Show $\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$ for $t \gt 0$ 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!
 A: Hint: $\frac{e^{-x}-e^{-xt}}{x} = \int\limits_{1}^{t}{e^{-xs}\text{ d}s}.$
A: Let $\displaystyle F(t) = \int_{0}^{\infty} \dfrac{e^{-x} - e^{-xt}}{x} dx$


*

*Show that $\displaystyle \dfrac{dF}{dt} = \dfrac{d}{dt}\left( \int_{0}^{\infty} \dfrac{e^{-x}-e^{-xt}}{x}dx \right)= \int_{0}^{\infty} \dfrac{d}{dt} \left(\dfrac{e^{-x}-e^{-xt}}{x} \right) dx = \dfrac{1}{t}$

*Show that $F(1)=0$ (the easy part)

*$F(t)$ satisfies the ODE $\dfrac{dF}{dt} = \dfrac{1}{t}$ with initial condition $F(1)=0$ (and $log(t)$ too, so they are equal)
A: Notice, Laplace transform 
$$L[1]=\int e^{-st}dt=\frac{1}{s}$$
Now, we have 
$$\int_{0}^{\infty}\frac{e^{-x}-e^{-xt}}{x}dx$$
$$=\int_{0}^{\infty}\frac{e^{-x}}{x}dx-\int_{0}^{\infty}\frac{e^{-xt}}{x}dx$$
$$=\int_{1}^{\infty}L[1]dx-\int_{t}^{\infty}L[1]dx$$
$$=\int_{1}^{\infty}\frac{1}{x}dx-\int_{t}^{\infty}\frac{1}{x}dx$$
$$=[\ln |x|]_{1}^{\infty}-[\ln |x|]_{t}^{\infty}$$
$$=\lim_{x\to \infty}\ln|x|-\ln 1-(\lim_{x\to \infty}\ln|x|-\ln |t|)$$ 
$$=\lim_{x\to \infty}\ln|x|-0-\lim_{x\to \infty}\ln|x|+\ln |t|$$ 
$$=\ln |t|$$$$=\ln(t) \ \ \ \ \forall\ \ \  t>0$$
A: Let $$I(t) = \int_0^\infty \dfrac{\exp(-x) - \exp(-xt)}{x} \, dx$$ Then
$$\dfrac{dI}{dt} = \int_0^{\infty} \exp(-xt) \, dx = \left. \dfrac{\exp(-xt)}{-t} \right \vert_0^\infty = \dfrac1t$$
Hence, $$I(t) = \ln(t) + c$$ But $$I(1) = \int_0^\infty \dfrac{\exp(-x) - \exp(-x)}{x} \, dx = \int_0^\infty 0 \, dx = 0 \implies c =0$$ Hence, $$I(t) = \ln(t)$$
A: Here is a method using Laplace transform. 
Fix positive $a,b$ and define function $\displaystyle f(x) = \frac{e^{-ax} - e^{-bx}}{x}$. 
Let $g(s) = $ Laplace transform of $f(x)$. Consider, 
$$ L[ xf(x) ] = -g'(s) \implies L[ e^{-ax} - e^{-bx}] = -g'(s)$$
Thus, we find that,
$$ g'(s) = \frac{1}{s+a} - \frac{1}{s+b} \implies g(s) = \log \left( \frac{s+a}{s+b} \right) + c $$ 
This constant $c=0$, because Laplace transform rule $g(\infty) = 0$. 
But now recall what Laplace transform of $f(x)$ even means by definition, 
$$ \int_0^{\infty} \frac{e^{-ax} - e^{-bx}}{x} \cdot e^{-sx} ~ dx = \log \left( \frac{s+a}{s+b} \right) $$
Set $s=0$ and get the answer. (I just realized I forgot to include the negative sign, but I am too lazy to fix it, so you need to adjust accordingly). 
A: Use of double integral
$$
\begin{aligned}
\int_0^{\infty} \frac{e^{-x}-e^{-x t}}{x} d x =& \int_0^{\infty} \int_1^t e^{-x y} d y d x \\
=& \int_1^t \int_0^{\infty} e^{-x y} d x d y \\
=& \int_1^t\left[\frac{e^{-x y}}{y}\right]_0^{\infty} d y \\
=& \int_1^t \frac{1}{y} d y \\
=& {[\ln y ]_1^t } \\
=& \ln t
\end{aligned}
$$
