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This question already has an answer here:

I am trying to find the Laplace Transform of $$\frac{1-\cos(ax)}{x}.$$ I have tried to directly compute the integral and also tried to see if I could use the convolution theorem in some way but both lead nowhere. Any help would be appreciated!

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marked as duplicate by Jack D'Aurizio Feb 11 '18 at 18:25

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This is a well known exercise, and the result is $$\frac{1}{2} \ln \left(\frac{a^2}{s^2}+1\right)$$

I am going to write further details now, but first of all I wanted you to know the result, directly. Now I will edit and prove the result.

Edits

The function is absolutely integrable hence we can use Leibniz rule.

Be

$$F(s) = \int_0^{+\infty} \frac{1 - \cos(a x)}{x} e^{-sx}\ dx$$

Hence

$$F'(s) = \int_0^{+\infty} -(1 - \cos(ax))e^{-sx}\ dx$$

This is a very trivial integral you can compute by your own, which leads to the result:

$$F'(s) = -\frac{1}{s}+\frac{s}{a^2+s^2}$$

Now:

$$F(s) = \int -\frac{1}{s}+\frac{s}{a^2+s^2}\ ds$$

Yet another very simple integral I leave you as an exercise, but write me if you have doubts.

The solution is then:

$$F(s) = -\ln (s)+\frac{1}{2} \ln \left(a^2+s^2\right)$$

Or as I wrote at the beginning

$$F(s) = \frac{1}{2} \ln \left(\frac{a^2}{s^2}+1\right)$$

Tricks for the future

If $~\mathcal L\left\{ f(t)\right\}=F(s)~$ and Laplace of the function $~g(t)=\dfrac{f(t)}{t}~$ exists, then

$$\mathcal L \left\{\dfrac{f(t)}{t}\right\}=\int\limits_s^{\infty}F(u)\,\mathrm du$$

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Here is an approach that does not use Feynman's trick, but rather uses integration by parts ("IBP") along with the well-known integral (See the NOTE at the end of THIS ANSWER$$\int_0^\infty \log(t)e^{-t}\,dt=-\gamma\tag1$$where $\gamma$ is the Euler-Mascheroni constant. It is easy to see from $(1)$ that for $\text{Re}(s)>0$ $$\int_0^\infty \log(t)e^{-st}\,dt=-\frac{\gamma-\log(s)}s\tag2$$


First, by enforcing the substitution $x= t/a$ reveals that

$$\int_0^\infty \frac{1-\cos(ax)}{x}e^{-sx}\,dx=\int_0^\infty \frac{1-\cos(t)}{t}e^{-(s/a)t}\,dt$$


Taking the Laplace Transform of $(1-\cos(t))/t$ gives

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


Finally, replacing $s$ with $s/a$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{1-\cos(ax)}{x}e^{-sx}\,dx=\frac12\log\left(1+\frac{a^2}{s^2}\right)}$$

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you could use the two properties :

  1. $\mathcal L \left\{x{f(x)}\right\}=-F'(s)\ $
  2. $F(s) \to 0 \text{ as } s \to \infty $ (Riemann–Lebesgue lemma)

we have $$xf(x) = 1 - \cos(ax) \implies F'(s) = \frac{s}{s^2+a^2} -\frac1s \implies F(s) = \ln(\frac{\sqrt{s^2+a^2}}{s}) +C$$

by the Riemann–Lebesgue lemma $C = 0$

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