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I need to evaluate this integral: $$ I(t,a) = \int_{-\infty}^{t} e^{-(\tau+a)^2} \mathrm{erf}(\tau) \ \mathrm{d}\tau $$ where the $\mathrm{erf}(\tau)$ is the error function.

I can prove that this integral converges. By employing the python library

import numpy as np
from scipy.special import erf
import matplotlib.pyplot as plt
dtau = 0.01;p=[]
trange = np.arange(-20,20,0.1)
for t in trange:
    tau = np.arange(-20,t,dtau)
    I = np.exp(-(tau+a)**2)* erf(tau)
    p.append(np.trapz(I,tau))
p=np.array(p)
plt.plot(trange,p);plt.show();

I got three graphs for different $a$ enter image description here

therefor one can speculate the behavior of the integral for $|a|\ll1$ is as a Bi-gaussian function and for $|a|\gg1$ is as an $\mathrm{erf}(t)$ function. therefore the answer is something like this $$ I(t,a) \sim \alpha(a) \ \mathrm{erf}(t+a) + \beta(a) \ e^{-(t\pm a)^2} $$

I would highly appreciate it if someone could help me to solve it.

Edit:

if $t \rightarrow \infty$, the $I(\infty, a)$ is given by $$ I(t \rightarrow \infty,a) = \sqrt{\pi} \ \mathrm{erf} \Big(\dfrac{a}{\sqrt{2}} \Big) $$ this might be helpful.

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If $a$ is small, we could expand the exponential as a Taylor series $$e^{-(\tau+a)^2} =e^{-\tau^2}\sum_{n=0}^p f_n(\tau)\,a^n$$ where the first coefficients are $$\left\{1,-2 \tau ,-1+2 \tau ^2,2 \tau -\frac{4 }{3}\tau ^3,\frac{1}{2}-2 \tau ^2+\frac{2 }{3}\tau ^4,-\tau +\frac{4 }{3}\tau ^3-\frac{4 }{15}\tau ^5,\cdots\right\}$$ and face a bunch of integrals $$I_n=\int_{-\infty}^t e^{-\tau ^2} \text{erf}(\tau ) \tau^n\,d\tau$$ which do not present much difficulties.

The first ones are $$I_0=\frac{1}{4} \sqrt{\pi } \left(\text{erf}(t)^2-1\right)$$ $$I_1=\frac{1}{4} \left(\sqrt{2} \left(\text{erf}\left(\sqrt{2} t\right)+1\right)-2 e^{-t^2} \text{erf}(t)\right)$$ $$I_2=\frac{1}{8} \left(-4 e^{-t^2} t \text{erf}(t)+\sqrt{\pi } \left(\text{erf}(t)^2-1\right)-\frac{2 e^{-2 t^2}}{\sqrt{\pi }}\right)$$ $$I_3=\frac{1}{16} \left(-8 e^{-t^2} \left(t^2+1\right) \text{erf}(t)+5 \sqrt{2} \left(\text{erf}\left(\sqrt{2} t\right)+1\right)-\frac{4 e^{-2 t^2} t}{\sqrt{\pi }}\right)$$

Limited to first order, this would give $$\frac{\sqrt{\pi }}{4} \left(\text{erf}(t)^2-1\right)+a \left(e^{-t^2} \text{erf}(t)+\frac{\text{erfc}\left(\sqrt{2} t\right)-2}{\sqrt{2}}\right)+O\left(a^2\right)$$ which perfectly matches your plot for $a=0.2$.

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  • $\begingroup$ thanks, @claude-leibovici. However, if someone gives a more impact solution, would be great, due to I should insert this solution in a big equation and solve that on. for large $a$, do you have any solution? $\endgroup$ – Ghoti Sep 22 '20 at 13:05
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$$I(t,a)=\int_{-\infty}^te^{-(\tau+a)^2}\operatorname{erf}(\tau)d\tau=\int_{-\infty}^{t+a}e^{-u^2}\operatorname{erf}(u-a)du$$ we know that: $$\operatorname{erf}(\tau)=\frac{2}{\sqrt{\pi}}\int_0^{\tau}e^{-\sigma^2}d\sigma$$ so we could break up the integral into: $$I(t,a)=\frac{2}{\sqrt{\pi}}\int_{-\infty}^0\int_0^{u-a}e^{-(u^2+\sigma^2)}d\sigma du+\frac{2}{\sqrt{\pi}}\int_0^{t+a}\int_0^{u-a}e^{-(u^2+\sigma^2)}d\sigma du$$ polar coordinates would be nice to use here but we have a rather ugly domain in the two regions: $$\exists S\in\{(u,\sigma):((-\infty,0]\times[0,u-a]\wedge[0,t+a]\times[0,u-a])\}$$ If we split up the first one and call it something else for ease: $$J_1=\int_{-\infty}^0\int_0^u e^{-(u^2+\sigma^2)}d\sigma du=\int_{5\pi/4}^{3\pi/2}\int_0^\infty re^{-r^2}drd\theta=\frac{\pi}{8}$$ $$J_2=\int_0^{t+a}\int_0^{u}e^{-(u^2+\sigma^2)}d\sigma du=\int_0^{\pi/4}\int_{\sec\theta}^{\sqrt{2}(t+a)}re^{-r^2}drd\theta$$ $$J_3=\int_{-\infty}^0\int_{u-a}^ue^{-(u^2+\sigma^2)}d\sigma du$$ $$J_4=\int_0^{t+a}\int_{u-a}^ue^{-(u^2+\sigma^2)}d\sigma du$$ Hopefully you will find a way of representing these nicely :)

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  • $\begingroup$ thanks, @Henry. I tried to do it but could not. I got an expression that was wrong in comparison to the numerical method. why in the bound there is $\sec(\theta)$? $\endgroup$ – Ghoti Sep 23 '20 at 16:43
  • $\begingroup$ @Ghoti I believe I may have missed a term with the $\sec\theta$ but basically if you draw the domain it is a triangle shape, then you just use trig :) $\endgroup$ – Henry Lee Sep 23 '20 at 16:51
  • $\begingroup$ $I = \int_{-\infty}^{0} e^{-u^2} \mathrm{erf}(u-a) du + \int_{0}^{t+a} e^{-u^2} \mathrm{erf}(u-a) du = \dfrac{\pi}{4} \mathrm{erfc}(\dfrac{a}{\sqrt{2}})^2 + \dfrac{2}{\sqrt{\pi}} \int_{0}^{t+a} \int_{0}^{u} e^{-u^2-\sigma^2} d\sigma du+ \dfrac{2}{\sqrt{\pi}} \int_{0}^{t+a} \int_{u}^{u-a} e^{-u^2-\sigma^2} d\sigma du = \dfrac{\pi}{4} \mathrm{erfc}[\dfrac{a}{\sqrt{2}}]^2 + \dfrac{\pi}{4} \mathrm{erf}[t+a]^2+ \dfrac{2}{\sqrt{\pi}} \int_{0}^{t+a} \int_{u}^{u-a} e^{-u^2-\sigma^2} d\sigma du $ $\endgroup$ – Ghoti Sep 30 '20 at 12:55
  • $\begingroup$ .@Henry. I could not find the answer of $J_2$ using polar coordinates but got it from a different way. The last integration ($J_4$) left, and the domain is very strange, and the integration gets harder to solve. In your calculations, what is the domain of the $J_4$? $\endgroup$ – Ghoti Oct 5 '20 at 8:46

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