Calculating $\lim_{x\rightarrow \infty}\int_{x}^{+\infty}\frac{e^{-t}}{t}dt$?

Question

I stumbled over this question: Calculate $$f(x)=\int_{x}^{+\infty}\frac{e^{-t}}{t}dt$$ when $x$ approaches $+\infty$. Which made wonder what exactly I am supposed to do. After all we have $$\lim_{x\rightarrow \infty}f(x)=\int_{+\infty}^{+\infty}\frac{e^{-t}}{t}dt$$ which doesn't make any sense to me. What exactly should I do?

Answer 1

For $t \ge 1$ it holds $$e^{-t} \ge \frac{e^{-t}}{t} \ge 0$$

So it follows:

$$0 \le \lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow \infty}\int_{x}^{+\infty}\frac{e^{-t}}{t}dt \le \lim_{x\rightarrow \infty}\int_{x}^{+\infty}e^{-t}dt = \lim_{x\rightarrow \infty}e^{-x} = 0$$

So $$\lim_{x\rightarrow \infty}f(x) = 0$$

Answer 2

Hint $t\mapsto \frac{e^{-t}}{t}$ is continuous and positive.

As $$\lim_{t\to +\infty}t^2\frac{e^{-t}}{t}=0,$$

for enough large $t$,

$$0<\frac{e^{-t}}{t}\leq \frac{1}{t^2}.$$

the integral $$\int_1^{+\infty}\frac{e^{-t}}{t}dt$$ is convergent and your limit is zero as a remainder.

Answer 3

First, $t \mapsto \frac{e^{-t}}{t}$ is an integrable function on $[1,\infty)$ because it is continuous, positive and bounded by $t \mapsto e^{-t}$ which is of course integrable.

When you write $\int_1^{+\infty} \frac{e^{-t}}{t}dt $, you are only considering the limit :

\begin{equation*} \underset{x \to \infty}{\lim} \int_1^x \frac{e^{-t}}{t}dt \end{equation*}

Then if $x$ is a real number greater than one, you get : \begin{equation*} \left| \int_1^{+\infty} \frac{e^{-t}}{t}dt - \int_1^x\frac{e^{-t}}{t}dt \right| = \int_x^{\infty} \frac{e^{-t}}{t}dt \quad \text{by Chasles' relation} \end{equation*}

Therefore since the term on the left converges to zero (by definition!), the term on the right must converge to zero as well.

Hope it clarifies your doubts.

Answer 4

Perhaps what is wanted is the Exponential integral:

https://en.wikipedia.org/wiki/Exponential_integral

In particular, there is an asymptotic estimate for $E_1(z) =\int_z^{\infty}\dfrac{e^{-t}dt}{t} $ of

$E_1(z) =\dfrac1{ze^z}\sum_{n=0}^{N-1}\dfrac{(-1)^nn!}{z^n} +O(N!z^{-N}) $

and the elementary bracketing

$\frac12 e^{-z}\ln(1+\frac{2}{x}) < E_1(z) < e^{-z}\ln(1+\frac{1}{x}) $.

Answer 5

We can in fact develop the large $x$ asymptotic expansion of $I(x)=\int_x^\infty \frac{e^{-t}}{t}\,dt$, $x>0$, by integrating by parts. Proceeding with $u=1/t$ and $v=-e^{-t}$ we find

$$I(x)=\frac{e^{-x}}{x}+\int_x^\infty \frac{e^{-t}}{t^2}\,dt\tag 1$$

It is easy to see that the integral on the right-hand side of $(1)$ is $O(e^{-x}/x^2)$ (simply note that the integrand is bounded by $e^{-t}/x^2$).

Hence, $$I(x)=\frac{e^{-x}}{x}+O\left(\frac{e^{-x}}{x^2}\right)$$Obviously, the limit of $I(x)$ as $x\to\infty$ is $0$.