Self Study Inversion Formula Characteric Function Given the characteristic function:
$$
\varphi_X(t) = \frac{e^{\frac{-t^2}{2}}}{1 + it}
$$
i am supposed to describe the corresponding distribution. By the inversion THM, we know that:
$$
f_X(x) = \lim_{T \rightarrow \infty} \frac{1}{2\pi}\int_{\pm T} e^{-itx} \varphi_X(t)dt = \lim_{T \rightarrow \infty} \frac{1}{2\pi} \int_{\pm \infty} e^{-itx} \frac{e^{\frac{-t^2}{2}}}{1 + it} dt = \lim_{T \rightarrow \infty} \frac{1}{2\pi} \int_{\pm T} \frac{e^{\frac{-2itx-t^2}{2}}}{1 + it} dt
$$
Here I am stuck. I don't know how to integrate that. U-sub did not work.
Another approach that i was considering is to think of this as a transformation problem:
$$
\varphi_X(t)(1 + it) = e^{\frac{-t^2}{2}}
$$
which is the characteristic function of the standard normal distribution.
 A: It is not hard to see that the CF in OP is the CF of the random variable $X-Y/2$ where $X \sim \mathcal{N}(0,1)$ and $Y \sim \chi^2(2) $ where $X,Y$ are independent
$$E[e^{it(X-Y/2)}]=E[e^{itX}]E[e^{-itY/2}]=e^{-t^2/2}(1-2i(-t/2))^{-1}=e^{-t^2/2}(1+it)^{-1}$$
Also, note that $-Y/2\sim -\Gamma(1,1)$ where $-\Gamma$ is a Gamma distribution on the negative line. Therefore we can compute the density explicitly by convolution:
$$\begin{aligned}\int_\mathbb{R}f_{-\Gamma (1,1)}(y)f_{\mathcal{N}(0,1)}(x-y)dy&=\int_{(-\infty,0]}e^{y}\frac{1}{\sqrt{2\pi}}e^{-(x-y)^2/2}dy=\\
&=\frac{e^{y+1/2}}{2}\bigg(1-\textrm{erf}\bigg(\frac{y+1}{\sqrt{2}}\bigg)\bigg)\end{aligned}$$
A: In principle, your "another approach" is almost what I had in mind:
Assume we have a nice (differentiable) density function $f(x)$ with
$$\int^\infty_{-\infty}f(x)\,e^{itx}\,dx=\frac{e^{-t^2/2}}{1+it},$$ with
$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0$. Then, by partial integration, we have
$$\int^\infty_{-\infty}f'(x)\,e^{itx}\,dx=-it\,\int^\infty_{-\infty}f(x)\,e^{itx}\,dx,$$ i.e.
$$\int^\infty_{-\infty}(f(x)-f'(x))\,e^{itx}\,dx=(1+it)\,\int^\infty_{-\infty}f(x)\,e^{itx}\,dx=e^{-t^2/2}.$$
Thus, we must have $$f(x)-f'(x)=\frac1{\sqrt{2\pi}}\,e^{-x^2/2}\tag1.$$
Multiplying (1) by $e^{-x}$ gives
$$(f(x)-f'(x))\,e^{-x}=-\frac{d}{dx}(f(x)\,\,e^{-x})=\frac1{\sqrt{2\pi}}\,e^{-x^2/2-x}=\frac{e^{1/2}}{\sqrt{2\pi}}\,e^{-(x+1)^2/2},$$
and integrating, we see that
$$f(x)\,e^{-x}=C-\frac{e^{1/2}}{\sqrt{2\pi}}\,\int^x_{-\infty}e^{-(u+1)^2/2}\,du=C-e^{1/2}\,\Phi(x+1),$$ where $\Phi$ is the cdf of the standard normal distribution. The integration constant $C$ can be determined from $\lim_{x\to\infty}f(x)=0$, we must have $C=e^{1/2}$, and thus
$$f(x)=e^{1/2}\,e^x\,(1-\Phi(x+1)).$$ It's clear that this $f(x)$ has all those nice properties, so our derivation is justified. Integrating (1), we can also obtain the cdf:
$$F(x)=f(x)+\Phi(x)=e^{1/2}\,e^x\,(1-\Phi(x+1))+\Phi(x).$$
