## Solution

By the definition of Laplace transform, we have:

\displaystyle \mathscr{L}\{\sin{at}\}=\int_{t=0}^{\infty}\sin{at}\cdot e^{-st}dt

This transform will be easy if we use Euler’s formula, i.e

e^{ix}=\cos{x}+i\sin{x}

Which means we can write \sin{x}=\mathcal{I}(e^{ix}). Here \mathcal{I} stands for imaginary part of e^{ix}. Then above equation can be written as:

\mathscr{L}\{\sin{at}\}=\mathscr{L}\{\mathcal{I}(e^{iat})\}=\mathcal{I}(\mathscr{L}\{e^{iat}\})

From the Laplace transform of the exponential function, we know that:

\displaystyle \mathscr{L}\{e^{iat}\} =\frac{1}{s-ia}

Replacing this in the previous equation yields:

\displaystyle \mathscr{L}\{\sin{at}\}=\mathcal{I}\left(\frac{1}{s-ia}\right)

To remove the i from the denominator just rationalize by multiplying s+ia top and bottom of the right hand side:

\displaystyle\mathscr{L}\{\sin{at}\}=\mathcal{I}\left(\frac{1}{s-ia}\times\frac{s+ia}{s+ia}\right)

By simplifying the denominator, we get s^2-i^2a^2 and we know i^2=-1. So, above equation leads to

\displaystyle \mathscr{L}\{\sin{at}\}=\mathcal{I}\left(\frac{s+ia}{s^2+a^2}\right)=\mathcal{I}\left(\frac{s}{s^2+a^2}+\frac{a}{s^2+a^2}i\right)

Taking the imaginary part of the above equation, the Laplace transform of sine function is given by:

\displaystyle\mathscr{L}\{\sin{at}\}=\frac{a}{s^2+a^2}

This ends the proof.

We reached the end of this exercise about calculating the Laplace transform of the sine function using the definition. In this link, we calculated the Laplace transform of the sine function with a different method!