Problem Definition
I need to calculate the convolution between a Gaussian function $g(x)$ and $d(x)$, the derivative of the inverse error function,
$$d(x) * g(x) = \int_{-\infty}^{\infty} d(\tau) g(x-\tau) \text{d}\tau = \int_{-\infty}^{\infty} d(x-\tau) g(\tau) \text{d}\tau $$
with
$$g(x) = \frac{1}{\sqrt{2 \pi}} \text{e}^{-\frac{1}{2} x^2},$$
$$ d(x) = \left\{ \begin{array}{ll} 0 & x < a \\ \frac{\text{d}}{\text{d}x} erf^{-1}(x) & -a\leq x\leq a \\ 0 & x > a \\ \end{array} \right. $$
with the constant $a<1$.
Since $d(x)$ is only non-zero on the interval $x \in [-a, a] $ I assume I can take these as limits for the convolution integral instead of $[-\infty, \infty]$ (please correct me here).
Attempted solution:
Since $d(x)$ is defined as the derivative of a function, I thought about using integration by parts:
$$\int_{-a}^{a} \frac{\text{d}}{\text{d}\tau} erf^{-1}(\tau) g(x-\tau) \text{d}\tau = \Big[ erf^{-1}(\tau) g(x-\tau) \Big]_{-a}^{a} - \int_{-a}^{a} erf^{-1}(\tau) \frac{\text{d}}{\text{d}\tau} g(x-\tau) \text{d}\tau $$
Then, using the convolution-derivative theorem (see below), we see that the last term is equal to the negative original term,
$$ \frac{\text{d}}{\text{d}x} erf^{-1}(x) * g(x) = erf^{-1}(x) * \frac{\text{d}}{\text{d}x} g(x)$$
Moreover, $erf^{-1}(-x) = - er^{-1}(x)$ (property of an odd function).
Rearranging:
$$ 2 \frac{\text{d}}{\text{d}\tau} erf^{-1}(x) * g(x) = 2 erf^{-1}(a)\Big[g(x-a) + g(x+a) \Big],$$
so, (surprisingly!) the result is a sum of two Gaussian functions,
$$d(x)*g(x) = erf^{-1}(a)\Big[g(x-a) + g(x+a) \Big]$$
I believe this can not be correct, as the numerical convolution doesn't look like a sum of Gaussians at all (see attached plots). There has to be a mistake in my math! Can anyone please point it out to me? Also, if you have suggestions on how to attack the problem, let me know!
Convolution-derivative theorem:
For any two functions $f_1(x)$ and $f_2(x)$, with ' denoting the derivative $\frac{\text{d}}{\text{d}x} $:
$$ ( f_1(x) * f_2(x) )' = f_1(x) * f_2'(x) = f_1'(x) * f_2(x) $$
If it helps, I found that the derivative of the inverse funciton is defined as follows:
$$\frac{\text{d}}{\text{d}u} erf^{-1}(u) = \frac{1}{\frac{\text{d}}{\text{d}u} erf\Big(erf^{-1}(u)\Big)}, $$
with
$$\frac{\text{d}}{\text{d}u} erf(u) = \frac{2}{\sqrt{\pi}} \text{e}^{-x^2}. $$
This would give
$$d(x) = \frac{\sqrt{\pi}}{2} \text{e}^{\Big(\text{erf}^{-1}(u) \Big)^2}, $$
and the following integral for the convolution,
$$ d(x) * g(x) = \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{e}^{\Big(\text{erf}^{-1}(\tau) \Big)^2} \text{e}^{-\frac{1}{2} (x-\tau)^2} \text{d}\tau. $$
It looks difficult to attach directly, which is why I'm guessing playing with the derivatives might be the way to go.