Asymptotic behavior of the sequence $u_n=\int_0^1\frac{e^{-t}}{t+\frac{1}{n}}dt$

Question

I am interested in the asymptotic behavior of the following sequence at infinity $$u_n=\int_0^1\frac{e^{-t}}{t+\frac{1}{n}}dt.$$ I proved that $$\frac{1}{e}\log(n+1)\leq u_n\leq \log(n+1).$$ But can we find an equivalent of this sequence: i.e a sequence $v_n$ such that $$\lim_{n\to\infty}\frac{u_n}{v_n}=1$$ ?

Answer 1

Integrating by parts once, $$ u_n = \left[ e^{-t} \log{(t+1/n)} \right]_0^1 + \int_0^1 e^{-t}\log{(t+1/n)} \, dt \\ = \log{n} + e^{-1}\log{(1+1/n)} + \int_0^1 e^{-t}\log{(t+1/n)} \, dt , $$ and the remaining integral is bounded between $\int_0^1 e^{-t}\log{t} \, dt $ and $\int_0^1 e^{-t} \log{(t+1)} \, dt $, which are both constant and so $o(\log{n})$. The middle term is also $O(1/n)$, whence we conclude that $u_n \sim \log{n}$.

Answer 2

Hint: The reason $u_n$ is large is because of what happens for $t$ close to $0$. So fix $\delta\in(0,1)$ and note that $$u_n\ge e^{-\delta}\int_0^\delta\frac{dt}{t+\frac 1n}.$$

Answer 3

Via a change of variables $u=t+\frac{1}{n}$, you can rewrite $$ u_n = e^{\frac{1}{n}}\int_{1/n}^{1+1/n} \frac{e^{-u}}{u}du\,.\tag{1} $$ The factor $e^{1/n}$ goes to $1$ as $n\to\infty$, so the equivalent will be determined by the integral. Further, as the integral $\int_1^{1+1/n}\frac{e^{-u}}{u}du$ is finite and bounded between $0$ and $\int_1^{2}\frac{e^{-u}}{u}du$, the equivalent will be determined by the rest of the integral, i.e., $$ u_n \operatorname*{\sim}_{n\to\infty} \int_{1/n}^{1} \frac{e^{-u}}{u}du\,.\tag{2} $$ Now, it not hard to show (e.g., by rewriting $\frac{e^{-u}}{u} = \frac{1}{u} - \sum_{k=0}^\infty \frac{(-1)^k u^{k}}{(k+1)!}$ and showing the second term is integrable on $[0,1]$ by theorems for power series) that this last integral is equivalent, when $n\to\infty$, to $\int_{1/n}^{1} \frac{du}{u}$, so that $$ u_n \operatorname*{\sim}_{n\to\infty} \int_{1/n}^{1} \frac{du}{u} = \ln n\,. \tag{3} $$ Therefore, you can take $\boxed{v_n = \ln n}$.

Answer 4

$$u_n = \int_{0}^{n}\frac{e^{-t/n}}{t+1}\,dt = \log(n+1)-\int_{0}^{n}\frac{1-e^{-t/n}}{t+1}\,dt $$ and it is not difficult to prove (by exploiting the convexity of the exponential function) that the inequality $$ 0\leq 1-e^{-t/n} \leq \frac{4}{3}\cdot\frac{t}{t+n} $$ holds for any $t\in[0,n]$. In particular $$0\leq \int_{0}^{n}\frac{1-e^{-t/n}}{t+1}\,dt \leq \frac{4}{3}\int_{0}^{n}\frac{t\,dt}{(t+1)(t+n)}\leq\frac{4}{3}\int_{0}^{n}\frac{dt}{t+n}=\frac{4}{3}\log 2\leq 1 $$ and $u_n=\log(n+1)+O(1)$. By the dominated convergence theorem $$ \lim_{n\to +\infty}\left(u_n-\log(n+1)\right)=-\int_{0}^{1}\frac{1-e^{-t}}{t}\,dt $$ hence $$\boxed{u_n = \log(n+1)+\sum_{m\geq 1}\frac{(-1)^m}{m\cdot m!}+o(1).}$$

Answer 5

By the MVT for definite integrals, there is $\xi\in(0,1)$ such that $$ u_n = e^{\xi}\int_{0}^{1} \frac{1}{t+\frac1n}dt=e^{\xi}\ln(n+1) $$ which implies $u_n=O(\ln n)$.