Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm solving a definite integral where one of the borne is infinity. When I try to evaluate the borne at infinity, I'm getting stuck, because I'm getting the undetermined infinity form $ 0 \cdot \infty $. Here is the integral I'm trying to evaluate (it's already solved, I just need to evaluate it).

$$\left[-\frac{te^{-st}}{s} - \frac{e^{-st}}{s^2}\right]_{0^+}^{\infty}$$

And when I try to evaluate it, I get :

$$\left(-\frac{\infty \cdot 0}{s}\right) + \frac{1}{s^2}$$

I know it's possible to modify the borne slightly to evaluate the integral, but I don't think it makes sense to evalute the integral at $\infty^-$.

Also, when I view the formula that I'm integrating, it clearly looks like it's going toward 0, so my feeling tells me that the result should be $\dfrac{1}{s^2}$, but since it's an homework I need to prove it.

share|cite|improve this question
$te^{-st}={t\over e^{st}}$. Use l'Hopital. The result will be as you stated for $s>0$. – David Mitra Dec 3 '11 at 20:27
@DavidMitra that does work, thanks a lot. – HoLyVieR Dec 3 '11 at 20:51
For future reference, French borne in this context is translated limit. – Brian M. Scott Dec 3 '11 at 20:55
@Brian Thanks for mentioning that, I was wondering... – David Mitra Dec 3 '11 at 21:02
@Brian, except that no one uses évaluer la borne à l'infini to say évaluer la limite à l'infini (to evaluate the limit at infinity). The appearance of borne (bound) here is odd. – Did Dec 3 '11 at 22:11
up vote 3 down vote accepted

When you have $\infty$ as one the limits in the integral as below $$\int_0^{\infty} f(t) dt$$ what the integral represents is the following limit $$\lim_{R \rightarrow \infty} \int_0^{R} f(t) dt.$$ Hence, in your case, $$ \begin{align} \int_{0}^{\infty} t e^{-st} dt & = \lim_{R \rightarrow \infty} \int_{0}^{R} t e^{-st} dt\\ \int_{0}^{R} t e^{-st} dt & = \left[ \frac{t e^{-st}}{-s} - \int \frac{e^{-st}}{-s} dt\right]_0^R\\ & = \left[ \frac{t e^{-st}}{-s} - \frac{e^{-st}}{s^2} \right]_0^R\\ & = \left[ \frac{R e^{-sR}}{-s} - \frac{e^{-sR}}{s^2} \right] - \left[ - \frac1{s^2} \right]\\ & = \frac1{s^2} - \frac{R e^{-sR}}{s} - \frac{e^{-sR}}{s^2}\\ \int_{0}^{\infty} t e^{-st} dt & = \frac1{s^2} - \lim_{R \rightarrow \infty} \left( \frac{R e^{-sR}}{s} + \frac{e^{-sR}}{s^2} \right) \end{align} $$ For $s > 0$, the term $\displaystyle \lim_{R \rightarrow \infty} \frac{e^{-sR}}{s^2} = 0$.

The other term can be obtained by l'Hopital as David Mitra has suggest in his comments (or) as follows. Note that when $sR >0$, $e^{sR} > 1 + sR + \frac{s^2R^2}{2}$. Hence, $$0 < e^{-sR} < \frac1{1 + sR + \frac{s^2R^2}{2}}$$ This gives us that $$0 < R e^{-sR} < \frac{R}{1 + sR + \frac{s^2R^2}{2}} = \frac1{\frac1R + s + \frac{s^2R}{2}} < \frac2{s^2R}$$ Hence, $$0 \leq \lim_{R \rightarrow \infty} R e^{-sR} < \lim_{R \rightarrow \infty} \frac2{s^2R} = 0$$ Hence, $\displaystyle \lim_{R \rightarrow \infty} R e^{-sR} = 0$.

In general, you can follow a similar argument as above to conclude that $\displaystyle \lim_{R \rightarrow \infty} R^n e^{-R} = 0$, for any $n \in \mathbb{R}$.

Hence you can conclude that $\displaystyle \int_{0}^{\infty} t e^{-st} dt = \frac1{s^2}$.

share|cite|improve this answer
I understand how you're using the squeeze theorem for the proof, but I don't understand how you get the initial condition that $e^{sR} > 1 + sR + \frac{s^2R^2}{2}$. – HoLyVieR Dec 3 '11 at 21:24
@HoLyVieR: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$ and if $x > 0$, throw away terms from the fourth term, (which are all positive) to get the desired inequality. – user17762 Dec 3 '11 at 21:29
I see now. Thank you that was interesting. – HoLyVieR Dec 3 '11 at 21:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.