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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?

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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 $$\int_0^{\infty} \dfrac{e^{-ax}-e^{-bx}}xdx = \ln(b/a)$$

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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 $$ \begin{align} \int_a^b\frac{f(x)-f(\lambda x)}{x}\,\mathrm{d}x &=\int_a^b\frac{f(x)}{x}\,\mathrm{d}x-\int_a^b\frac{f(\lambda x)}{x}\,\mathrm{d}x\\ &=\int_a^b\frac{f(x)}{x}\,\mathrm{d}x-\int_{\lambda a}^{\lambda b}\frac{f(x)}{x}\,\mathrm{d}x\\ &=\int_a^{\lambda a}\frac{f(x)}{x}\,\mathrm{d}x-\int_b^{\lambda b}\frac{f(x)}{x}\,\mathrm{d}x\\[9pt] &\to v_0\log(\lambda)-v_\infty\log(\lambda)\\[6pt] \int_0^\infty\frac{f(x)-f(\lambda x)}{x}\,\mathrm{d}x &=(v_0-v_\infty)\log(\lambda) \end{align} $$

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