# Evaluating an integral that gives an undetermined infinity form

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 $s>0$. – 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

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 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