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$\newcommand{\erf}{\operatorname{erf}}$

I have the following integral and I need to simplify the solution. I have written first two steps. I don't know what is the value of

$$ \erf(\infty) $$

I believe it is 1.

\begin{align*} \int_{l/{2t^{1/2}}}^\infty \exp \left(-\left(\frac{ml^2}{4\lambda^2}+\lambda^2\right)\right)\,d\lambda = \frac{1}{4}\sqrt{\pi} \left[e^{-lm^{1/2}}\left(\erf\left(x-\frac{lm^{1/2}}{2x}\right)+1\right)\right. \\ {}+ \left.e^{lm^{1/2}} \left(\erf\left(x+\frac{lm^{1/2}}{2x}\right)\right)-1)\right]_{l/{2t^{1/2}}}^\infty \end{align*}

Still the equation becomes complex and I got error. So, any assistance is very helpful.

Thanks.

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