TITLE: Show $\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$ for $t \gt 0$
QUESTION [14 upvotes]: The problem is to show
$$\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$$
for $t \gt 0$.
I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution $u=e^{-x}$ and came to a new integral (hopefully after no mistakes)
$$ \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, $$
but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself.
Thanks a lot!

REPLY [2 votes]: Notice, Laplace transform 
$$L[1]=\int e^{-st}dt=\frac{1}{s}$$
Now, we have 
$$\int_{0}^{\infty}\frac{e^{-x}-e^{-xt}}{x}dx$$
$$=\int_{0}^{\infty}\frac{e^{-x}}{x}dx-\int_{0}^{\infty}\frac{e^{-xt}}{x}dx$$
$$=\int_{1}^{\infty}L[1]dx-\int_{t}^{\infty}L[1]dx$$
$$=\int_{1}^{\infty}\frac{1}{x}dx-\int_{t}^{\infty}\frac{1}{x}dx$$
$$=[\ln |x|]_{1}^{\infty}-[\ln |x|]_{t}^{\infty}$$
$$=\lim_{x\to \infty}\ln|x|-\ln 1-(\lim_{x\to \infty}\ln|x|-\ln |t|)$$ 
$$=\lim_{x\to \infty}\ln|x|-0-\lim_{x\to \infty}\ln|x|+\ln |t|$$ 
$$=\ln |t|$$$$=\ln(t) \ \ \ \ \forall\ \ \  t>0$$