Odd Inverse of a Characteristic function I saw this formula today:
$$ \mathbb{P} \left[ X > K \right] = 
 \frac{1}{2} + \frac{1}{\pi} \int_0^\infty Re\left( \frac{e^{-iuK}\varphi(u)}{iu} \right) du $$
Where $\varphi(u)$ is the characteristic function of the random variable $X\in L^1$.
I must definitely be missing something since I can only get to:
$$
\mathbb{P} \left[ X > K \right] = 
 \frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{-iuK}\varphi(u)}{iu}  du 
$$
Do any of you out there know how to show this?
Any help would be appreciated!
 A: I will just consider the case when $X$ has a pdf.
First let me define the Fourier transform of a function $f(x)$ to be:
$$
\hat{f}(w) = \int_{\mathbb{R}} f(x)e^{-iwx}dx
$$
Note that we have the identity:
$$
\langle f, g \rangle = \frac{1}{2\pi}\langle \hat{f}, \hat{g} \rangle
$$
This is one version of Parseval's identity which assumes $f,g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ (at least in the version I'm familiar with).
Ok, now assume that $X$ has some density function $p(x)$. We're looking at:
\begin{align*}
\int_K^{\infty} p(x) dx &= \int_{\mathbb{R}} \chi_{[K,\infty)}(x) p(x) dx \\
&= \int_{\mathbb{R}} \chi_{[0,\infty)}(x-K) p(x) dx
\end{align*}
Where $\chi$ is the usual indicator function so that $\chi_{[0,\infty)}$ is the Heaviside step function. Using the above identity, we have:
\begin{align*}
\int_{\mathbb{R}} \chi_{[K,\infty)}(x) p(x) dx &= \frac{1}{2\pi} \int_{\mathbb{R}} \left( \frac{1}{iw} + \frac{1}{\pi}\delta(w) \right) e^{-iKw} \hat{p}(w) dw \\
&= \frac{1}{2} + \frac{1}{2\pi} \int_{\mathbb{R}} \frac{e^{-iKw} \hat{p}(w)}{iw} dw
\end{align*}
Note that the Fourier transform of the step is a distribution (see here for instance), and $\chi$ is certainly not in $L^1$ or $L^2$. So this isn't as airtight as it could be at this point.
Finally, consider the second term:
\begin{align*}
\int_{\mathbb{R}} \frac{e^{-iKw} \hat{p}(w)}{iw} dw &= \int_0^{\infty} \frac{e^{iKw}\hat{p}(-w)}{-iw} + \frac{e^{-iKw}\hat{p}(w)}{iw} dw \\
&= 2\int_0^{\infty} Re\left( \frac{e^{-iKw}\hat{p}(w)}{iw} \right) dw
\end{align*}
Since $\hat{p}(-w) = \hat{p}(w)^*$, the complex conjugate of $\hat{p}(w)$.
