I have seen that:
$\mathfrak{F} \{e^{i\omega_0x}\} = \sqrt{2 \pi} \delta(\omega_0+\omega)$, using the convention: $\mathfrak{F} \{f(x)\} = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} f(x) e^{i\omega x} dx$
I understand this result by using the property:
$ 2\pi \, \delta(\alpha) = \int_{-\infty}^{\infty} e^{i\alpha x} dx$
But I used the following procedure and got a different answer:
$\mathfrak{F} \{e^{i\omega_0x}\} = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{i\omega_0x+i\omega x} dx =\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{ix(\omega +\omega_0)}dx$
Using: $u = - ix(\omega + \omega_0)$ and $du = -i (\omega + \omega_0) dx$ I got:
$\mathfrak{F}\{e^{i\omega_0x}\} =\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-u} \frac{1}{- i(\omega + \omega_0)}du = \frac{1}{\sqrt{2 \pi}}\frac{1}{- i(\omega + \omega_0)} \int_{-\infty}^{\infty} e^{-u} du = \frac{1}{\sqrt{2 \pi}}\frac{1}{- i(\omega + \omega_0)} \sqrt{\pi} $
So my result is: $\mathfrak{F}\{e^{i\omega_0x}\} = \frac{i}{\sqrt{2} (\omega_0 +\omega)}$
Can anyboday tell me if I am wrong in some step of my procedure or if my result is equivalent to the answer in the first equation.
There are several problems. Here are three:
1) The integral you try to compute is not convergent. You must use distributions, and I suggest you to learn to use them correctly, not via some $\int_{-\infty}^{+\infty}e^{i\alpha x}\,dx=2\pi\delta(\alpha)$ that does note really make sense.
2) You do a change of variables $u=-ix(\omega+\omega_0)$. Complex changes of variables do not work that way. When you have complex functions you should rather think of the integrals as path integrals. So the "change of variables" will typically mean that you also change the contour you integrate along.
3) The integral $\int_{-\infty}^{+\infty}e^{-u}\,du$ is also divergent. The function is very large for (large) negative values of $u$.