Integral $\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx$

Question

I am working on the improper integral:

$$\int_0^{\infty}\frac{e^{-x}-e^{-2x}}{x}dx$$

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:

$$f'(t)=\int_0^{\infty}\frac{-xe^{-xt}+2xe^{-2xt}}{x}dx=\int_0^{\infty}-e^{-xt}+2e^{-2xt}dx=0$$

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?

Answer 1

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

Answer 2

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

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

Answer 3

We are going to use Feynman’s Integration Technique with the integral $$ I(a)=\int_0^{\infty} \frac{e^{-x}-e^{-a x}}{x} d x $$ Differentiating w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} e^{-a x} d x \\ & =\left[\frac{e^{-a x}}{-a}\right]_0^{\infty} \\ & =\frac{1}{a} \end{aligned} $$ Integrating back $1$ to $2$ yields $$ \begin{aligned} I & =I(2)-I(1) \\ & =\int_1^2 I^{\prime}(a) d a \\ & =\int_1^2 \frac{1}{a} d a \\ & =\ln 2 \end{aligned} $$ In general, replacing $1,2$ by $b,a$ gives $$\int_0^{\infty} \frac{e^{-bx}-e^{-a x}}{x} d x=\ln \left|\frac{a}{b}\right| $$