Integration with infinity and exponential How is
$$\lim_{T\to\infty}\frac{1}T\int_{-T/2}^{T/2}e^{-2at}dt=\infty\;?$$
however my answer comes zero because putting limit in the expression, we get:
$$\frac1\infty\left(-\frac1{2a}\right) [e^{-\infty} - e^\infty]$$ which results in zero?
I think I am doing wrong. So how can I get the answer equal to $\infty$
Regards

Assume wlog. $a>0$.
By integration,
$$\frac{1}{T}\int^{T/2}_{-T/2} e^{-2at} dt = \frac{1}{2aT}e^{aT} - \frac{1}{2aT}e^{-aT}$$
Note that the second summand goes to zero as $T\rightarrow\infty$ because $1/T$ and $e^{-aT}$ each go to zero. 
However, the first summand goes to infinity, as can be seen most intuitively when looking at the Taylor expansion of the exponential function:
$$\frac{1}{T} e^{aT} = \frac{1}{T} \left( 1 + aT + \frac{(aT)^2}{2} + \cdots \right) =  \frac{1}{T} + a + \frac{a^2T}{2} + \cdots$$ 
