# Integral involving the error function II

I would like to know if the following integral has an explicit expression or not. $$F(r)=\int_1^r e^{t^2/2} \operatorname{erf}\left(\frac{t}{\sqrt{2}}\right)dt$$

where the error function inside is defined as

$$\operatorname{erf}\left(\frac{t}{\sqrt{2}}\right)=\frac{1}{\sqrt{\pi}}\int_0^t e^{-x^2/2}dx.$$

Thus, $$F(r) = \frac{1}{\sqrt{\pi}}\int_{1}^{r}\int_0^{t}e^{\frac{t^2-x^2}{2}}dt dx.$$

Are there any known explicit expressions, series expansion or asymptotic known for such integrals?

• There is a result in terms of the Owen T function, see eg here
– Sal
Jun 25 at 16:13

$$F(r) = \frac{1}{\sqrt{\pi}}\int_{1}^{r}\int_{0}^{1}t\exp\left(\frac{t^2}{2}(1-x^2)\right)\,dx\,dt =\frac{1}{\sqrt{\pi}}\int_{0}^{1}\frac{\exp\left(\frac{r^2}{2}(1-x^2)\right)-\exp\left(1-x^2\right)}{1-x^2}\,dx$$ equals $$\frac{1}{\sqrt{\pi}}\int_{0}^{1}\frac{\exp\left(\frac{r^2}{2}(1-x)\right)-\exp\left(1-x\right)}{2\sqrt{x}(1-x)}\,dx=\frac{1}{\sqrt{\pi}}\int_{0}^{1}\frac{\exp\left(\frac{r^2}{2}x\right)-\exp\left(x\right)}{2x\sqrt{1-x}}\,dx$$ or $$\frac{1}{2\sqrt{\pi}}\sum_{n\geq 1}\frac{1}{n!}\int_{0}^{1}\frac{\left(\frac{r^{2n}}{2^n}-1\right)x^n}{x\sqrt{1-x}}\,dx=\frac{1}{2\sqrt{\pi}}\sum_{n\geq 1}\frac{1}{n!}\cdot\frac{4^n}{n\binom{2n}{n}}\cdot\left(\frac{r^{2n}}{2^n}-1\right)$$ which is a hypergeometric function of the $$\phantom{}_2 F_2$$ kind.

There is an antiderivative $$\int e^{\frac{t^2}{2}} \text{erf}\left(\frac{t}{\sqrt{2}}\right)\,dt=\frac{t^2 }{\sqrt{2 \pi }}\, _2F_2\left(1,1;\frac{3}{2},2;\frac{t^2}{2}\right)=\sum_{n=1}^\infty\frac {t^{2n} } {2^{n+\frac{1}{2}}\ n\ \Gamma \left(n+\frac{1}{2}\right) }$$

$$\color{blue}{F(r)=\sum_{n=1}^\infty\frac {r^{2n}-1 } {2^{n+\frac{1}{2}}\ n\ \Gamma \left(n+\frac{1}{2}\right) }}$$

Computational aspects

For the computation of the initial sum, it is quite simple since, if $$a_n=\frac {t^{2n} } {2^{n+\frac{1}{2}}\ n\ \Gamma \left(n+\frac{1}{2}\right)}$$

$$a_{n+1}=\frac{n\, t^2}{(n+1) (2 n+1)}\, a_n \qquad \text{with}\qquad a_1=\frac{t^2}{\sqrt{2 \pi }}$$

On the other hand, if we search for $$n$$ such that $$a_n \leq 10^{-k}$$, a slight overestimate is given by $$n \sim \frac 12 t^2\,e^{1+W(b)}-1 \qquad \text{with}\qquad b=\frac{1}{e t^2}\log \left(\frac{10^{2 k}}{\pi }\right)$$ $$W(.)$$ being Lambert function.

For a test with $$k=20$$ and $$t=4$$, this gives, as a real, $$n=51.0778$$ while the exact solution is $$49.5921$$.

• As soon as I saw the $a_n$ and $b_n$ I thought it was you. Jun 26 at 3:57
• @martycohen. When I see a specific typing, I know that it is from you ! Cheers :-) Jun 26 at 4:01