Integrate $\frac{e^{itx}}{1-it}$ I need this integral to find the probability density function from the characteristic function but I don't know how to find it, every method that I try fails. I tried integration by parts but I didn't get the result that I need to. 
The integral is 
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{e^{itx}}{1-it}dt}.$$
The result I need to get by my computations is $e^{-x}$, but I dont know how.
 A: First we rewrite the expression and interpret it as a Fourier transform:
$$
\int_{-\infty}^{\infty} \frac{e^{itx}}{1-it} \, dt
= \int_{-\infty}^{\infty} \frac{1+it}{1+t^2} e^{itx} \, dt
= \mathcal{F}\left\{ \frac{1}{1+t^2} \right\} + \mathcal{F}\left\{\frac{-it}{1+t^2} \right\},
$$
where
$$
\mathcal{F}\{f(t)\} = \int_{-\infty}^{\infty} f(t) \, e^{-ixt} \, dt
$$
Then, according to rule 208,
$$
\mathcal{F}\{ e^{-|t|} \} = \frac{2}{1+x^2}
$$
so by the inversion theorem, rule 105,
$$
\mathcal{F}\left\{ \frac{2}{1+t^2} \right\} = 2\pi \, e^{-|x|}.
$$
Thus,
$$
\mathcal{F}\left\{ \frac{1}{1+t^2} \right\} = \pi \, e^{-|x|}.
$$
For the other term we use rule 107 giving
$$
\mathcal{F}\left\{ \frac{-it}{1+t^2} \right\}
= -i \mathcal{F}\left\{ t \frac{1}{1+t^2} \right\}
= -i \cdot i \left( \pi \, e^{-|x|} \right)' = -\pi \, \operatorname{sign}(x) e^{-|x|}.
$$
Thus,
$$
\int_{-\infty}^{\infty} \frac{e^{itx}}{1-it} \, dt
= \pi \, e^{-|x|} - \pi \, \operatorname{sign}(x) e^{-|x|}
= \pi \left( 1 - \operatorname{sign}(x) \right) e^{-|x|}
= \pi H(-x) e^{-|x|},
$$
where $H$ is the Heaviside function:
$$
H(x) = \begin{cases}
0, & (x<0) \\
1, & (x\geq 0)
\end{cases}
$$
A: $$I_n=\frac 1{2\pi}\int_{-n}^n\frac{e^{itx}}{1-it}dt$$
now let $u=1-it$ $du=-idt\Rightarrow dt=idu$
$$I_n=i\int_{1+in}^{1-in}\frac{e^{(1-u)x}}{u}du=ie^x\int_{1+in}^{1-in}\frac{e^{-ux}}{u}du$$
now we will try and manipulate this integral a bit:
$$\int_{1+in}^{1-in}\frac{e^{-ux}}{u}=\Gamma(0,x(1-in))-\Gamma(0,x(1+in))$$
and your integral is this as $n\to\infty$ with the additional constants at the beginning. Notice that this is undefined/divergent and so we cannot properly express the answer although there way be other interpretations in terms of Fourier
