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In my research I came across the following integral: \begin{equation} \int_{-\infty}^{+\infty}\frac{\partial{p(t)}}{\partial{t}}\frac{1}{4}\Big(1-\operatorname{erf}\Big(\frac{t-a}{\sigma\sqrt{2}}\Big)\Big)\Big(1+\operatorname{erf}\Big(\frac{t-c}{\sigma\sqrt{2}}\Big)\Big)\,dt \end{equation}


\begin{equation} p(t)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(t-b)^2}{2{\sigma}^2}} \end{equation}

\begin{equation} \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-s^2}ds \end{equation}

\begin{equation} a>b>c>0 \end{equation}

Does anyone know if this is solvable? I can tolerate a solution in terms of erf functions of $a,b,c,\sigma$. Thank you!

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You can get the right size for a pair of parentheses by preceding them with \left and \right, respectively. You can get proper formatting (font and spacing) for function names like $\operatorname{erf}$ using \operatorname{erf}. – joriki Sep 20 '12 at 7:22
Have you tried integrating by parts $\int u\,\mathrm dv$ where $u$ is the expression involving the error functions and $v = p(t)$? This will leave a second term (still to integrate) involving the product of $p(t)$ and derivatives of the error function, that is, a bunch of terms of the form $\exp(\text{quadratic function of}~t)$ which can be evaluated by completing the square in the exponent and writing the terms as normal density functions times a constant. – Dilip Sarwate Sep 20 '12 at 14:10

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