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Given the integral: $$\int_{0}^{+\infty}dz\, e^{-z^2} \text{erf}(z+\beta)=\frac{\sqrt{\pi}}{2}\left(1-\frac{1}{2}\text{erfc}^2\left(\frac{\beta}{\sqrt{2}}\right)\right),$$ where $\text{erf}(z)$ and $\text{erfc}(z)$ are respectively the Gauss error function and the complementary error function, I need to solve the following: $$\int_{0}^{+\infty}dz\, e^{-\color{red}\alpha z^2} \text{erf}(z+\beta).$$ On the tables where all possible integrals involving error functions are given, I cannot find an appropriate formula which could help me to evaluate my integral.

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    $\begingroup$ Why not go through the same steps you used to derive the first one? What difficulties do you run into with the $\alpha$ ? $\endgroup$ Commented Nov 26, 2021 at 11:56
  • $\begingroup$ You're right, but I need to prove also the first result, unfortunately I don't know the steps. $\endgroup$
    – MariNala
    Commented Nov 26, 2021 at 15:58
  • $\begingroup$ I believe the trick is to turn the integral into a double integral by the definition of error function, then reverse the order of integration. $\endgroup$ Commented Nov 26, 2021 at 18:41

1 Answer 1

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You can obtain a result for the second integral in terms of Owen's T-function which is defined as $$ T(x, p) = \frac{1}{2\pi}\int_{0}^{p} \frac{e^{-\frac{1}{2} x^2 (1+t^2)}}{1+t^2} \, dt \quad \left(-\infty < x, p < +\infty\right). $$ From this definition, a simple calculation using Feynman's trick tells us that $\require{cancel}$ \begin{align*} \frac{d}{d\beta} 2\sqrt{\frac{\pi}{\alpha}}T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)& = \cancel{2}\sqrt{\frac{\pi}{\alpha}} \frac{1}{\cancel{2}\pi}\int_{0}^{ \frac{1}{\sqrt{\alpha}}}\frac{\partial}{\partial \beta} \frac{e^{-\frac{1}{\cancel{2}}\frac{\cancel{2}\alpha}{\alpha+1} \beta^2 (1+t^2)}}{1+t^2} dt\\ & = \frac{1}{\sqrt{\pi \alpha}}\int_{0}^{ \frac{1}{\sqrt{\alpha}}} \frac{-2\frac{\alpha}{\alpha+1}\cancel{(1+t^2)}\beta e^{-\frac{\alpha}{\alpha+1}\beta^2 (1+t^2)}}{\cancel{1+t^2}} dt\\ & = -\frac{2\beta}{\sqrt{\pi }} \frac{\sqrt{\alpha}}{\alpha+1}e^{-\frac{\alpha}{\alpha+1}\beta^2}\int_{0}^{ \frac{1}{\sqrt{\alpha}}} e^{-\left(\sqrt{\frac{\alpha}{\alpha+1}}\beta t\right)^2} \, dt\\ & \overset{\color{blue}{u = \sqrt{\frac{\alpha}{\alpha+1}}\beta t}}{=} -\frac{\color{green}{2}\cancel{\beta}}{\color{green}{\sqrt{\pi }}} \frac{\cancel{\sqrt{\alpha}}}{\alpha+1}e^{-\frac{\alpha}{\alpha+1}\beta^2}\frac{\color{blue}{\sqrt{\alpha+1}}}{\color{blue}{\cancel{\beta}}\cancel{\color{blue}{\sqrt{\alpha}}}} \color{green}{\int_{0}^{\frac{\beta}{\sqrt{\alpha+1}}} e^{-u^2} \, du}\\ & = - \frac{1}{\sqrt{\alpha+1}}e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2}\color{green}{\text{erf}\left( \frac{\beta}{\sqrt{\alpha+1}}\right)} \tag{1} \end{align*}

Now let's define a function $I(\beta)$ in terms of the integral we want $$ I(\beta) := \int_{0}^{\infty} e^{-\alpha x^2} \text{erf}(x+\beta) \, dx $$ by again using Feynman's trick we get \begin{align*} \frac{d}{d\beta}I(\beta) = \int_{0}^{\infty}e^{-\alpha x^2} \frac{\partial}{\partial \beta }\text{erf}(x+\beta) \, dx = \int_{0}^{\infty}e^{-\alpha x^2} \frac{2}{\sqrt{\pi}}e^{-(x+\beta)^2} \, dx = \frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\color{blue}{\sqrt{\alpha+1}}\right)^2x^2 +2\color{purple}{\beta}x +\color{green}{\beta^2}\right)} \, dx \end{align*} where we used the fundamental theorem of calculus to differentiate the error function, along with its definition $\text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} \, dt$. Now, from equation $4$ in section $3.2$ of the tables you linked we know that $$ \int_{0}^{\infty} e^{-(\color{blue}{a}^2 t^2 + 2\color{purple}{z} t + \color{green}{c})} \, dt = \frac{\sqrt{\pi}}{2a}\text{erfc}\left( \frac{z}{a}\right)e^{-\left(c-\frac{z^2}{a^2}\right)} = \frac{\sqrt{\pi}}{2a}e^{-\left(c-\frac{z^2}{a^2}\right)}-\frac{\sqrt{\pi}}{2a}\text{erf}\left( \frac{z}{a}\right)e^{-\left(c-\frac{z^2}{a^2}\right)} $$ using that $\text{erfc}(x) = 1 - \text{erf}(x)$ by definition. We thus get that $$ I'(\beta) = \frac{1}{\sqrt{\alpha+1}}e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2}- \frac{1}{\sqrt{\alpha+1}}\text{erf}\left( \frac{\beta}{\sqrt{\alpha+1}}\right)e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2} \tag{2} $$ Using the definition of the error function, notice that $$ \frac{d}{d\beta}\left[ \frac{1}{2} \sqrt{\frac{\pi}{\alpha}}\text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) \right]= \frac{1}{\sqrt{\alpha+1}}e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2} $$ And thus, we can combine the previous equation with equations $(1)$ and $(2)$ to get that $$ \frac{d}{d\beta}I(\beta) = \frac{d}{d\beta}\left[\frac{1}{2} \sqrt{\frac{\pi}{\alpha}}\text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) + 2\sqrt{\frac{\pi}{\alpha}}T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)\right] $$ But since we have two functions whose derivatives are equal, the functions being differentiated must be equal up to some constant $\color{blue}{C}$: $$ I(\beta) = \sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2} \text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) + 2T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)\right) + \color{blue}{C} $$ Evaluating at $\beta =0$ we get $$ I(0) = \sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2} \text{erf}\left(0\right) + 2T\left(0, \frac{1}{\sqrt{\alpha}} \right)\right) +C = \sqrt{\frac{\pi}{\alpha}}\left( \frac{\text{arctan}\left(\frac{1}{\sqrt{\alpha}} \right)}{\pi}\right) +C \tag{3} $$ where we used the fact that $T(0, x)= \frac{1}{2\pi}\int_{0}^{x} \frac{1}{1+t^2} \, dt= \frac{\arctan(x)}{2\pi}$. If we now evaluate $I(0)$ using our original integral definition we can obtain the value for $C$ and complete the problem. We see that $$ I(0) = \int_{0}^{\infty} e^{-\sqrt{\alpha}^2 x^2} \text{erf}(x) \, dx $$ and using equation $2$ from section $4.3$ of the tables you linked we get \begin{align} I(0) = \frac{\sqrt{\pi}}{2\sqrt{\alpha}} - \frac{1}{\sqrt{\pi\alpha}}\arctan(\sqrt{\alpha}) = \cancel{\frac{\sqrt{\pi}}{2\sqrt{\alpha}}}-\frac{1}{\sqrt{\pi\alpha}}\left(\cancel{\frac{\pi}{2}} - \arctan\left(\frac{1}{\sqrt{\alpha}}\right) \right) =\frac{\arctan\left(\frac{1}{\sqrt{\alpha}}\right)}{\sqrt{\pi \alpha}} \tag{4} \end{align} where we used that $\arctan(x) + \arctan\left( \frac{1}{x}\right)= \frac{\pi}{2}$ since $ \alpha > 0$ in order for the integral to converge. Combining equations $(3)$ and $(4)$ we get that $C=0$, so we can finally conclude that for $a>0$ and $b \in \mathbb{R}$: $$ \boxed{\int_{0}^{\infty} e^{-\alpha x^2} \text{erf}(x+\beta) \, dx = \sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2} \text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) + 2T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)\right)} $$


The integral you give in your question can be derived from the general case above. First, notice that $$ \frac{1}{\pi}\int_{0}^{1} \frac{e^{-\frac{x^2}{2} (1+t^2)}}{1+t^2} \, dt=2T\left(x,1 \right) = \frac{1}{4} \text{erfc}\left( \frac{x}{\sqrt{2}}\right) \left(2- \text{erfc}\left( \frac{x}{\sqrt{2}}\right) \right) $$ which can be proven by differentiating both sides of the equation by $x$ and using Feynman's trick. We thus get that \begin{align} \int_{0}^{\infty} e^{- x^2} \text{erf}(x+\beta) \, dx &= \sqrt{\frac{\pi}{1}}\left(\frac{1}{2} \text{erf}\left(\sqrt{\frac{1}{1+1}} \beta\right) + 2T\left(\sqrt{\frac{\cancel{2}}{\cancel{1+1}}}\beta, \frac{1}{\sqrt{1}} \right)\right)\\ &= \sqrt{\pi}\left(\frac{1}{2} \text{erf}\left( \frac{\beta}{\sqrt{2}}\right) + 2T\left(\beta, 1\right)\right)\\ &= \sqrt{\pi}\left(\frac{1}{2} \left(1-\text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)\right) + \frac{1}{4} \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right) \left(2- \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right) \right)\right)\\ &= \sqrt{\pi}\left(\frac{1}{2}-\cancel{\frac{1}{2}\text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)} + \cancel{\frac{1}{2} \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)}- \frac{1}{4} \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)\text{erfc}\left( \frac{\beta}{\sqrt{2}}\right) \right)\\ &= \frac{\sqrt{\pi}}{2}\left(1-\frac{1}{2}\text{erfc}^2\left(\frac{\beta}{\sqrt{2}}\right)\right) \end{align}


As a final note, a nice complimentary identity I found whilst solving this problem is the integral $$ \int_{-\infty}^{\infty} e^{-\alpha x^2} \text{erf}(\gamma x + \beta) \, dx = \sqrt{\frac{\pi}{\alpha}} \text{erf}\left( \frac{\beta \sqrt{\alpha}}{\sqrt{\gamma^2 +\alpha}}\right) $$ which might also be useful to you.

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  • $\begingroup$ Very well explained. Thank you! $\endgroup$
    – MariNala
    Commented Nov 27, 2021 at 9:29
  • $\begingroup$ Hi, Robert! Very nice answer! Is that possible to express the last integral in terms of special functions if one of the integration limits is finite, and when $\alpha,\beta,\gamma$ are complex? (considering the integral is converging) $\endgroup$
    – Sl0wp0k3
    Commented Apr 25, 2023 at 9:09
  • $\begingroup$ Here is a very related question, would be glad if you can come up with the idea as it is an integral quite similar to the lase one math.stackexchange.com/questions/4686979/… $\endgroup$
    – Sl0wp0k3
    Commented Apr 27, 2023 at 9:14

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