This is laplace transform of $\dfrac{1-e^{-t}}{t}$ and the integral exists according to wolfram

Do i get any help/hints about how to work this ? I have been trying integration by parts with different combinations for u and dv but none of them are working. Any help is appreciated thanks!



$$I(s) = \int_0^\infty \frac{1-e^{-t}}{t} e^{-st}\,dt$$

for $\operatorname{Re} s > 0$. Compute $I'(s)$ (by differentiating under the integral sign), and from that obtain $I(s)$ by noting that $\lim\limits_{\operatorname{Re} s \to +\infty} I(s) = 0$ by the dominated convergence theorem.

  • $\begingroup$ (+1) In the last step, we get $$I(s)=\log\left(1+\frac{1}{s}\right).$$ $\endgroup$ – Jack D'Aurizio Nov 16 '14 at 16:00
  • $\begingroup$ Thank you so much :) I am trying this on my notes $\endgroup$ – rsadhvika Nov 16 '14 at 16:01
  • $\begingroup$ I am getting $$I'(s) = -\dfrac{1}{s+1}$$ integrating both sides gives $$I(s) = \ln\left(\dfrac{1}{s+1}\right)$$ $\endgroup$ – rsadhvika Nov 16 '14 at 16:44
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    $\begingroup$ That's not correct as you probably know. You should get $\frac{1}{s+1} - \frac{1}{s}$. What integral did you get by differentiating under the integral sign? $\endgroup$ – Daniel Fischer Nov 16 '14 at 16:47
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    $\begingroup$ You don't get a factor of $s$ down by differentiating. $$\frac{\partial}{\partial s} e^{-st} = -t\cdot e^{-st}.$$ So it's $$I'(s) = \int_0^\infty (e^{-t}-1)\cdot e^{-st}\,dt.$$ $\endgroup$ – Daniel Fischer Nov 16 '14 at 16:54

Use the fact that

$$\frac{1-e^{-t}}{t} = \int_0^1 du \, e^{-t u}$$

So the integral in question is

$$\int_0^{\infty} dt \, \int_0^1 du \, e^{-t (u+s)} $$

Because all integrals here converge absolutely, we may switch the order of integration and get

$$\int_0^1 du \, \int_0^{\infty} dt \, e^{-t (u+s)} = \int_0^1 \frac{du}{u+s} = \log{\left ( 1+\frac1{s} \right )}$$


A last way similar to both Ron Gordon and Daniel Fischer is to use an usefull proposition related to the laplace transform.

$$\mathcal{L}\left( \frac{f(t)}{t}\right) = \int_s^\infty F(\sigma) \,\mathrm{d}\sigma$$

Given that the integral exists, eg that there exists some $M$ such that $f(t)e^{-t} < M$ for all $t$. Here we have $$f(t) = 1-e^{-t} \quad \text{ and } \quad F(t) = \frac1\sigma - \frac1{1+\sigma}$$ Which you should know how to compute by now, or look it up in a standard laplace table. Hence $$ \begin{align*} \int_0^{\infty} \frac{1-e^{-t}}{t}e^{-st}\,\mathrm{d}t & = \mathcal{L}\left(\frac{1-e^{-t}}{t}\right) \\ & = \int_s^\infty \mathcal{L}\left( 1-e^{-t}\right) \mathrm{d}\sigma \\ & = \int_s^\infty \frac1\sigma - \frac1{1+\sigma}\mathrm{d}\sigma \\ & = \log (1+s) - \log s = \log \left( 1 + \frac{1}{s} \right) \end{align*}$$ And we are done $\square$


An easy way to evaluate the integral is using Frullani's theorem $$\int_0^\infty \frac{f(at)-f(bt)}{t}\,dt=\bigg[f(0)-f(\infty)\bigg]\ln\left(\frac{b}{a}\right)$$ Taking $f(t)=e^{-t},\, a=s,$ and $b=1+s$ then the integral is simply evaluated to $$\ln\left(\frac{1+s}{s}\right)$$ which matches up other's answers.

  • $\begingroup$ This is the approach I would have taken. +1 $\endgroup$ – Mark Viola Jan 29 '16 at 21:20

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#66f}{\large\int_{0}^{\infty}{1 - \expo{-t} \over t}\,\expo{-st}\,\dd t} =\int_{0}^{\infty}\bracks{\expo{-st} - \expo{-\pars{s + 1}t}}\ \overbrace{\int_{0}^{\infty}\expo{-tx}\,\dd x\,\dd t}^{\ds{=\ \color{#c00000}{1 \over t}}} \\[5mm]&=\int_{0}^{\infty}\int_{0}^{\infty} \bracks{\expo{-\pars{s + x}t} - \expo{-\pars{s + 1 + x}t}} \,\dd t\,\dd x =\int_{0}^{\infty}\pars{{1 \over x + s} - {1 \over x + s + 1}}\,\dd x \\[5mm]&=\left.\ln\pars{x + s \over x + s + 1} \right\vert_{\, x\ =\ 0}^{\, x\ \to\ \infty} =-\ln\pars{s \over s + 1}=\color{#66f}{\large\ln\pars{1 + {1 \over s}}} \end{align}

Another way to evaluate the integral is: \begin{align} &\color{#66f}{\large\int_{0}^{\infty}{1 - \expo{-t} \over t}\,\expo{-st}\,\dd t} =s\int_{0}^{\infty}\ln\pars{t}\expo{-st}\,\dd t -\pars{s + 1}\int_{0}^{\infty}\ln\pars{t}\expo{-\pars{s + 1}t}\,\dd t \\[5mm]&=\int_{0}^{\infty}\ln\pars{t \over s}\expo{-t}\,\dd t -\int_{0}^{\infty}\ln\pars{t \over s + 1}\expo{-t}\,\dd t =\ln\pars{s + 1 \over s}\ \underbrace{\int_{0}^{\infty}\expo{-t}\,\dd t}_{\ds{=\ \color{#c00000}{1}}} \\[5mm]&=\color{#66f}{\large\ln\pars{1 + {1 \over s}}} \end{align}


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