I am trying to find the following definite integral:
\begin{equation} I = \int_{0}^{b} Q\left((b-x)\, a \right)\,\frac{x}{\sigma^2}\,\exp\left(-\frac{x^2}{2\sigma^2}\right)\,dx, \end{equation} where $a$, $b$, $\sigma^2$ are some positive constants, and $Q(u)\stackrel{\triangle}{=} \int_{u}^{+\infty}\frac{\exp(-t^2/2)}{\sqrt{2\pi}}\,dt$ is the Gaussian Q function.
I have tried to use integration by parts and use some table of integrals to solve it but in vain. Any help or hint would be highly appreciated!
Using the integration by part, the solution to the above integral denoted by $I$ is as follows: \begin{align} I &= \int_{0}^{b}\underbrace{Q\left((b-x)\,a\right)}_{u}\,\underbrace{\left[-\frac{d}{dx}\left(\exp(-x^2/2\sigma^2)\right)\right]\,dx}_{dv}\\ &= Q(ab) - \frac{1}{2}\exp(-b^2/2\sigma^2) + \frac{a}{\sqrt{2\pi}}\exp(-a^2b^2/2 )\int_{0}^{b}\exp(-\beta^2 x^2 + a^2 b\, x)\,dx \\ &= Q(ab) - \frac{1}{2}\exp(-b^2/2\sigma^2) + \frac{\sqrt{\pi}}{2\beta}\exp(a^4y^2/4\beta^2)\left\{\text{erf}(a^2y/2\beta) + \text{erf}((\beta - a^2/2\beta)y)\right\}. \end{align} where $\beta \stackrel{\triangle}{=} a^2/2 + 1 / 2\sigma^2$, and $\text{erf}(x) \stackrel{\triangle}{=} \frac{2}{\sqrt{\pi}} \int_{0}^{x}\exp(-t^2)\,dt$.