Show $\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$ for $t \gt 0$

Question

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!

Answer 1

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, $$I(t) = \ln(t)$$

Answer 2

Hint: $\frac{e^{-x}-e^{-xt}}{x} = \int\limits_{1}^{t}{e^{-xs}\text{ d}s}.$

Answer 3

Let $\displaystyle F(t) = \int_{0}^{\infty} \dfrac{e^{-x} - e^{-xt}}{x} dx$

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

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

3. $F(t)$ satisfies the ODE $\dfrac{dF}{dt} = \dfrac{1}{t}$ with initial condition $F(1)=0$ (and $log(t)$ too, so they are equal)

Answer 4

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

$$=\lim_{x\to \infty}\ln|x|-0-\lim_{x\to \infty}\ln|x|+\ln |t|$$

$$=\ln |t|$$$$=\ln(t) \ \ \ \ \forall\ \ \ t>0$$

Answer 5

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).

Answer 6

Use of double integral $$ \begin{aligned} \int_0^{\infty} \frac{e^{-x}-e^{-x t}}{x} d x =& \int_0^{\infty} \int_1^t e^{-x y} d y d x \\ =& \int_1^t \int_0^{\infty} e^{-x y} d x d y \\ =& \int_1^t\left[\frac{e^{-x y}}{y}\right]_0^{\infty} d y \\ =& \int_1^t \frac{1}{y} d y \\ =& {[\ln y ]_1^t } \\ =& \ln t \end{aligned} $$

Answer 7

This is a Frullani type integral. Let $f(x)$ be a continuous function whose limit to infinity and its value at $x=0$ exists. Then we have $$\int_0^\infty\frac{f(ax)-f(bx)}xdx=(f(\infty)-f(0))\ln\left(\frac ab\right)$$Where $f(\infty)=\lim_{x\rightarrow\infty}f(x)$. Substitute $f(x)=e^{-x}$, $a=1$, and $b=t$ to get the answer.