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.
 A: 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 $$\ln \left[ \frac{x+1}{\sqrt{1+(x+1)^2}} \right]$$
A: We can use the property 
$$
\mathcal{L}\left[\frac{f(t)}{t}\right](s)=\int_s^\infty \mathcal{L}[f(t)](s')ds'
$$
so we have
$$
\int_0^\infty\frac{1-\cos t}{t}e^{-t}dt
  =\mathcal{L}\left[\frac{1-\cos t}{t}\right](1)=\\
\int_1^\infty\mathcal{L}[1-\cos t](s)ds
  =\int_1^\infty\left(\frac{1}{s}-\frac{s}{1+s^2}\right)ds\\
\left.\left(\log s-\frac{1}{2} \log(1+s^2)\right)\right|_{1}^\infty
  =\left.\log\frac{s}{\sqrt{(1+s^2)}}\right|_{1}^\infty=-\log\left(\frac{1}{\sqrt{2}}\right)
$$
