How to apply fourier transform to \$0.5^n u(n)\$
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} =&gt; (\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.