# B How can I find this characteristic function

1. Sep 8, 2016

### S_David

Hello all,

I'm trying to find the characteristic function of the random variable $X$ whose PDF is $f_X(x)=1/(x+1)^2$ where $X\in[0,\,\infty)$. I started like this:

$$\phi_X(j\nu)=E\left[e^{j\nu X}\right]=\int_0^{\infty}\frac{e^{j\nu X}}{(x+1)^2}\,dx$$

where $j=\sqrt{-1}$. I searched the Table of Integrals book, and found the formula attached. However, according to the conditions, the real part of $\mu$ must be positive. In my case it's 0. So, I think I cannot use this formula. How then can I evaluate the above integral? I tried integration by parts by letting $u=e^{j\nu x}$ and $dv=dx/(x+1)^2$, which gave me:

$$\int_0^{\infty}\frac{e^{j\nu X}}{(x+1)^2}\,dx=\left. \frac{-e^{j\nu x}}{x+1}\right|_0^{\infty}+j\nu\int_0^{\infty}\frac{e^{j\nu x}}{x+1}\,dx$$

Is this the right way? If yes, then what? I have the same problem again in the right hand side (I mean the integral), besides the first term that I'm not sure if it's 0 or infinity at $x=\infty$.

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2. Sep 12, 2016

### Ssnow

I think you can use a similar formular in your link, if you set in your integral $x+1=-w$ you have:

$=-\int_{-1}^{-\infty}\frac{e^{j\nu-j\nu w}}{w^2}dw=e^{j\nu}\int_{-\infty}^{-1}\frac{e^{-j\nu w}}{w^2}dw$

now if you put this in wolfram Mathematica you will obtain:

$-\frac{j\nu w E_{i}(-j\nu w)+e^{-j\nu w}}{w}$

as primitive. Rewrite with $x$ and take the limit ... ($E_{i}$ is another expression for the exponential integral)