So I just watched this video about the Laplace transformation and I wanted to follow along on the steps. Timestamp is $3:14$.
In the beginning he starts out with a fourier transformation of this function: $$f(t) = e^{-t} \cdot \sin(t)$$
He goes on to say that this is pretty easy using integration by parts and viewing $i$ as a constant. So far so good.
This is the expected outcome of the Fourier transform:
$$ X(\omega) = \frac{1}{1+(1+i\omega)^2} $$
So I try the fourier transform like this: $$X(\omega) = \int_0^\infty e^{-t} \cdot \sin(t) \cdot e^{-i\omega t} dt$$ simplify $$X(\omega) = \int_0^\infty e^{-i\omega t - t} \cdot \sin(t) dt$$ first integration by parts $$X(\omega) = e^{-i\omega t - t} \cdot (-\cos(t)) - (i \omega +1) \int_0^\infty e^{-i\omega t - t} \cdot \cos(t) dt$$ So far so good, lets go for the second round $$X(\omega) = e^{-i\omega t - t} \cdot (-\cos(t)) - (i \omega +1) \cdot ( e^{-i\omega t - t} \cdot (-\sin(t)) -(i\omega+1)\int_0^\infty e^{-i\omega t - t} \cdot \sin(t) dt)$$ Now the last part is the same as the Equation we started out with and we get $$X(\omega) = \frac{-e^{-i\omega t - t} \cdot (\cos(t)) + (i \omega +1) \cdot ( e^{-i\omega t - t} \cdot (\sin(t))}{(i\omega+1)^2}$$
How do I get from this to here $$ X(\omega) = \frac{1}{1+(1+i\omega)^2}~? $$
I mean the $(i\omega+1)^2$ part looks promising but I am quite rusty in calculus. Is it a special property of the Fourier transform i have to use? Because lots of fourier transforms have 1's in them. from what i have gathered. I tried playing around with eulers identity but so far no luck.
Help me, please and thank you.
EDIT : The Function $f(t)$ is only defined from $0$ to $\infty$
EDIT 2 : So i guess it should go something like this:
$$X(\omega) = \frac{[-e^{-i\omega t - t} \cdot (\cos(t))]_0^\infty + (i \omega +1) \cdot [( e^{-i\omega t - t} \cdot (\sin(t))]_0^\infty}{(i\omega+1)^2}$$
How am I supposed to evaluate $cos (\infty)$, that is non existant.
EDIT 3: Ok, so the first [] is $$ [-e^{-i\omega t - t} \cdot (\cos(t))]_0^\infty $$ makes $$ 0 - 1 = -1 $$ The second [] is just 0 which means that the solo $(1+i\omega)$ vanishes
So the whole thing just becomes: $$ \frac{1}{(i\omega+1)^2} $$
EDIT 4:
If I go from here: $$ X(\omega) = [e^{-i\omega t - t} \cdot (-\cos(t))]_0^\infty - (i \omega +1) \cdot [( e^{-i\omega t - t} \cdot (-\sin(t)) ]_0^\infty-(i\omega+1)\int_0^\infty e^{-i\omega t - t} \cdot \sin(t) dt)$$ which is to make it a little clearer $$X(\omega) = 1 - (i \omega +1) \cdot 0 -(i\omega+1)^2\int_0^\infty e^{-i\omega t - t} \cdot \sin(t) dt$$
which is $$X(\omega) = 1 - (i \omega +1) \cdot 0 -(i\omega+1)^2 X(\omega)$$
So I bring $-(i\omega+1)^2 X(\omega)$ to the other side of the = I factorize and bring anything that is not $X(\omega)$ back again which gives me
$$X(\omega) = \frac{1-0}{(i\omega+1)^2 } $$
EDIT5: I'm dumb and made a mistake in my head in the last step. Good night yall.
$$X(\omega) = \int_0^\infty e^{-t} \cdot \sin(t) \cdot e^{-i\omega t} dt$$ Integrating by part: $$X(\omega) = [-e^{-i\omega t - t} \cdot (\cos(t))]_0^\infty - (i \omega +1)( \int_0^\infty e^{-i\omega t - t} \cdot \cos(t) dt)$$ $$ [-e^{-i\omega t - t} \cdot (\cos(t))]_0^\infty=0\color{red}{+1} $$ The evaluation in the first by part integration is $+1$ not $-1$ $$X(\omega) = 1 - (i \omega +1)( \int_0^\infty e^{-i\omega t - t} \cdot \cos(t) dt)$$ Since as you wrote in the comment: $$ [e^{-i\omega t - t} \cdot (\sin(t))]_0^\infty=0 $$ We have that: $$X(\omega) = 1 - (i \omega +1)^2 \cdot ( \int_0^\infty e^{-i\omega t - t} \cdot \sin(t) dt)$$
$$X(\omega) = 1 - (i \omega +1)^2 X(\omega) $$ Put the integral on the left side and factorize:
$$X(\omega)(1+(i \omega +1)^2 ) = 1 $$ $$\implies X(\omega)=\frac 1 {1+(i \omega +1)^2 }$$