10
$\begingroup$

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!

$\endgroup$
15
$\begingroup$

Write

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

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.

$\endgroup$
  • $\begingroup$ (+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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
14
$\begingroup$

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

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

$\endgroup$
9
$\begingroup$

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 $\square$

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is the approach I would have taken. +1 $\endgroup$ – Mark Viola Jan 29 '16 at 21:20
4
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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