Integrals of Gradshteyn and Ryzhik: 3.322

We define the error function as:
\begin{equation}
\mathrm{erf}(z) = \frac{2}{\sqrt{\pi}} \int\limits_{0}^{z} \mathrm{e}^{-x^2} dx
\end{equation}

3.322.1

\begin{align}
\int\limits_{u}^{\infty} \mathrm{e}^{-2ax-x^2} dx
&= \mathrm{e}^{a^2} \int\limits_{u}^{\infty} \mathrm{e}^{-(x+a)^2} dx
= \mathrm{e}^{a^2} \int\limits_{u+a}^{\infty} \mathrm{e}^{-y^2} dy \\
&= \frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \mathrm{erf}(y) \Bigg|_{u+a}^{\infty} \\
&= \frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \left[1-\,\mathrm{erf}(u+a)\right],\,\, u \gt 0
\end{align}
We completed the square in the argument of the exponential function and then used
the substitution \(y=x+a\).

3.322.2

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-2ax-x^2} dx =
\frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \left[1-\,\mathrm{erf}(a)\right]
\end{equation}
follows from 3.322.1.

3.322.3

\begin{equation}
\mathrm{PV} \int\limits_{0}^{\infty} \mathrm{e}^{\pm i\lambda x^2} dx
= \frac{1}{\sqrt{\lambda}} \int\limits_{0}^{\infty} \mathrm{e}^{\pm i y^2} dy
= \frac{1}{2}\sqrt{\frac{\pi}{\lambda}} \mathrm{e}^{\pm i \pi /4}
\end{equation}
we let \(y^2=\lambda x^2\) and this problem was solved here.

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