Integration of $|e^{-(2+j)t}|^2$ The integration of $|e^{-(2+j)t}|^2$ from zero to infinity is $1/4$ when I separate above as $|e^{-2t}|^2 \cdot |e^{-2jt}|^2$ and integrate. $|e^{-2jt}|$ was taken as $1$.  But when I integrate the problem taking $|e^{-(2+j)t}|^2 = e^{-(4+2j)t}$ the answer for the integral is $(2-j)/10$. Can someone tell what is the correct method to integrate?
 A: Use the fact that:
$$\| z \|^2 = z\bar{z}$$
$$\| e^{-(2 + j)t} \|^2 = e^{-(2 + j)t}e^{-(2 - j)t} = e^{-4t}$$
Now for the integration:
$$\int_0^\infty \| e^{-(2 + j)t} \|^2 dt = \int_0^\infty e^{-4t} dt = -\frac{1}{4}e^{-4t} \Bigg|_0^\infty = \frac{1}{4}$$
A: $$\lim_{n\to\infty}\int_{0}^{n}\left|e^{-\left(2+i\right)t}\right|^2\space\text{d}t=$$
$$\lim_{n\to\infty}\int_{0}^{n}\left|e^{-2\left(2+i\right)t}\right|\space\text{d}t=$$
$$\lim_{n\to\infty}\int_{0}^{n}\left|e^{\left(-4-2i\right)t}\right|\space\text{d}t=$$
$$\lim_{n\to\infty}\int_{0}^{n}e^{-4t}\space\text{d}t=$$

Substitute $u=-4t$ and $\text{d}u=-4\space\text{d}t$:

$$-\frac{1}{4}\lim_{n\to\infty}\int_{0}^{-4n}e^{u}\space\text{d}u=$$
$$-\frac{1}{4}\lim_{n\to\infty}\left[e^u\right]_{0}^{-4n}=$$
$$-\frac{1}{4}\lim_{n\to\infty}\left(e^{-4n}-1\right)=$$
$$\frac{1}{4}-\frac{1}{4}\lim_{n\to\infty}e^{-4n}=$$
$$\frac{1}{4}-\frac{1}{4}\lim_{n\to\infty}\frac{1}{e^{4n}}=$$
$$\frac{1}{4}-\frac{1}{4}\cdot 0=$$
$$\frac{1}{4}-0=$$
$$\frac{1}{4}$$
