Bilateral Laplace transformation of $sin(t)*(H(t)-H(t -\pi))$

Question

As the title says, I'm trying to get the Laplace transformation of $sin(t)(H(t)-H(t -\pi))$, which I would say is $\frac 1{s^2+1}(1-e^{-\pi t})$. But when I put it into Wolfram Alpha it says it's $\frac 1{s^2+1}(1+e^{-\pi t})$. Why does that minus sign turn into a plus sign?

Note: $H(t)$ is the Heaviside function.

Answer 1

Well, we have that:

$$\mathscr{B}_t^{-1}\left[\sin\left(t\right)\left(\theta\left(t\right)-\theta\left(t-\pi\right)\right)\right]_{\left(\text{s}\right)}:=\int_{-\infty}^\infty\sin\left(t\right)\left(\theta\left(t\right)-\theta\left(t-\pi\right)\right)\cdot e^{-\text{s}t}\space\text{d}t$$

Now, for the heaviside step function:

$$\theta\left(t\right)-\theta\left(t-\pi\right)=\begin{cases} 0\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space t<0\space\wedge\space t\ge\pi\\ \\ 1\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space 0\le t<\pi \end{cases}$$

So, we get:

$$\mathscr{B}_t^{-1}\left[\sin\left(t\right)\left(\theta\left(t\right)-\theta\left(t-\pi\right)\right)\right]_{\left(\text{s}\right)}=\int_0^\pi\sin\left(t\right)\left(\theta\left(t\right)-\theta\left(t-\pi\right)\right)\cdot e^{-\text{s}t}\space\text{d}t$$

Multiply everything out:

• $$\int_0^\pi\sin\left(t\right)\theta\left(t\right)e^{-\text{s}t}\space\text{d}t=\int_0^\pi\sin\left(t\right)e^{-\text{s}t}\space\text{d}t=\frac{1+e^{-\pi\text{s}}}{1+\text{s}^2}$$
• $$\int_0^\pi\sin\left(t\right)\theta\left(t-\pi\right)e^{-\text{s}t}\space\text{d}t=0$$