I am reading a paper about tail probability approximation. However, I got into trouble at the very first formula.
The background setting and formula goes like this:
$\bar X=\frac{1}{n}\Sigma_{i=1}^nX_i$ where $X_i$'s are i.i.d.
Hence, following fourier transform, the density of $\bar X$ could be written as:
$f(\bar x)=\frac{n}{2\pi}\int_{-\infty}^\infty e^{n(K(iu)-iux)}du$, where $K(u)$ is the cumulant generating function defined as $K(u)=lnEe^{uX}$.
The tail probability $P(\bar X\geq x_0)= \int_{x_0}^\infty f(\bar x)dx$.
Reversing the order of integration gives $P(\bar X\geq x_0)=\frac{n}{2\pi i}\int_{c-\infty}^{c+\infty} e^{nK(u)-ux}du$.
My problem:
Without rigorous proof, I can follow all the steps except the last one. I did a quick study on fourier transform and with what I learned, I can only get $P(\bar X\geq x_0)=0.5+\frac{n}{2\pi i}\int_{c-\infty}^{c+\infty} e^{nK(u)-ux}du$. Here the difference is that my result is 0.5 greater than the result in the paper.
If it is legit to reverse the order of integration, as the paper suggested, my steps are:
$P(\bar X\geq x_0)=\frac{n}{2\pi}\int_{-\infty}^\infty e^{nK(iu)}\int_{-\infty}^\infty H(x-x_0)e^{-iux}dxdu$ $=\frac{n}{2\pi}\int_{-\infty}^\infty e^{n(K(iu)-iux_0)}(\frac{\pi}{n}\delta(u)+\frac{1}{inu})du$. $\delta(x)$ is the dirac function and $H(x)=\delta([x,+\infty))$.
As you can see, this result has a same part as of that in the paper, but the leftover, the part with $\delta(u)$ yields 0.5.
My question is, where i got wrong?