Fourier coefficients of 1/(1+it)
Assuming that you want the Fourier Transform of this signal, you can easily obtain it by using a very useful property and a well known FT pair in continuous time. The pair is

$$x(t) = e^{-at}u(t), \: \: a&gt;0 \longleftrightarrow X(f) = \frac{1}{a+j2\pi f}$$

and the so-called duality property says that if $$x(t) \longleftrightarrow X(f)$$ is an FT pair, then:

$$X(t) \longleftrightarrow x(-f)$$

For $a=1$, the first equation gives $$x(t)=e^{-t}u(t) \longleftrightarrow X(f)= \frac{1}{1+j2\pi f}$$

Now you have to apply the duality property to get your result quickly and nicely. That is, since $$x(t) = e^{-at}u(t) \longleftrightarrow X(f) = \frac{1}{a+j2\pi f}$$ the duality property suggests that $$X(t) = \frac{1}{a+j2\pi t} \longleftrightarrow x(-f) = e^{af}u(-f)$$ From that, we can extract the signal we're interested in:
$$X(t)= \frac{\frac{1}{2\pi}}{\frac{a}{2\pi} + jt}= \frac{1}{2\pi}\frac{1}{\frac{a}{2\pi} + jt}\longleftrightarrow x(-f)=e^{af}u(-f)$$ and then $$\frac{1}{\frac{a}{2\pi} + jt} \longleftrightarrow 2\pi e^{af}u(-f)$$ by multiplying both terms by $2\pi$.
Finally, setting $a=2\pi$ yields $$\frac{1}{1+jt} \longleftrightarrow 2\pi e^{2\pi f}u(-f)$$