How to prove the limit of this integral?

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.

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


\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.