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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.

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    $te^{-st}={t\over e^{st}}$. Use l'Hopital. The result will be as you stated for $s>0$.2011-12-03
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    @DavidMitra that does work, thanks a lot.2011-12-03
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    For future reference, French *borne* in this context is translated *limit*.2011-12-03
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    @Brian Thanks for mentioning that, I was wondering...2011-12-03
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    @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.2011-12-03
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    @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.2011-12-03
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    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.2011-12-03
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    @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.2011-12-03

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

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    I understand how you're using the squeeze theorem for the proof, but I don't understand how you get the initial condition that $e^{sR} > 1 + sR + \frac{s^2R^2}{2}$.2011-12-03
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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.2011-12-03
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    I see now. Thank you that was interesting.2011-12-03