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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.

Asaf Karagila
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HoLyVieR
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    $te^{-st}={t\over e^{st}}$. Use l'Hopital. The result will be as you stated for $s>0$. – David Mitra Dec 03 '11 at 20:27
  • @DavidMitra that does work, thanks a lot. – HoLyVieR Dec 03 '11 at 20:51
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    For future reference, French borne in this context is translated limit. – Brian M. Scott Dec 03 '11 at 20:55
  • @Brian Thanks for mentioning that, I was wondering... – David Mitra Dec 03 '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 03 '11 at 22:11
  • @Didier: I wondered a bit, since so far as I know borne is only limit in the sense of boundary, but I couldn’t imagine what else it could be. I’m really quite curious at this point. – Brian M. Scott Dec 03 '11 at 22:19
  • Well we are using the term borne d'intégration (in French and in Quebec) to say the value at which we are evaluating it. I'm not sure if it was the exact way to say it, but that's the way it's written my textbook. – HoLyVieR Dec 03 '11 at 22:20
  • @HoLyVieR, exactly. Here the borne d'intégration is infinity and one evaluates a function at infinity, so there is no way one can evaluate the borne itself. – Did Dec 03 '11 at 22:28

1 Answers1

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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 $\displaystyle \int_{0}^{\infty} t e^{-st} dt = \frac1{s^2}$.

  • 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 03 '11 at 21:24
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    @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. –  Dec 03 '11 at 21:29
  • I see now. Thank you that was interesting. – HoLyVieR Dec 03 '11 at 21:36