How to prove the limit for fucntion $f = 0 $ by Lebesgue Dominated Convergence Theorem:

$$f = \lim_{n\to \infty}\int_0^1 \frac{e^{-nt}-(1-t)^n}{t}\, dt=0$$ I believe I can use the equality: $1-e^{-nt}(1-t)^n=\int_0^t {ne^{-n\tau}\tau(1-\tau)^{n-1}}\, dt$; but after I do this equality n times, I find I get $$\lim_{n\to \infty}\int_0^1 \frac{e^{-nt}}{t} dt$$ However, this will explode when t = 0 but I do not know how to find the function g $\geq \lvert f\rvert$ that to apply the Lebesgue Dominated Convergence Theorem to switch the $\lim$ and $\int$ position.


I thought that it might be of interest to post a solution without appealing to the Dominated Convergence Theorem. To that end, we now proceed.

Enforcing the substitution $t\mapsto t/n$ in the integral of interest, we find that

$$\begin{align} \int_0^1 \frac{e^{-nt}-(1-t)^n}{t}\,dt&=\int_0^n \frac{e^{-t}-(1-t/n)^n}{t}\,dt\\\\ &=\int_0^n \frac{e^{-t}}t \left(1-e^t(1-t/n)^n\right)\,dt\tag1 \end{align}$$

Next, we have the estimates

$$\begin{align} \left|1-e^t(1-t/n)^n\right|&\le 1-(1-t^2/n^2)^n\\\\ &\le t^2/n\tag2 \end{align}$$

Using the estimate from $(2)$ in $(1)$ reveals

$$\left|\int_0^n \frac{e^{-t}}t \left(1-e^t(1-t/n)^n\right)\,dt\right|\le \frac1n \int_0^n te^{-t}\,dt\le \frac1n$$

And we are done!

To use the Dominated Convergence Theorem, simply note that

$$\left|\frac {e^{-t}-(1-t/n)^n}{t} \xi_{[0,n]}(t)\right|\le \frac{e^{-t}-(1-t)\xi_{[0,1]}(t)}t$$, which is absolutely integrable.

  • $\begingroup$ Thanks a lot for your answer! Is amazing! $\endgroup$
    – Horo
    Sep 18 '20 at 20:37
  • $\begingroup$ Nice answer Mark +1 $\endgroup$
    – MtGlasser
    Sep 18 '20 at 20:42
  • 1
    $\begingroup$ @aryadeva Thank you my friend. Much appreciated. $\endgroup$
    – Mark Viola
    Sep 18 '20 at 20:49
  • $\begingroup$ @horo You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Sep 18 '20 at 20:49
  • $\begingroup$ Could I know how do you get the first Inequality at Step (2)? Thanks! $\endgroup$
    – Horo
    Sep 18 '20 at 21:04

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

\begin{align} &\bbox[5px,#ffd]{\lim_{n \to \infty}\int_{0}^{1} {\expo{-nt} - \pars{1 - t}^{n} \over t}\,\dd t} \\[5mm] = &\ \lim_{n \to \infty}\bracks{% \int_{0}^{1}{\expo{-nt} - 1 \over t}\,\dd t + \int_{0}^{1}{1 - \pars{1 - t}^{n} \over t}\,\dd t} \\[5mm] = &\ \lim_{n \to \infty}\bracks{% -\int_{0}^{n}{1 - \expo{-t} \over t}\,\dd t + \int_{0}^{1}{1 - t^{n} \over 1 - t}\,\dd t} \\[5mm] = &\ \lim_{n \to \infty}\bracks{-\operatorname{Ein}\pars{n} + H_{n}} \end{align}

$\ds{\operatorname{Ein}}$ is the Complementary Exponential Integral And $\ds{H_{z}}$ is a Harmonic Number. $$ \mbox{As}\ n \to \infty,\quad \left\{\begin{array}{lcll} \ds{\operatorname{Ein}\pars{n}} & \ds{\sim} & \ds{\ln\pars{n} + \gamma + {\expo{-n} \over n}} & \ds{\color{red}{\large\S}} \\ \ds{H_{n}} & \ds{\sim} & \ds{\ln\pars{n} + \gamma + {1 \over 2n}} & \ds{\color{blue}{\large\#}} \end{array}\right. $$ $$ \mbox{such that}\quad\bbox[5px,#ffd]{\int_{0}^{1} {\expo{-nt} - \pars{1 - t}^{n} \over t}\,\dd t} \sim {1 \over 2n}\quad\mbox{as}\quad n \to \infty $$ \begin{align} & \mbox{} \\ &\ \implies \bbx{\bbox[5px,#ffd]{\lim_{n \to \infty}\int_{0}^{1} {\expo{-nt} - \pars{1 - t}^{n} \over t}\,\dd t} = 0} \\ & \end{align}

$\ds{\color{red}{\large\S}}$: See this link and this one.

$\ds{\color{blue}{\large\#}}$: The asymptotic $\ds{H_{z}}$ behavior is given in the above cited link.


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.