How to find the Laplace Transform of $\frac{1-\cos(t)}{ t}$?

Question

What I tried is to find the transform of $f(t) = \frac{1-\cos(t)}{t}$ with $$\int_{s}^{\infty }\frac{1}{u} du + \int_{s}^{\infty }\frac{u}{u^2+1}du$$

$\int_{s}^{\infty }\frac{1}{u} du = \ln(\infty)-\ln(s)$. Is it a valid integral if I get an infinite value as a result?

$$\int_{s}^{\infty }\frac{u}{u^2+1}du = \frac{1}{2} \ln(\infty) - \frac{1}{2} \ln(s^2+1)$$

If it is right to have infinite values in the result of the integral, $\ln(\infty)$ cancels with $\frac{1}{2} \ln(\infty)$

and $F(s) = \ln(\infty) - \ln(s) - \frac{1}{2} \ln(\infty) + \frac{1}{2} \ln(s^2+1) $

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

Answer 1

Define the Laplace transform, $$f(s) = \int_{0}^{\infty} e^{- s t} f(t) \, dt,$$ with the short notation $f(t) \doteqdot f(s)$.

The long method: \begin{align} \frac{1 - \cos(a t)}{t} &\doteqdot \int_{0}^{\infty} e^{- s t} \, \frac{1 - \cos(a t)}{t} \, dt \\ &\doteqdot \int_{s}^{\infty} \int_{0}^{\infty} e^{- u t} \, (1 - \cos(at)) \, dt \, du \\ &\doteqdot \int_{s}^{\infty} \left(\frac{1}{u} - \frac{u}{u^{2} + a^{2}} \right) \, du \\ &\doteqdot \frac{1}{2} \, \left[ \ln(u^{2}) - \ln(u^{2}+a^{2}) \right]_{s}^{\infty} \\ &\doteqdot - \frac{1}{2} \, \ln\left(1 + \frac{a^{2}}{\infty^{2}}\right) + \frac{1}{2} \, \ln\left(\frac{s^{2} + a^{2}}{s^{2}}\right) \\ &\doteqdot \ln\left(\frac{\sqrt{s^{2}+a^{2}}}{s}\right). \end{align}

Answer 2

Here is a different approach:

Note that $f(t)={1 - \cos t \over t} = -\int_0^1 \sin (tx) dx$.

Note that ${\cal L} (t \mapsto \sin (xt))(s) = {x \over x^2+s^2}$.

Hence $\hat{f}(s) = \int_0^1 {\cal L} (t \mapsto \sin (xt))(s) dx = -\int_0^1 { x\over x^2+s^2} dx = -{1 \over 2} \log(x^2+s^2) \mid_0^1 = \log \sqrt{1+ {1 \over s^2}}$.