Laplace Tansform of $\frac{1-\cos(2t)}t$

Question

What is the Laplace Transform of $\frac{1-\cos(2t)}t$ and is there a general formula for such situations with $\cos$, $\sin$ and others?

Answer 1

Note that $$\begin{align}\frac{1-\cos\omega t}{t}&=\int_0^{\omega}\sin\omega't\,\mathrm{d}\omega' \\ \frac{\sin\omega t}{t}&=\int_0^{\omega}\cos\omega't\,\mathrm{d}\omega' \end{align}$$ so that, once you've justified switching the order of integration, you can get these functions' Laplace transforms by integrating those for $\cos\omega t$ and $\sin \omega t$ with respect to $\omega$.

Answer 2

We will use the identity

$$\mathscr{L}\{\log (t)\}(s)=\frac{-\gamma-\log(s)}{s}\tag1$$

where $\mathscr{L}\{\log (t)\}(s)$ is the Laplace Transform of $\log (t)$ and where $\gamma$ is the Euler-Mascheroni constant. We will prove this identity at the end of this writing.

Now, taking the Laplace Transform of $(1-\cos(2t))/t$ gives

$$\begin{align} \mathscr{L} \left( \frac{1- \cos (2t)}{t}\right)(s)&=\int_0^{\infty} \left(\frac{1- \cos (2t)}{t}\right)e^{-st}\,dt\\\\ &\overbrace{=}^{t\mapsto t/2}\int_0^{\infty} \left(\frac{1- \cos (t)}{t}\right)e^{-st/2}\,dt\\\\ &=\int_0^{\infty} (1- \cos (t))\left(\frac{d\log(t)}{dt}\right)e^{-st/2}\,dt\\\\ &=\int_0^{\infty} \left(e^{-st/2}- \frac12 e^{-(s/2+i)t}-\frac12 e^{-(s/2-i)t}\right)\left(\frac{d\log(t)}{dt}\right)\,dt\\\\ &\overbrace{=}^{IBP}-\int_0^{\infty} \log(t)\frac{d}{dt}\left(e^{-st/2}- \frac12 e^{-(s/2+i)t}-\frac12 e^{-(s/2-i)t}\right) \,dt\\\\ &=\frac s2\int_0^{\infty} \log(t)e^{-st/2}\,dt\\\\ &-\frac12(s/2+i)\int_0^{\infty} \log(t)e^{-(s/2+i)t}\,dt\\\\ &-\frac12(s/2-i)\int_0^{\infty} \log(t)e^{-(s/2-i)t}dt\\\\ &\overbrace{=}^{\text{Using}\,(1)}s\left(\frac{-\gamma-\log(s)}{s}\right)\\\\ &-\frac12 (s/2+i)\left(\frac{-\gamma-\log(s/2+i)}{s/2+i}\right)\\\\ &-\frac12 (s/2-i)\left(\frac{-\gamma-\log(s/2-i)}{s/2-i}\right)\\\\ &=\frac12 \log\left(\frac{s^2+4}{s^2}\right) \end{align}$$

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Proof of the identity

$$\mathscr{L}\{\log (t)\}(s)=-\frac{\gamma-\log(s)}{s}$$

The Gamma Function $\Gamma$ is defined as

$$\Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t}\,dt$$

for $\text{Re}\{z\}>0$.

We can write this integral representation as a Laplace Transform by letting $t \to st$. Then, we have

$$\begin{align} \Gamma(z) &= \int_0^{\infty} (st)^{z-1}e^{-st}s\,dt\\\\ &=s^z\int_0^{\infty} t^{z-1}e^{-st}\,dt \end{align}$$

The derivative of the Gamma Function follows directly as

$$\begin{align} \Gamma'(z) &= s^z\log (s)\int_0^{\infty} t^{z-1}e^{-st}dt+s^z\int_0^{\infty} t^{z-1}e^{-st}\log (t) \,dt \end{align}$$

Note that $\Gamma'(1)=\log(s)+s\mathscr{L}\{\log (t)\}(s)$, where

$$\mathscr{L}\{\log (t)\}(s)=\int_0^{\infty}\log(t) e^{-st}\,dt$$

is the Laplace Transform of $\log (t)$. Solving for $\mathscr{L}\{\log (t)\}(s)$ yields

$$\mathscr{L}\{\log (t)\}(s)= \frac{-\gamma-\log(s)}{s}$$

where we have noted that $\Gamma'(1)=-\gamma$, is the Euler-Mascheroni constant.