The question is:
this integral is finite or unlimited? $$ I=\int^{\infty}_{-\infty}|f(t)|^2 dt $$ where $$ f(t) = u(t)e^{-t}\cos(2\pi t) $$
I calculate the Fourier transform of $f(t)$:
$$F(t) = \frac{1+jw}{4\pi^2 + (1+jw)^2}\ $$
I want to use the Parseval theorem to understand where this integral is finite or unlimited.
but when I take its size with a lot of calculations, It is difficult to integrate it, and I can't find the answer.
Thank you if you can tell me another way.
Parseval's is overkill here, we can simply integrate the function as follows:
$$\int_{-\infty}^{\infty} |u(t)e^{-t}\cos(2\pi t)|^2 dt = \int_0^{\infty}|e^{-t}\cos(2\pi t)|^2 dt + \int_{-\infty}^0 0 dt$$
$$\int_0^{\infty} |e^{-t}\cos(2\pi t)|^2 dt = \int_0^\infty |e^{-2t}\cos^2(2\pi t)| dt = \int_0^\infty e^{-2t}\cos^2(2\pi t) dt$$
From here we know the integral has to converge because $0 \leq \cos^2(2\pi t) \leq 1$ and $e^{-2t} > 0$ for all real $t$ give us that $0 \leq e^{-2t}\cos^2(2\pi t) \leq e^{-2t}$, so $0 \leq \int_0^{\infty} e^{-2t}\cos^2(2 \pi t) dt \leq \int_0^{\infty} e^{-2t} dt$, where $\int_0^{\infty} e^{-2t} dt$ converges and equals $\frac{1}{2}.$
Now we can use the definition of $\cos x$ yielded from Euler's formula to simplify. (and just for you, I'll use $j$)
$$\int_0^\infty e^{-2t}\left(\frac{e^{2\pi jt} + e^{-2\pi j t}}{2}\right)^2 dt = \frac{1}{4} \int_0^{\infty} e^{-2t}(e^{4\pi j t}+2+e^{-4\pi j t}) dt$$
Distributing and breaking up the integral we now get:
$$\frac{1}{4}\Big(\int_0^{\infty} 2e^{-2t} dt + \int_0^{\infty} e^{(4\pi j - 2)t} dt + \int_0^{\infty} e^{(-4\pi j - 2)t} dt\Big)$$
Notice that all of the integrals converge because the integrands are of the form $e^{at}$ with $Re(a) < 0.$
Taking the integrals we now have:
$$\frac{1}{4}\Big(1 + \frac{1}{2-4\pi j}+ \frac{1}{2 + 4\pi j}\Big) = \frac{1}{4}\Big(1 + \frac{2(2)}{2^2 + (4\pi)^2}\Big) = \frac{1}{4} + \frac{1}{4 + 16\pi^2}$$