Find the Laplace transform of the sine function (Laplace transform of sin(t)).

To calculate the Laplace transform of sin(t), we take the integral on 0 to infinity of e^-st*sin(t), where s is a constant with respect to the t integral.

This Laplace transform combines improper integrals with the integration by parts looping trick, which made it a perfect bonus problem for my last Calculus II exam!

To proceed with the integral, we apply integration by parts, letting u=e^-st and dv=sin(t)dt. This results in a second integral with a cosine instead of a sine, and we let u=e^-st and dv=cos(t)dt. After cleaning things up, we find a copy of the original integral on the right hand side of our work.

We evaluate the leftover terms from 0 to infinity, noting that e^(-infinity) is unambiguously zero, so the integral converges if s is greater than 0. These leftover terms yield a constant 1 as the result.

Now we close the deal on the looping trick: we gather both copies of the original integral on the left hand side, and factor out the integral. Using the name F(s) for the integral, we find F(s)(1+s^2)=1 or F(s)=1/(1+s^2), and that's the Laplace transform of sin(t).

We comment at the end on the fact that Laplace transforms are very useful in solving differential equations, where we transform an entire differential equation to Laplace transform s-space and find the solution using simple algebra. Then we transform the solution back to t space! This is a great prototype for mathematical physics in general: the idea that a shift in perspective to an abstract space can render all the mathematics trivial, then you solve the problem in the abstract space, then you take the solution back to the real world. That's a beautiful mathematical idea that we see all the time in mathematical physics!