Getting probability density function from a complex characteristic function

Question

I have been trying to convert the characteristic function of the chi-squared distribution:

$$\phi(t) = (1-2it)^{-k/2}$$

to its probability density function using the following equation:

$$f(x) = \frac{1}{2\pi} \int \limits_{-\infty}^{\infty} e^{-itx}(1-2it)^{-k/2}dt$$

The way I do it is as follows:

$$f(x) = \frac{1}{2\pi} \left( \frac{2}{x} \right)^{-k/2} \int \limits_{-\infty}^{\infty} e^{-itx}\left(\frac{x}{2}-itx\right)^{-k/2}dt$$

I make a substitution:

$$ a = \frac{x}{2}; u = -tx; t = -\frac{u}{x}; dt = -\frac{du}{x} $$

$$ f(x) = -\frac{1}{2\pi} \left( \frac{2}{x} \right)^{-k/2} \frac{1}{x} \int \limits_{-\infty}^{\infty} e^{iu}\left(a+iu\right)^{-k/2}du $$

$$ f(x) = -\frac{1}{2\pi} \left( \frac{2}{x} \right)^{-k/2} \frac{1}{x} e^{-a} \int \limits_{-\infty}^{\infty} e^{a+iu}\left(a+iu\right)^{-k/2}du $$

In the book "A course of modern analysis" by Whittaker and Watson I found (chapter 12) that

$$ \frac{1}{\Gamma(z)} = \frac{1}{2\pi} \int \limits_{-\infty}^{\infty} e^{a+iu}(a+iu)^{-z}du $$

which is exactly the integral in my equation with $z = k/2$. Then,

$$ f(x) = -\left( \frac{2}{x} \right)^{-k/2} \frac{1}{x} e^{-a} \frac{1}{\Gamma(k/2)} = -\frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma(k/2)} $$

The obtained result is pretty much the same as the chi-squared PDF but I have the minus sign in front. This minus sign comes from the substitution $dt = -du/x$.

I have tried to understand what the problem could be and it seems to be related to the fact that the characteristic function is complex. If I convert a real characteristic function like the one of the normal distribution then I get the right result. But for the complex characteristic function this is not the case.

In the very end of the following link https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory) I found another formula to calculate a PDF from a characteristic function which is

$$ f(x) = \frac{1}{2\pi} \int_R e^{itx} \overline{\phi(t)} dt $$

where the line over $\phi(t)$ indicates the complex conjugation. If I use this formula I get the right result without the minus sign. But unfortunately there is no reference given on the Wikipedia page and I could not find this formula in any book. Thus, here are my questions:

1. I would highly appreciate if somebody could explain me why I erroneously get the minus sign in my calculations.
2. Is the formula given in the end of the Wikipedia page valid? If so, why do I get different results by applying these two equations:

$$f(x) = \frac{1}{2\pi} \int \limits_{-\infty}^{\infty} e^{-itx}(1-2it)^{-k/2}dt$$ $$ f(x) = \frac{1}{2\pi} \int_R e^{itx} \overline{\phi(t)} dt $$

Answer 1

Using $u=-tx$, with $x>0$, means that when $t$ goes from $-\infty$ to $+\infty$ the substitution variable $u$ goes from $+\infty$ to $-\infty$ so you need to reverse your limits and that will eliminate the minus sign.