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

$I=\int_{-T/2}^{T/2}e^{-2at}dt=\left.\frac{-1}{2a}e^{-2at}\right|_{-T/2}^{T/2}=-\frac{e^{-aT}+e^{aT}}{2a}$
Despite that $a$ positive or negative, one of the exponents will tend to zero at our limit, so we can rewrite it as :
$\lim_{T\rightarrow\infty}\frac{I}{T}=\lim_{T\rightarrow\infty}\left(-\frac{e^{-aT}+e^{aT}}{2aT}\right)=\lim_{T\rightarrow\infty}\left(-\frac{e^{\left|a\right|T}}{2aT}\right)
 $
But becuase exponental functions are growing much faster than $T$ , this makes the limit is always infinity, thus finaly we have:
$\lim_{T\rightarrow\infty}\frac{I}{T}=\begin{cases}
-\infty & a>0\\
+\infty & a<0
\end{cases}$
