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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!

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You have in fact a more general formula, see – Davide Giraudo Jun 29 '12 at 11:09
It may use Lebesgue integration... huh. – Paul Plummer Aug 12 '15 at 5:14
I thought it because it was in just after a theorem about lebesgue integration. – André Gomes Aug 12 '15 at 5:15
up vote 7 down vote accepted

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

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Well, I was wanting a hint, but that's okay. Thank you. – mathmath8128 Jun 28 '12 at 23:23
@platinumtucan I am sorry. I usually just look for the problem and go about solving it. I didn't read that you just wanted a hint. – user17762 Jun 28 '12 at 23:28

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

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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).

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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)

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

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

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

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