Convergence of $\int_{0}^{\infty }{\frac{e^{-x}-e^{-2x}}{x}}dx $

Question

I should find the result of $$\int_{0}^{\infty }{\frac{e^{-x}-e^{-2x}}{x}}dx $$ So I have to study the convergence In 0, i have $$\frac{e^{-x}-e^{-2x}}{x}=\frac{1-x-1+2x+\mathcal O(x^2)}{x}\stackrel{x\to0}\longrightarrow1$$,why can I say that if $$\lim_{x\to0}{\frac{e^{-x}-e^{-2x}}{x}}=1 $$so the integral is convergent in 0?Shouldn't I calculate the primitive firstly then I'll find the limit? Thank you for replying

Answer 1

It is enough to show the function is continuous and bounded in a region close to zero to say that it converges as $x\to0$. To be more precise, one could establish bounds:

$$0<\int_0^1\frac{e^{-x}-e^{-2x}}x~\mathrm dx<\int_0^11~\mathrm dx=1$$

$$0<\int_1^\infty\frac{e^{-x}-e^{-2x}}x~\mathrm dx<\int_1^\infty e^{-x}~\mathrm dx=\frac1e$$

So it converges and is bounded by

$$0<\int_0^\infty\frac{e^{-x}-e^{-2x}}x~\mathrm dx<1+\frac1e$$

Specifically, we may use Frullani's integral to see the given integral is $\ln(2)$.

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To solve the integral in an elementary fashion, consider the more general integral:

$$I(t)=\int_0^\infty\frac{e^{-x}-e^{-(t+1)x}}x~\mathrm dx$$

Let $u=e^{-x}$ to get

$$I(t)=\int_0^1\frac{u^t-1}{\ln(u)}~\mathrm du$$

Now differentiate w.r.t. $t$ to get

$$I'(t)=\int_0^1u^t~\mathrm du=\frac1{t+1}$$

Integrate back to get

$$I(t)-I(0)=\int_0^t\frac1{x+1}~\mathrm dx=\ln(t+1)$$

Since it should be trivial that $I(0)=0$, we find that

$$I(1)=\ln(2)$$

As claimed.

Answer 2

$$ \begin{align} \int_0^\infty\frac{e^{-x}-e^{-2x}}x\,\mathrm{d}x &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-x}-e^{-2x}}x\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}x\,\mathrm{d}x -\int_\epsilon^\infty\frac{e^{-2x}}x\,\mathrm{d}x\right]\\ &=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}x\,\mathrm{d}x -\int_{2\epsilon}^\infty\frac{e^{-x}}x\,\mathrm{d}x\right]\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^{2\epsilon}\frac{e^{-x}}x\,\mathrm{d}x\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^{2\epsilon}\frac{1+O(x)}x\,\mathrm{d}x\\[6pt] &=\lim_{\epsilon\to0^+}(\log(2)+O(\epsilon))\\[9pt] &=\log(2) \end{align} $$

Answer 3

Writing the integral as a double integral then interchanging limits:

$$\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}}{x} dx$$

$$=\int_{0}^{\infty} \int_{-2x}^{-x} e^y\frac{1}{x} dy dx$$

$$=\int_{-\infty}^{0} \int_{-\frac{y}{2}}^{-y} e^y \frac{1}{x} dx dy$$

$$=\int_{-\infty}^{0} e^{y} \ln 2 dy$$

$$=\ln 2$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\expo{-x} - \expo{-2x} \over x}\,\dd x & = \int_{0}^{\infty}\int_{1}^{2}\expo{-xt}\,\dd t\,\dd x = \int_{1}^{2}\int_{0}^{\infty}\expo{-xt}\,\dd x\,\dd t = \int_{1}^{2}{1 \over t}\,\dd t = \bbx{\ln\pars{2}} \end{align}