Integral equation solve using Laplace transform How can I solve this integral equation using Laplace transform?
$${\int\limits_0^{\infty}\ }\frac{e^{-t}(1-\cos t)}{t}\operatorname  d\!t$$
Knowing that $$ \mathcal{L}\{\cos t\} = \frac{s}{s^2+1} $$
I think I can start by taking limits:
$$\lim_{b \rightarrow \infty} {\int\limits_0^{b}\ }\frac{e^{-t}(1-\cos t)}{t}\operatorname  d\!t$$
ant then apply the shortcut of $$\mathcal{L}\{\cos t\}$$
but I don't know how to continue. Any help will be appreciated.

To be honest I'm not sure what you meant by 'apply the shortcut of ...', but one way to do it is by writing
 $$1/t = \int_0^{\infty}e^{-tx}\, dx $$
so that 
$$\int_0^{\infty} \frac{e^{-t}(1-\cos t)}{t} dt = \int_0^{\infty} e^{-t}(1-\cos t)  \int_0^{\infty} e^{-tx} \, dx \,  dt \\
= \int_0^{\infty} \int_0^{\infty}   e^{-(x+1)t}(1-\cos t)   \,  dt \, dx$$
Then apply your knowledge of the Laplace transform of $\cos$ to calculate the inner integral. The outer integral can then be evaluated by elementary methods. 
In somewhat more detail, you obtain
$$
\int_0^{\infty}\frac{1}{x+1}-\frac{x+1}{1+(x+1)^2} \, dx
$$
which you can integrate using the primitive $$\ln \left[ \frac{x+1}{\sqrt{1+(x+1)^2}} \right]$$
