Query concerning $\int\limits_0^\infty {{e^{\left( {it - 1} \right)x}}dx} = \frac{1} {{1 - it}}$ $$ \int\limits_0^\infty  {{e^{\left( {it - 1} \right)x}}dx}  = \frac{1}
{{1 - it}}.$$
The above came up in a probability question and I was fairly happy it is true, I just don't really feel I understand properly when I go through step by step as follows.
$$ \int\limits_0^\infty  {{e^{\left( {it - 1} \right)x}}dx}  = 
\frac{1}{it-1} \left[ e^{\left( {it - 1} \right)x} \right]_0^\infty =
\frac{1}{it-1} \left( e^{\left( {it - 1} \right)\infty} - e^{\left( {it - 1} \right)0} \right)
$$
Once we get here we seem to say that $(it-1)\leq 1$, and I get that the real part is, because $e^{it}$ goes round the unit circle, but are we just ignoring the non-real part of it? It is a long time since I have done any complex analysis. I was looking at a characteristic function so $t\in\mathbb{R}$ and we are only looking at $x$ over $\mathbb{R}_+$. Thanks for any insight.
 A: The fundamental theorem of calculus proper deals only with integrals over a finite interval. When we use it to evaluate an integral over an infinite interval, say
$$\int_0^\infty f(x)\,dx = F(\infty) - F(0),$$
where $F$ is a primitive of $f$, which we assume sufficiently regular, then this has to be interpreted as a limit,
$$\int_0^\infty f(x)\,dx = \lim_{R\to\infty} \int_0^R f(x)\,dx = \lim_{R\to\infty} F(R) - F(0),\tag{$\ast$}$$
where the integral exists - as an improper Riemann integral - by definition if and only if the first limit exists [and $f$ is Riemann integrable over every finite interval $[0,R]$; but to have a reasonable primitive such that the FTC makes sense, that is necessary]. Since the equality
$$\int_0^R f(x)\,dx = F(R) - F(0)$$
holds by the fundamental theorem of calculus, the first limit in $(\ast)$ exists if and only if the second limit exists. In that case, we interpret $F(\infty)$ as $\lim\limits_{R\to\infty} F(R)$.
For the given integrand $e^{(it-1)x}$ here, we have $F(x) = \frac{1}{it-1}e^{(it-1)x}$, and we have $\lim\limits_{R\to\infty} F(R) = 0$ since $\lvert e^z\rvert = e^{\operatorname{Re} z}$, and the real part of $(it-1)R$ tends to $-\infty$.
