Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working on the improper integral:


This function does not have an elementary anti-derivative, so here is what I did: define:

$$f(t):=\int_0^{\infty}\frac{e^{-xt}-e^{-2xt}}{x}dx,\quad t>0$$

Then differentiation gives:


this means $f$ is constant. I feel something is wrong here because $f$ should depend on $t$. Where am I wrong and what is the right way to do this?

share|cite|improve this question
In the integral that is $f(t)$, make the substitution $u=xt$. What do you find out? – Steve Kass Dec 7 '13 at 5:43
@SteveKass Oops! I see the problem, $f(t)$ is actually the same as the original integral – Wayne Dec 7 '13 at 5:45
The antiderivative write ExpIntegralEi[-x] - ExpIntegralEi[-2 x] and the integral is Log[2] as elegantly shown by user17762 – Claude Leibovici Dec 7 '13 at 5:48
This is a special case of Frullani integrals. Let $f : [0,\infty) \to \mathbb{R}$ be any continuously differentiable function such that $f(\infty) = \lim\limits_{x\to\infty} f(x)$ exists. Then for any $a, b \in (0,\infty)$, we have $$\int_0^\infty \frac{f(ax)-f(bx)}{x} dx = ( f(0) - f(\infty) ) \log\frac{b}{a}$$ For proof and generalization of these, see answers of the question – achille hui Dec 7 '13 at 8:54
@achillehui: I had just added that to my answer :-) – robjohn Dec 7 '13 at 8:58
up vote 10 down vote accepted

Note that $$e^{-x} - e^{-2x} = x\int_{1}^{2}e^{-xt}dt$$ Hence, $$\int_0^{\infty} \dfrac{e^{-x}-e^{-2x}}xdx = \int_0^{\infty} \int_{1}^{2}e^{-xt}dtdx = \int_1^2 \int_0^{\infty}e^{-xt}dxdt = \int_1^2\dfrac{dt}t = \ln(2)$$ In general, by similar idea, we have $$\int_0^{\infty} \dfrac{e^{-ax}-e^{-bx}}xdx = \ln(b/a)$$

share|cite|improve this answer
Awesome! What a clever way it is! – Wayne Dec 7 '13 at 5:48
Nice (+1). When I saw this question, I thought in a completely different direction (as I posted). This is an interesting idea. – robjohn Dec 7 '13 at 8:24

$$ \begin{align} \int_a^b\frac{e^{-x}-e^{-2x}}{x}\,\mathrm{d}x &=\int_a^b\frac{e^{-x}}{x}\,\mathrm{d}x-\int_a^b\frac{e^{-2x}}{x}\,\mathrm{d}x\\ &=\int_a^b\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{2a}^{2b}\frac{e^{-x}}{x}\,\mathrm{d}x\\ &=\int_a^{2a}\frac{e^{-x}}{x}\,\mathrm{d}x-\int_b^{2b}\frac{e^{-x}}{x}\,\mathrm{d}x\\[9pt] &\to\log(2)-0 \end{align} $$ as $a\to0$ and $b\to\infty$ since, for any $c\gt0$, $$ e^{-2c}\log(2) \le\int_c^{2c}\frac{e^{-x}}{x}\,\mathrm{d}x \le e^{-c}\log(2) $$

There is nothing special about $e^{-x}$ here. As long as $\lim\limits_{x\to0}f(x)=v_0$ and $\lim\limits_{x\to\infty}f(x)=v_\infty$, then $$ \begin{align} \int_a^b\frac{f(x)-f(\lambda x)}{x}\,\mathrm{d}x &=\int_a^b\frac{f(x)}{x}\,\mathrm{d}x-\int_a^b\frac{f(\lambda x)}{x}\,\mathrm{d}x\\ &=\int_a^b\frac{f(x)}{x}\,\mathrm{d}x-\int_{\lambda a}^{\lambda b}\frac{f(x)}{x}\,\mathrm{d}x\\ &=\int_a^{\lambda a}\frac{f(x)}{x}\,\mathrm{d}x-\int_b^{\lambda b}\frac{f(x)}{x}\,\mathrm{d}x\\[9pt] &\to v_0\log(\lambda)-v_\infty\log(\lambda)\\[6pt] \int_0^\infty\frac{f(x)-f(\lambda x)}{x}\,\mathrm{d}x &=(v_0-v_\infty)\log(\lambda) \end{align} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.