How to prove the limit of this integral?

Question

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.

Answer 1

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.

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

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

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Next, we have the estimates

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

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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!

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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.

Answer 2

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

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\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}

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

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