Characteristic function for density $f(x) = \dfrac{1}{2(1 + \vert x \vert)^2}, x \in \mathbb{R}$?

Question

Let $X$ be a random variable on $\mathbb{R}$ with the density $$ f_X(x) = \dfrac{1}{2(1 + \vert x \vert)^2} $$ I want to find the characteristic function $\varphi_X(t)$. Here is what I've done: $$ \begin{align*} \varphi_X(t) = \mathbb{E}\left[e^{itX}\right] &= \int_{\mathbb{R}} \dfrac{e^{itx}}{2(1 + \vert x \vert)^2}dx\\ &= \int_{0}^\infty \dfrac{e^{itx}}{2(1 + x)^2}dx + \int_{-\infty}^0 \dfrac{e^{itx}}{2(1 - x)^2}dx\\ &= \left[\int_0^\infty \dfrac{\cos(tx)}{2(1 + x)^2}dx + \int_{-\infty}^0 \dfrac{\cos(tx)}{2(1 - x)^2}dx\right] + i \left[\int_0^\infty \dfrac{\sin(tx)}{2(1 + x)^2}dx + \int_{-\infty}^0 \dfrac{\sin(tx)}{2(1 - x)^2}dx\right]\\ &= \int_0^\infty \dfrac{\cos(tx)}{(1 + x)^2}dx \end{align*} $$ If $t = 0$, it is easy to check that then $\varphi_X(t) = 1$. Assume $t > 0$, then, by letting $v = tx$, the integral above becomes: $$ \int_0^\infty \dfrac{\cos(tx)}{(1 + x)^2}dx = t \int_0^\infty \dfrac{\cos v}{(v + t)^2}dv $$ And here is where I'm getting stuck at. Any hints would be appreciated, thank you!

Answer 1

We can use integration by parts to express $\int_{0}^{\infty} \frac{\cos(tx)}{(1+x)^{2}} \, \mathrm dx $ in terms of the sine integral function $\operatorname{Si}(x)$ and the cosine integral function $\operatorname{Ci}(x)$.

Assume that $t > 0$.

Then we have

$$\begin{align} \int_{0}^{\infty} \frac{\cos(tx)}{(1+x)^{2}} \, \mathrm dx &= \int_{1}^{\infty} \frac{\cos \left(t(y-1 \right)}{y^{2}} \, \mathrm dy \\ &= - \frac{1}{y} \, \cos \left(t(y-1) \right) \Bigg|_{1}^{\infty}- t \int_{1}^{\infty} \frac{\sin \left(t(y-1) \right)}{y} \, \mathrm dy \\& = 1 - t \int_{1}^{\infty} \frac{\sin \left(t(y-1) \right)}{y} \, \mathrm dy \\ & =1 - t \cos(t)\int_{1}^{\infty} \frac{\sin(ty)}{y} \, \mathrm dy +t \sin(t) \int_{1}^{\infty} \frac{\cos(ty)}{y} \, \mathrm dy \\ & = 1 - t \cos(t) \int_{t}^{\infty} \frac{\sin (u)}{u} \mathrm du + t\sin(t) \int_{t}^{\infty} \frac{\cos (u)}{u} \, \mathrm du \\ &=1 - t \cos(t) \left(\frac{\pi}{2} - \operatorname{Si}(t) \right) - t\sin(t) \operatorname{Ci}(t). \end{align}$$

This is the same result that Wolfram Alpha returns.