I have problem solving Laplace transform of
$$\frac{\sin\alpha t\cos\beta t}t$$
I have no idea on how it arrived at this answer:
$$\frac{1}{2} \arctan{\frac{\alpha+\beta}{s}}+\frac {1}{2}\arctan{\frac{\alpha-\beta}{s}}$$
Please I want the proof using the definition of Laplace Transform.
Sorry for the format, just a new account here. Thanks in advance.
On
Let $$F(s) = \int_0^\infty \frac{\sin\alpha t \cos \beta t}{t} e^{-st}dt$$ $t$ that appears in the denominator suggest to differentiate $F$. Indeed, differentiating we get $$F'(s) = \int_0^\infty \sin\alpha t\cos\beta t~e^{-st} dt$$ This integral is easier to compute, we have $$F'(s) = \frac{1}{2}\left(\frac{\alpha + \beta}{(\alpha + \beta)^2 + s^2} - \frac{\alpha - \beta}{(\alpha - \beta)^2 + s^2}\right)$$
Now integrate this function to arrive at the answer.
On
Easiest way for me is to start with $$f(t)=\frac{\sin t}t$$ Then $$F(s)=\int_0^{\infty}\frac{\sin t}te^{-st}dt$$ $$F^{\prime}(s)=-\int_0^{\infty}\sin te^{-st}dt=\left.\frac{s\sin te^{-st}+\cos te^{-st}}{1+s^2}\right|_0^{\infty}=-\frac1{1+s^2}$$ Then $$F(s)=-\int\frac{ds}{1+s^2}=-\tan^{-1}s+C$$ $$\lim_{s\rightarrow\infty}F(s)=-\frac{\pi}2+C=0$$ So $$F(s)=\frac{\pi}2-\tan^{-1}s$$ Then $$\int_0^{\infty}\frac{\sin\gamma t}te^{-st}dt=\int_0^{\infty}\frac{\sin u}ue^{-\frac s{\gamma}u}du=F\left(\frac s{\gamma}\right)=\frac{\pi}2-\tan^{-1}\left(\frac s{\gamma}\right)$$ So $$\begin{align}\int_0^{\infty}\frac{\sin\alpha t\cos\beta t}te^{-st}dt&=\frac12\left[\int_0^{\infty}\frac{\sin(\alpha+\beta)t}te^{-st}dt+\int_0^{\infty}\frac{\sin(\alpha-\beta)t}te^{-st}dt\right]\\ &=\frac12\left[\frac{\pi}2-\tan^{-1}\left(\frac s{\alpha+\beta}\right)+\frac{\pi}2-\tan^{-1}\left(\frac s{\alpha-\beta}\right)\right]\end{align}$$
The result follows from the fact that $$\mathcal{L} \left\{ \sin(\alpha t) \cos(\beta t) \right\}(s)=\frac{\alpha \left(\alpha ^2-\beta ^2+s^2\right)}{\left((\alpha -\beta )^2+s^2\right) \left((\alpha +\beta )^2+s^2\right)}, $$ and frequency-domain integration using partial fractions.