1}{2}e^{-st}dt + \int_{0^{+}} ^{\infty} e^{-st}dt$$

The first integral will be zero, so only the second survives! That is why the $\frac{1}{2}$ is not forgotten, but its integral does not survive. Indeed, the $\frac{1}{2}$ is "between" the zero, and is integral will be zero. Your $t_0$ is the point where you function starts, that in you case is $0^{-}$.