I am trying to calculate the indefinite integral:

$\displaystyle\int dx \exp \left(-\dfrac{(a x + b)^2}{2}\right) {\rm erf} \left(\dfrac{x}{\sqrt{2}} \right), \quad x \in \mathbb{R}$

Context. This integral arises in the integration of a multivariate Normal distribution with finite integration extremes. Let be $I(x)$ the integral, so far I have been able to show that, if $b=0$:

$I(x) = \dfrac{1}{a}\sqrt{\dfrac{\pi}{2}} {\rm erf} \left( \dfrac{x}{\sqrt{2}} \right) {\rm erf}\left( \dfrac{a x}{\sqrt{2}} \right) + \\ \quad \dfrac{2}{a \sqrt{\pi}} \exp\left(-\dfrac{x^2}{2} \right) \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k!(2k+1)} \left(\dfrac{a}{\sqrt{2}}\right)^{2k+1} \displaystyle\sum_{j=0}^{k}\dfrac{2k!!}{2j!!}x^{2j}$

I have done this by expanding the exponential in the integrand in series, then calculating the first four of the resulting integrals, then figuring out the recurrence formulas in k and j. Part of the series can be summed to ${\rm erf}$, but the second term has proven hard to simplify. Even though the formula above looks like a step in the right direction (to me), I have no idea how to even reach this point if $b\neq0$.


For $b=0$, the integral can be evaluated exactly in terms of Owen's T function (https://en.wikipedia.org/wiki/Owen%27s_T_function), \begin{equation} I = -\frac{\sqrt{\pi}}{\sqrt{2}a} \left[1+4 T\left(ax,\frac{1}{a}\right) \right]. \end{equation} For $b\ne0$, no idea...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.