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Mathematica will gladly tell me that the integral

$$ I\left[y,a\right]=\int_{y}^{\infty}dx\,e^{-x^{2}}\mathrm{erf}\left(ax\right)$$

where $\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}dt\,e^{-t^{2}}$ is the error function, can be written as

$$ I\left[y,a\right]=-\frac{1}{2} \sqrt{\pi } \left(4 T\left(\sqrt{2} a y,\frac{1}{a}\right)+\mathrm{erf}(y)\, \mathrm{erf}(a y)-1\right)$$

where $$T(x,a) =\frac{1}{2 \pi }\int_0^a \frac{e^{-\left(t^2+1\right) x^2/2}}{ t^2+1} \, dt$$ is Owens T-function.

How is this derived? And more importantly: Can a similar result be derived for multiple error functions like $$ I\left[y;a_{1},\ldots a_{n}\right]=\int_{y}^{\infty}dx\,e^{-x^{2}}\prod_{j=1}^{n}\mathrm{erf}\left(a_{j}x\right) $$

My end goal is to compute integrals like $$ G_{n}=\int_{-\infty}^{\infty}dx_{0}\,\prod_{j=1}^{n}\int_{x_{j-1}}^{\infty}dx_{j}\,e^{-\sum_{j=0}^{n}x_{j}^{2}} \prod_{j=0}^n x_j^{p_j}$$ where the $p_j$ are "not-too-large" non-negative integers. In these integrals, multiple error functions naturally pop up.

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    $\begingroup$ I think the result follows from differentiation with respect to the integral sign w.r.t. $a$. $\endgroup$
    – projectilemotion
    Apr 29, 2021 at 13:54
  • $\begingroup$ @projectilemotion Thank you. I now also realize that mathematica is (as usual?) not giving me the "simplest" for of the T-function. $\endgroup$
    – Mikael Fremling
    Apr 29, 2021 at 14:15
  • $\begingroup$ Indeed, I obtain the result $2\sqrt{\pi}\cdot T(\sqrt{2}y,a)$, which agrees numerically with the result you wrote. $\endgroup$
    – projectilemotion
    Apr 29, 2021 at 14:41
  • $\begingroup$ Also, I don't think a "nice" result for multiple error functions can be derived for multiple error functions. Even the relatively simple case $y=0$ has complicated results for $n\geq 2$: An integral involving error functions and a Gaussian. $\endgroup$
    – projectilemotion
    Apr 29, 2021 at 15:23
  • $\begingroup$ @projectilemotion Indeed it get's pretty ugly very quickly. It seems however that one can keep the number or error functions down by converting them to T-functions, whether that helps in the long run remains to be seen... $\endgroup$
    – Mikael Fremling
    Apr 30, 2021 at 8:35

1 Answer 1

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This is not a full answer, more of a note. Using integration by parts, notice that: $$\int_0^\infty e^{-x^2}\operatorname{erf}(ax)\,dx=\frac{\sqrt{\pi}}{2}-\int_0^\infty e^{-x^2}\operatorname{erf}\left(\frac xa\right)\,dx$$ and you can try to split your integral up into: $$\int_y^\infty=\int_0^\infty-\int_0^y$$

Addressing what others have said: $$I(y,a)=\int_y^\infty e^{-x^2}\operatorname{erf}(ax)\,dx$$ $$\frac{\partial I(y,a)}{\partial a}=\int_y^\infty e^{-x^2}\frac{\partial}{\partial a}\left[\operatorname{erf}(ax)\right]\,dx=\frac{2}{\sqrt{\pi}}\int_y^\infty xe^{-(a^2+1)x^2}dx$$ $$=\frac{1}{\sqrt{\pi}}\frac{e^{-(a^2+1)y^2}}{(a^2+1)}$$ now you need to integrate wrt $a$, which you will see brings in our $T$ function

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  • $\begingroup$ Thanks for noticing. The partial integration does however not really do the trick since that just transform my integral into a integral of the same form , but with a different $y$ and $a$. $\endgroup$
    – Mikael Fremling
    Apr 29, 2021 at 14:05
  • $\begingroup$ true but the first part of your question was where did the $T$ function come from, and that is the answer $\endgroup$
    – Henry Lee
    Apr 29, 2021 at 19:01
  • $\begingroup$ I acknowledge that the comment's about integration under the integral sign got me on the right track. I had been spending the entire day playing the various forms of partial integration and not getting anywhere... $\endgroup$
    – Mikael Fremling
    Apr 30, 2021 at 8:33

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