$ L\left[1-\cos t\right]=L (1) - L(\cos t)\[2ex]

\displaystyle= \ \frac{1}{s}-\ \frac{s}{s^2+1}\[2ex]

\displaystyle L\ \left{\frac{1- \cos t}{t}\right} = \int_s^{\infty{}}\left[\frac{1}{s}-\ \frac{s}{s^2+1}\right]\ ds\[2ex]

\displaystyle= {\left[\log{s-\ \frac{1}{2}\log{(s^2+1)}}\right]}_s^{\infty{}}\[2ex]

\displaystyle= \frac{1}{2} {\left[\log{\frac{s^2}{s^2+1}}\right]}_s^{\infty{}}\[2ex]

\displaystyle= -\frac{1}{2}\log{\frac{s^2}{s^2+1}}\[2ex]

\displaystyle= \frac{1}{2}\log{\frac{s^2+1}{s^2}}\[2ex]

\displaystyle\therefore{} L\ \left{\frac{1- \cos t}{t^2}\right} = \frac{1}{2}{\left[\log{\left(\frac{s^2+1}{s^2}\right).s-\int s .\frac{s^2}{s^2+1}\left(\frac{s^2.2s-(s^2+1)2s}{s^4}\right)ds}\right]}_s^{ \infty{} }\[2ex]

\displaystyle= \frac{1}{2}\ {\left[s \log{\left(\frac{s^2+1}{s^2}\right)+2\int\frac{ds}{s^2+1}ds}\right]}_s^{\infty{}}\[2ex]

\displaystyle= \frac{1}{2}\ {\left[\ s \log{\left(\frac{s^2+1}{s^2}\right)+2{\tan}^{-1}s}\right]}_s^{\infty{}}\[2ex]

\displaystyle= \frac{1}{2}\ \left[\ 0+2.\frac{\pi{}}{2}-s \ log{\left(\frac{s^2+1}{s^2}\right)-2{\tan}^{-1}s}\right]\[2ex]

\displaystyle= \frac{\pi{}}{2}-\ \frac{s}{2} \log{\frac{s^2+1}{s^2}} - {\tan}^{-1}s\[2ex]

\displaystyle\therefore{}\

L\ \left{\frac{1-\cos t}{t^2}\right} = \frac{\pi{}}{2}-\ \frac{s}{2}\log{\frac{s^2+1}{s^2}} - {\tan}^{-1}s\[2ex] $