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