The characteristic function of a random variable $X$ is given by $$\Phi_X(\omega) = \mathbb{E}e^{j\omega X}=\int_{-\infty}^\infty e^{j\omega x}f_X(x) dx.$$ One can easily capture the similarity between this integral and the Fourier transform.

For a standard normal random variable, the characteristic function can be found as follows: $$\Phi_X(\omega)= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}e^{j\omega x} dx = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(x^2-2j\omega x)}{2}\right)dx $$

I know that the answer must be $\Phi_X(\omega) = \exp(-\omega^2/2)$, but can you explain how to evaluate the integral with a complex number in the exponent?

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    $\begingroup$ You can take the derivative with respect to $\omega$, and integrate by parts to get a differential equation. $\endgroup$ Nov 27, 2011 at 13:32
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    $\begingroup$ Is it a good way to use the fact that "If $f$ and $g$ are real functions then $\int (f + i g) = \int f + i \int g$" math.stackexchange.com/q/85941/1281 ? $\endgroup$
    – Tim
    Nov 27, 2011 at 13:40

2 Answers 2


I will give two answers:

Do it without complex numbers, notice that

$$ \begin{eqnarray} \mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x \\ &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{-j \omega x} \mathrm{d} x \\ &=& 2 \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x \end{eqnarray} $$ Now, compute $\mathcal{F}^\prime(\omega)$, and integrate by parts: $$\begin{eqnarray} \mathcal{F}^\prime(\omega) &=& -\frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} x \sin(\omega x) \mathrm{d} x = \frac{2}{\sqrt{2\pi}} \int_0^\infty \sin(\omega x) \mathrm{d} \left( \mathrm{e}^{-\frac{x^2}{2}} \right) \\ &=& \frac{2}{\sqrt{2\pi}} \left. \mathrm{e}^{-\frac{x^2}{2}} \sin(\omega x) \right|_0^\infty - \frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} \omega \cos(\omega x) \mathrm{d} x \\ &=& - \omega \mathcal{F}(\omega) \end{eqnarray} $$ The solution to so obtained ODE, $\mathcal{F}^\prime(\omega) = - \omega \mathcal{F}(\omega)$ is $\mathcal{F}(\omega) = c \exp\left( - \frac{\omega^2}{2} \right)$, and the integration constant is seen to be one from normalization requirement $\mathcal{F}(0)=1$ of the Gaussian probability density.

Complex integration: As you have started, complete the square: $$ \left( -\frac{x^2}{2} + j \omega x \right) = \left( -\frac{x^2}{2} + j \omega x + \frac{\omega^2}{2} \right) - \frac{\omega^2}{2} = -\frac{1}{2} \left( x - j \omega \right)^2 - \frac{\omega^2}{2} $$ We then have: $$ \mathcal{F}(\omega) = \mathrm{e}^{-\frac{\omega^2}{2}} \cdot \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x $$ The integral $I = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x$ is indeed $1$. To see this, consider $$ \begin{eqnarray} I_L &=& \int_{-L}^L \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x = \int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\ &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\ &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \mathcal{I}_L \end{eqnarray} $$ Here we denoted $\mathcal{I}_L = \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z$. Notice that $\lim\limits_{L \to \infty} \mathcal{I}_L = 1$.

Consider a complex contour $\mathcal{C}$, $ -L \to L \to L - j \omega \to -L - j \omega \to -L$: $$ \begin{eqnarray} I_L - \mathcal{I}_L &=&\left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) \\ &=& -\int_\mathcal{C} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \end{eqnarray} $$ The integral over $\mathcal{C}$ is zero, since the integrand is holomorphic. Therefore: $$ I-1 = \lim_{L \to \infty} (I_L-\mathcal{I}_L) = \lim_{L \to \infty} \left( - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \right) $$ And the limit above is easily seen to vanish. Indeed: $$ \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(-L - j \omega t)^2}{2}} \right| = \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(L^2 + \omega^2 t^2)}{2}} \right| =0. $$

  • $\begingroup$ +1. Thanks! Why instead of directly computing $\int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x $ (btw, I don't know how to directly integrate it), you solved it indirectly by constructing a ODE? $\endgroup$
    – Tim
    Nov 27, 2011 at 14:14
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    $\begingroup$ @Tim It seemed like the simplest approach. The only two other approaches I am aware of are 1) to write $\cos(\omega x)$ into Taylor series and integrate term-wise, 2) use Mellin-convolution theorem. Neither is simpler. $\endgroup$
    – Sasha
    Nov 27, 2011 at 14:24
  • $\begingroup$ Sasha: the $+1$ term which appears at the third line in the computation of $I_L$ should read $1+o(1)$, I believe. $\endgroup$
    – Did
    Dec 10, 2011 at 21:35
  • $\begingroup$ @DidierPiau Welcome back and thank you for the feedback. I have edited the post to make the equation precise. $\endgroup$
    – Sasha
    Dec 10, 2011 at 22:17
  • $\begingroup$ how in the first method, is moving d/dw into the integral justified? $\endgroup$
    – DinkyDoe
    Nov 13, 2012 at 0:19

There is an alternative approach which is based on the relation between moment-generating function and characteristic function. More accurately, assume that $X$ is a random variable such that there exists $\delta\in(0,\infty]$ for which

$$M_X(t)\equiv Ee^{tX}<\infty\ , \ \forall t\in(-\delta,\delta)\,.$$

Then, it is known (e.g., see my answer to How to prove: Moment Generating Function Uniqueness Theorem) that

$$\phi_X(z)\equiv Ee^{zX}\ \ , \ \ \forall z\in \Omega\equiv\left\{x+iy\ ;\ x\in(-\delta,\delta)\right\}$$

is an analytic continuation of $M_X(\cdot)$ to the domain $\Omega$. This implies that for every $t\in\mathbb{R}$ $$M_X(t)=\phi_X(t)$$ and $$\psi_X(t)\equiv Ee^{itX}=\phi_X(it)\,.$$

Now, it is straightforward that for every $z\in\Omega$ it is possible to derive an expresssion for $\phi_X(iz)$ by computing a formula for $\phi_X(z)$ and then insert to this formula $iz$ instead of $z$. Thus, as a special case, in order to derive an expression for $\psi_X(t)$ at some point $t\in(-\delta,\delta)$, it is possible to compute $M_X(t)$ and then to insert $it$ instead of $t$ into the resulting expression.

In particular, if $X\sim N(0,1)$, then it is straightforward to show that for every $t\in\mathbb{R}$ $$M_X(t)=e^{\frac{t^2}{2}}<\infty$$ (that is $\delta=\infty$) and hence, replacing $t$ by $it$ implies that

$$\psi_X(t)= e^{-\frac{t^2}{2}}\ , \ \forall t\in\mathbb{R}\,.$$

  • $\begingroup$ Thanks for the nice proof. One question though: In Sasha's answer using the first method, they obtain $c \exp(-t^2/2)$ as the characteristic function, and the integration constant $c$ would be $1/\sqrt{2\pi}$ if I am not mistaken. Why does your answer not have the factor of $1/\sqrt{2\pi}$? $\endgroup$
    – Leonidas
    Dec 20, 2022 at 16:05

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