I want to prove $\displaystyle\int_0^\infty e^{-t*x}\frac{\cos{(x)}-1)}{x}dx=\ln{\frac{t}{\sqrt{1+t^2}}}$ for $t>0$.
What happens to this integral as $t\rightarrow 0^+?$
My attempt:-
Let $F(t)=\displaystyle\int_0^\infty e^{-tx} \frac{(\cos{(x)}-1)}{x}dx$
If we compute $F'(t)$ for $t>0$ using differentiation under integral sign,we get,
$$F'(t)=\displaystyle\int_0^\infty\frac{\partial}{\partial t}\left(e^{-tx}\frac{\cos{(x)}-1}{x}\right)dx=\displaystyle\int_0^\infty-e^{-tx}(\cos{(x)}-1)dx$$
and
$$F''(t)=\displaystyle\int_0^\infty x*e^{-tx}(\cos{(x)}-1)dx$$
Now , how to utilise these results to answer the question?