Compute the limit of $\displaystyle\lim_{n \to \infty} \int_0^1 \frac{e^{-nt}}{\sqrt{t}} \, dt$

Question

I am attempting to compute the limit of

$$\displaystyle\lim_{n \to \infty} \int_0^1 \frac{e^{-nt}}{\sqrt{t}} \, dt$$

---

Proof Attempt

Recall that $\int_a^b f(x) \, dx = \displaystyle\lim_{n \to \infty} \int_a^b f_n(x) \, dx$ provided that each $f_n$ is continuous and $f_n \to f$ converges uniformly on $[a,b]$ (Thomson & Bruckner, p. 388).

Denote $\langle f_n \rangle$ be a sequence of functions defined by $f_n(x)= \frac{e^{-nt}}{\sqrt{t}}$ where $n \in \mathbb{N}$ and $t \in [0,1]$. Note that each $f_n$ is continuous on $[0,1]$.

Consider, the fact that $f_n \to f$ pointwise, but does not converge to the function $f$ uniformly on $[0,1]$, however. This is because $t=0$ is undefined for the function $f_n \hspace{0.3cm} \forall n \in \mathbb{N}$. But notice that $f_n \to f$ uniformly on $[a,1]$ for $a>0$. So Theorem 9.26 is applicable for $t \in [a,1]$. Thus, consider the following:

$$(A.) \hspace{0.5cm} \displaystyle\lim_{n \to \infty} \int_0^1 f_n(x) = \displaystyle\lim_{n \to \infty} \int_{\frac{1}{n}}^1 f_n(x)$$

since, for $n$ large enough, $\frac{1}{n} < \varepsilon$ for any $\varepsilon > 0$ (Archimedean Property). Since $\displaystyle\lim_{n \to \infty} \langle \frac{1}{n}: n \in \mathbb{N} \rangle = 0$, the above statement $(A.)$ holds.

---

NOTE: This proof is not complete.

The basic idea of my proof is to observe that $t=0$ breaks the functions $f_n$. So the intuition is to split up $\int f_n \, dt$ into some other integrals that satisfy uniform convergence so that we may use the result from Thomson & Bruckner.

Does this motivation for a proof seem in the right ballpark? Or is there a better means of computing this limit?

Answer 1

You are on the right track! Firstly, I want to note that $$ \lim_{n\to \infty} \int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt$$ does exist. Indeed, on $[0,1]$, $e^{-nt} \geqslant e^{-(n+1)t} $ so $$\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt $$ is monotone decreasing and bounded from below by $0$.

---

Now fix $0<\varepsilon<1$ and write $$\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt = \int_0^\varepsilon \frac{e^{-nt}}{\sqrt{t}}dt +\int_\varepsilon^1 \frac{e^{-nt}}{\sqrt{t}}dt . $$ As you have observed, $f_n $ converges uniformly on $[\varepsilon,1]$. In fact, $f_n \to 0$ uniformly on this interval so $$ \lim_{n \to \infty} \int_\varepsilon^1 \frac{e^{-nt}}{\sqrt{t}}dt = 0 $$ for each $\varepsilon$. Hence, $$\lim_{n \to \infty}\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt = \lim_{n \to \infty}\int_0^\varepsilon \frac{e^{-nt}}{\sqrt{t}}dt. $$ Next, $e^{-nt} \leqslant 1$ so $$\int_0^\varepsilon \frac{e^{-nt}}{\sqrt{t}}dt \leqslant \int_0^\varepsilon \frac1{\sqrt{t}}dt =2 \sqrt{\varepsilon} .$$ Thus, $$ 0 \leqslant \lim_{n \to \infty}\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt \leqslant 2 \sqrt{\varepsilon}. $$ Since this estimate holds for all $\varepsilon>0$ we may take $\varepsilon \to 0^+$ to obtain $$\lim_{n \to \infty}\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt =0. $$

Answer 2

Let me first point out that in this case, one CAN interchange the limit with the integral, so the answer is $\int_0^10\,dt=0$. However, a direct justification for this uses Lebesgue's dominated convergence theorem, which you may not want to invoke if you're trying to solve this problem using only the tools available to a first-year calculus student.

Your general idea of isolating the origin is a good one. However, the way you've done it doesn't seem ideal because your lower bound of integration depends on $n$. The equation you labeled (A) is certainly true (by DCT), however I don't see how it's useful for explicitly evaluating the integral. Instead, what I suggest is the following: fix $0<a<1$ and $n\in\Bbb{N}$. Then, \begin{align} \left|\int_0^1\frac{e^{-nt}}{\sqrt{t}}\,dt\right|&= \int_0^a\frac{e^{-nt}}{\sqrt{t}}\,dt +\int_a^1\frac{e^{-nt}}{\sqrt{t}}\,dt\\ &\leq \int_0^a\frac{1}{\sqrt{t}}\,dt+\int_a^1\frac{e^{-nt}}{\sqrt{t}}\,dt\\ &=2\sqrt{a}+\int_a^1\frac{e^{-nt}}{\sqrt{t}}\,dt \end{align} Now, since $0<a<1$, the function converges to $0$ uniformly on $[a,1]$ as $n\to\infty$, so by taking the limit superior, we have \begin{align} \limsup\limits_{n\to\infty}\left|\int_0^1\frac{e^{-nt}}{\sqrt{t}}\,dt\right|&\leq 2\sqrt{a}+ \int_a^1\limsup_{n\to\infty}\frac{e^{-nt}}{\sqrt{t}}\,dt\\ &=\sqrt{a}+0 \end{align} Finally, since $0<a<1$ was arbitrary, we can let $a\to 0^+$ to deduce \begin{align} \limsup\limits_{n\to\infty}\left|\int_0^1\frac{e^{-nt}}{\sqrt{t}}\,dt\right|&=0. \end{align} Since the $\limsup$ is $0$, the limit itself exists and is also $0$. Hence, \begin{align} \lim_{n\to\infty}\int_0^1\frac{e^{-nt}}{\sqrt{t}}\,dt&=0. \end{align}

Answer 3

The sequence of functions $f_n(t)=e^{-tn}$ converges to $0$ uniformly on any closed subinterval $[a,b]$ with $0<a<b<\infty$. Also, $0<e^{-nt}\leq 1$ for all $0<t<\infty$. Given $\varepsilon>0$, take $0<\delta<\frac{\varepsilon^2}{16}$ so that $$0<\int^\delta_0e^{-nt}\frac{1}{\sqrt{t}}\,dt\leq\int^\delta_0\frac{1}{\sqrt{t}}\,dt=2\sqrt{\delta}<\frac{\varepsilon}{2} $$ On $[\delta,1]$, $f_n$ converges uniformly to $0$. Hence, there is $N$ such that for $n\geq N$, $$0<e^{-nt}<\frac{\varepsilon}{4(1-\sqrt{\delta})}$$ Then, for all $n\geq N$ $$\int^1_\delta e^{-nt}\frac{1}{\sqrt{t}}\,dt\leq \frac{\varepsilon}{4(1-\sqrt{\delta})}2(1-\sqrt{\delta})=\frac{\varepsilon}{2} $$ Putting things together, we obtain that for all $n\geq N$ $$0<\int^1_0 e^{-nt}\frac{1}{\sqrt{t}}\,dt<\varepsilon$$ This shows that $$\lim_{n\rightarrow\infty}\int^1_0 e^{-nt}\frac{1}{\sqrt{t}}\,dt=0$$

Answer 4

Υοu can find the limit without using uniform convergence and so many estimations.

Just make the change of variables $y=\sqrt{t}$ and another change of variables and using the fact that $e^{-y^2}$ is positive, the integral $I_n$ becomes: $$0 \leq I_n=\frac{2}{\sqrt{n}}\int_0^{\sqrt{n}}e^{-y^2}dy \leq \frac{2}{\sqrt{n}}\int_0^{\infty}e^{-y^2}dy \to 0$$

You just have to use that the integral $\int_0^{\infty}e^{-y^2}dy$ converges (and we also know its value)

Answer 5

Let $f_n(t) = \dfrac{e^{-nt}}{\sqrt t}.$ Then $f_n(t) \to 0$ for each $t\in (0,1).$ Furthermore $|f_n(t)|\le \dfrac{1}{\sqrt t}$ on $(0,1), n=1,2,\dots$ Since $\dfrac{1}{\sqrt t}\in L^1(0,1),$ the DCT gives $\int_0^1 f_n(t)\,dt \to 0.$