Evaluate the limit $\lim_{t\rightarrow\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)$

Question

$$\lim_{t\rightarrow\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)$$ I have no idea where to start. Any help will be appreciated!

Answer 1

Rewrite the limit as a $\frac{0}{0}$ indeterminate form and use L'Hopital.

$$\begin{align} \lim_{t\rightarrow\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)&=\lim_{t\rightarrow\infty}\frac{\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s}{e^{-t}/t}\\ &=\lim_{t\rightarrow\infty}\frac{-e^{-t}/t}{-\left(\frac{e^{-t}(t+1)}{t^2}\right)}\\ &=\lim_{t\rightarrow\infty}\frac{t}{t+1} \end{align}$$

Answer 2

We can use the Incomplete Gamma function just for a little fun.

$$ \Gamma(a,z)=\int_z^{\infty}t^{a-1}e^{-t}dt $$

$$\lim_{t\to\infty}\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}ds\right)= \lim_{t\to\infty}\left(te^t\Gamma(0,t)\right) $$

Now the series expansion of the limit...

$$\lim_{t\to\infty}\left[ te^te^{-t}\left(\frac{1}{t}-\frac{1}{t^2}+\frac{2}{t^3}-\frac{6}{t^4}+\mathcal{O}\left(\frac{1}{t^5}\right)\right)\right] $$

$$\implies \lim_{t\to\infty}\left[ 1-\frac{1}{t}+\frac{2}{t^2}-\frac{6}{t^3}+\mathcal{O}\left(\frac{1}{t^4}\right) \right]=\dots $$

Answer 3

Integrate by parts :$$te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s=1-te^t\int_t^\infty\frac{e^{-s}}{s^2}ds$$

It remains to see that $$\displaystyle \int_t^\infty\frac{e^{-s}}{s^2}ds = o\left(\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)$$

Lemma

Let $f: [a,\infty(\to \mathbb R$ and $\phi : [a,\infty(\to \mathbb R^+$ be such that $f=o_\infty(\phi)$

If $\int_a^\infty\phi$ converges then $\int_x^\infty f(t)dt = o\left(\int_x^\infty \phi(t) dt\right)$

Using the lemma here yields $$te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s +o\left(te^t\int_t^{\infty}\frac{e^{-s}}{s}\text{d}s\right)=1$$

Hence the limit is $1$.