Trouble with Laplace Transform

Question

Can anyone help me with this Laplace Transform $$\mathcal{L}[(1-\cos(u))/u] ?$$

Thanks in advance

Answer 1

Notice that $$ \frac{1-\cos{u}}{u} = \int_0^1 \sin{u} \, da. $$ Interchanging the order of integration, $$ \mathcal{L}\left( \frac{1-\cos{u}}{u} \right)(s) = \int_0^1 \int_0^{\infty} e^{-su} \cos{au} \, du \, da. $$ Now, we know the Laplace transform of $\cos{au}$, or at least we can get it using complex exponentials or integration by parts: it's $\frac{a}{a^2+s^2}$. Therefore the transform we need to evaluate is given by $$ \int_0^1 \frac{a}{a^2+s^2} \, da. $$ But this is integral is easy: it's just $$\left[ \frac{1}{2} \log{(a^2+s^2)} \right]_{a=0}^1 = \frac{1}{2} \log{\left( 1 + \frac{1}{s^2} \right)}. $$

Answer 2

Here is another way to proceed. 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 $(1-\cos(t))/t$ gives

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

---

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.