Integrating the errorfunction using the method of undetermined coefficients

Question

$\DeclareMathOperator\erf{Erf}$ I am trying to solve using the method of undetermined coefficients:

$$\int\erf(x)dx$$

With the method of undetermined coefficients one would start by "simply" guessing a general solution to the integral.

In the case of $\int\erf(x)dx$ my calculusbook suggests we try:

$$\int\erf(x)=P(x)\erf(x)+Q(x)e^{-x^2}+C$$

Here P and Q are polynomials to be determined.

Question: Where on earth does this guess come from? I don't intuit it at all. Any help would be appreciated.

Answer 1

What could be on the right hand side of $$\int f(x)\,\mathrm dx = \ldots? $$ After differentiating, the left hand side becomes $f(x)$, of course. So we want something that produces - initially perhaps with some additional "waste" - $f(x)$. By the product rule, $xf(x)$ does produce $f(x)$ (and also $xf'(x)$, with which we have to deal as well), so allowing $xf(x)$ would be a good idea. A slight generalization is to allow not just $x$ times $f(x)$, but in fact $P(x)f(x)$ for arbitrary polynomial $P$, as well as $Q(x)f'(x)$ in order to somehow hopefully cancel the "waste". We may want to continue with polynomial multiples of $f''(x)$ etc., but fortunately in our specific example, $f''$ is already a polynomial multiple of $f'$ ...