I think you'd better not split the functions. If you want to calculate the fourier transform of \$ 0.5^t u(t)\$, then you might put all that into the integral. Therefore, your integral limits will become \$0\$ to \$+\infty\$ because of \$u(t)\$. And your integral result should not be infinity because your function \$0.5^t\$ goes to zero.

Solving the integral:

\$\large\int_0^\infty e^{(-jwt)} . 0.5^t dt\$

the inside part could be rewrite:

\$\huge\frac{e^{(-jwt)}}{2^t} => (\frac{e^{(-jw)}}{2})^t\$

You could name \$\large\frac{e^{(-jw)}}{2} = u\$
so you get integral of \$\large u^t dt = \frac{u^t}{ln(u)} + C\$

Then you get (replacing u):

\$\huge\frac{(\frac{e^{-jw}}{2})^t}{ ln(e^{-jw}/2)}\$

When t goes to \$+\infty\$ you get zero. When it goes to zero you get:

\$\large\frac{1}{ln(e^{-jw}/2)} \$

which left us:

\$\large\frac{1}{-jw - ln(2)}\$

I'm not sure I did not get lost in calculations but I believe its correct.

There is an important trick to solve such problems. If you have a standard transform pair and you just interchange the variables \$t\$ and \$\omega\$, then you can just use your table if you know the following:

$$\mathcal{F}\{f(t)\}=F(\omega)\Longrightarrow\mathcal{F}\{F(t)\}=2\pi f(-\omega)\tag{1}$$

This is a consequence of the fact that the Fourier transform and the inverse transform are essentially identical, apart from the factor \$2\pi\$ and a minus sign in the exponent. This is exactly what you see in (1): you get a factor of \$2\pi\$ and you have to invert the frequency variable \$\omega\$.

So in your case the standard transform pair is

$$\mathcal{F}\{e^{-t}u(t)\}=\frac{1}{1+j\omega}$$

from which you get using (1)

$$\mathcal{F}\left\{\frac{1}{1+jt}\right\}=2\pi e^{\omega}u(-\omega)\tag{2}$$

PS: The mistake in your calculations is in the last derivative. You compute the derivative as if you considered the function \$e^{-\omega}\$ instead of \$e^{-|\omega|}\$. The correct derivative is

$$\pi\text{sign}(\omega)e^{-|\omega|}$$

which gives the answer

$$X(\omega)=\pi e^{-|\omega|}(1-\text{sign}(\omega))=2\pi e^{-|\omega|}u(-\omega)=2\pi e^{\omega}u(-\omega)$$

which is of course identical to (2).

## Best Answer

By definition of the impulse distribution, we have

$$\int_{-\infty}^{\infty}g(\omega)\delta(\omega)d\omega = g(0)$$

The Fourier synthesis equation is

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}d\omega$$

These are all you need to get the result.