Proof that $\int_0^{\infty} e^{-st} (\frac{1-\cos t}{t^2})dt = \frac{\pi}{2} - s\ln(\frac{s^2}{s^2+1}) + 2\tan^{-1}(s)$ I'm trying to solve this Laplace problem from Laplace Transform by Schaum's Outline, chapter 1, math 167(a). But no matter what approach I follow, I end up with this line:
$L(\frac{1-\cos t}{t^2}) = \frac{\pi}{2} + \frac{1}{2}s\ln(\frac{s^2}{s^2+1}) - \tan^{-1}(s)$
And by definition, $L(\frac{1-\cos t}{t^2}) = \int_0^{\infty} e^{-st} (\frac{1-\cos t}{t^2})dt$
So I've no idea why and how the problem expects me to end up from $\frac{\pi}{2} + \frac{1}{2}s\ln(\frac{s^2}{s^2+1}) - \tan^{-1}(s)$ to $\frac{\pi}{2} - s\ln(\frac{s^2}{s^2+1}) + 2\tan^{-1}(s)$
Looking for your kind help.
 A: You have to use
$$ L\bigg\{\frac{f(t)}{t}\bigg\}=\int_s^\infty F(\tau)d\tau $$
twice, where $F(s)=L\{f(t)\}$.
Let $f(t)=\frac{1-\cos t}{t}$ and then
\begin{eqnarray}
F(s)=L\{f(t)\}&=&\int_s^\infty L\{1-\cos t\}(\tau)d\tau=\int_s^\infty\bigg(\frac1\tau-\frac{\tau}{\tau^2+1}\bigg)d\tau\\
&=&\frac12\ln\bigg(\frac{\tau^2}{\tau^2+1}\bigg)\bigg|_s^\infty=-\frac12\ln\bigg(\frac{s^2}{s^2+1}\bigg). 
\end{eqnarray}
So
\begin{eqnarray}
L\bigg\{\frac{1-\cos t}{t^2}\bigg\}&=&L\bigg\{\frac{f(t)}{t}\bigg\}=\int_s^\infty F(\tau)d\tau\\
&=&-\frac12\int_s^\infty \ln\bigg(\frac{\tau^2}{\tau^2+1}\bigg)d\tau\\
&=&\cdots
\end{eqnarray}
You finish the rest.
A: Well, first thing I thaught of was to use the fact that : $$ \left(\forall x\in\mathbb{R}^{*}\right),\ \frac{1-\cos{x}}{x^{2}}=\int_{0}^{1}{\left(1-y\right)\cos{\left(xy\right)}\,\mathrm{d}y} $$
Then, with Fubini's theorem will, I shall find the result.
Let $ s>0 $, we have : \begin{aligned}\int_{0}^{+\infty}{\mathrm{e}^{-sx}\left(\frac{1-\cos{x}}{x^{2}}\right)\mathrm{d}x}&=\int_{0}^{+\infty}{\mathrm{e}^{-sx}\int_{0}^{1}{\left(1-y\right)\cos{\left(xy\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\left(1-y\right)\int_{0}^{+\infty}{\mathrm{e}^{-sx}\cos{\left(xy\right)}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=s\int_{0}^{1}{\frac{1-y}{s^{2}+y^{2}}\,\mathrm{d}y}\\ &=\int_{0}^{1}{\frac{\frac{\mathrm{d}y}{s}}{1+\left(\frac{y}{s}\right)^{2}}}-\frac{s}{2}\int_{0}^{1}{\frac{\frac{2y\,\mathrm{d}y}{s^{2}}}{1+\left(\frac{y}{s}\right)^{2}}}\\ &=\int_{0}^{\frac{1}{s}}{\frac{\mathrm{d}x}{1+x^{2}}}-\frac{s}{2}\int_{0}^{\frac{1}{s^{2}}}{\frac{\mathrm{d}x}{1+x}}\\ \int_{0}^{+\infty}{\mathrm{e}^{-sx}\left(\frac{1-\cos{x}}{x^{2}}\right)\mathrm{d}x}&=\arctan{\left(\frac{1}{s}\right)}-\frac{s}{2}\ln{\left(1+\frac{1}{s^{2}}\right)}\end{aligned}
