Let's say $x \in \mathbb{R}$ is normally distributed so that $x \sim \mathcal{N}(\mu,\sigma^2)=p(x)$. I want to simplify the following integral, which includes arbitrary, but fixed integration limits $a\in \mathbb{R}$ and $b\in \mathbb{R}$.

$I := \int_{a}^{b} x^2 p(x)dx$.

In the case of $a=-\infty, b=\infty$ the integral $I$ equals $\sigma^2$, right? Does anyone have an idea how to simplify $I$ with different bounds?

  • 1
    $\begingroup$ With bounds of $\pm\infty$, and assuming by "$p$" you mean the normal density function, this integral equals $\mu^2+\sigma^2$. For arbitrary bounds there are two basic techniques that will work: integrate by parts or use the substitution $y=x^2$. The latter will exhibit the solution as proportional to the sum of two incomplete Gamma integrals. $\endgroup$
    – whuber
    Feb 14, 2018 at 23:11
  • $\begingroup$ Thanks for your comment @whuber. I did forget that $\sigma^2 = \text{var}[x] = \mathbb{E}[x^2]-\mathbb{E}[x]^2 = I_{-\infty,\infty} - \mu^2$, such that $I_{-\infty,\infty} = \sigma^2 + \mu^2$. $\endgroup$
    – Looper
    Feb 16, 2018 at 9:06

2 Answers 2


Hint: Since $\frac{\mathrm d}{\mathrm dx}\phi(x) = -x\phi(x)$ where $\phi(x)$ is the pdf of the standard normal variable, it is possible to find the anti-derivative of $x^2\phi(x)$ by writing it as $x\cdot (x\phi(x))$ and then using the integration by parts formula: $$\int u\, \mathrm dv = uv - \int v \,\mathrm du$$ with $u = x$ and $v = \phi(x)$. Your mission, if you choose to accept it, is to figure out how this idea can be adapted for use with $p(x)$, the pdf of an arbitrary normal random variable. Good luck! This answer will self-destruct in thirty days.


Thanks for the hint @sarwarte. Here's what I got so far:

$ I := \int_a^bx^2p(x)dx = \frac{1}{\sqrt{2\pi\sigma^2}} \int_a^bx^2\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx.$

Sidenote: Using the Substituation $z := \frac{x-\mu}{\sigma}$, and therefore $dx = \sigma dz$.

$I = \frac{1}{\sqrt{2\pi}} \int_{(a-\mu)/\sigma}^{(b-\mu)/\sigma}(z\sigma+\mu)^2\exp(-\frac{z^2}{2})dz.$

Sidenote: Define $\hat{a} := (a-\mu)/\sigma$ and $\hat{b} := (b-\mu)/\sigma$.

$I = \frac{1}{\sqrt{2\pi}} ( \sigma^2\int_{\hat{a}}^{\hat{b}}z^2\exp(-\frac{z^2}{2})dz + 2\mu\sigma \int_{\hat{a}}^{\hat{b}}z\exp(-\frac{z^2}{2})dz + \mu^2 \int_{\hat{a}}^{\hat{b}}\exp(-\frac{z^2}{2})dz)$

Calculating the antiderivative for each of these integrals seperatly.

$(i): \int{z^2\exp(-\frac{z^2}{2})dz} = -z\exp(-\frac{z^2}{2}) + \int{\exp(-\frac{z^2}{2})}dz \\ \qquad \qquad \qquad \qquad = -z\exp(-\frac{z^2}{2}) + \sqrt{\pi/2} \, \text{erf}(z/\sqrt{2}),$

where $erf(\cdot)$ is the error function. Here I used the hint with the partiell integration.

$(ii): \int z\exp(-\frac{z^2}{2})dz = -\exp(-\frac{z^2}{2}).$

$(iii): \int \exp(-\frac{z^2}{2})dz = \sqrt{\pi/2} \, \text{erf}(z/\sqrt{2}).$

Now putting things togehter:

$I = \frac{1}{\sqrt{2\pi}} ( \sigma^2[-z\exp(-\frac{z^2}{2}) + \sqrt{\pi/2} \, \text{erf}(z/\sqrt{2})]_{\hat{a}}^{\hat{b}} + 2\mu\sigma[-\exp(-\frac{z^2}{2})]_{\hat{a}}^{\hat{b}} \qquad + \mu^2 [\sqrt{\pi/2} \, \text{erf}(z/\sqrt{2})]_{\hat{a}}^{\hat{b}}) \\ \qquad = \frac{1}{\sqrt{2\pi}} ( \exp(-\hat{a}^2/2)(\hat{a}\sigma^2+2\mu\sigma) - \exp(-\hat{b}^2/2)(\hat{b}\sigma^2+2\mu\sigma) \\ \qquad + \sqrt{\pi/2} (\sigma^2+\mu^2) (\text{erf}(\hat{b}/\sqrt{2}) - \text{erf}(\hat{a}/\sqrt{2})) ).$

And thats my final result; Mission accomplished?

edit: Result verified.

  • 1
    $\begingroup$ The overall form is about right but in detail there are multiple errors. This is made immediately clear by applying dimensional analysis: for instance, the factor $\hat a + 2\mu$ is the sum of a dimensionless constant $\hat a$ and a constant $\mu$ that has the same units of measurement as $z$ itself. $\endgroup$
    – whuber
    Feb 16, 2018 at 15:21

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