Integrals of Gradshteyn and Ryzhik: 3.321

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}

3.321.2

\begin{equation}
\int\limits_{0}^{u} \mathrm{e}^{-q^{2}x^{2}} dx = \frac{1}{q} \int\limits_{0}^{qu} \mathrm{e}^{-z^{2}} dz
= \frac{\sqrt{\pi}}{2q} \mathrm{erf}(z) \Big|_{0}^{qu}
= \frac{\sqrt{\pi}}{2q} \mathrm{erf}(qu),\,\,q \gt 0
\end{equation}
using the substitution \(z=qx\)

3.321.3

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-q^{2}x^{2}} dx = \frac{\sqrt{\pi}}{2q},\,\,q \gt 0
\end{equation}
follows from 3.321.2

3.321.4

\begin{equation}
I = \int\limits_{0}^{u} x\mathrm{e}^{-q^{2}x^{2}} dx
= \frac{\sqrt{\pi}}{2q} \,x\, \mathrm{erf}(qx) \,- \frac{\sqrt{\pi}}{2q} \int \mathrm{erf}(qx) dx
\end{equation}
using integration by parts.
\begin{align}
\int \mathrm{erf}(qx) dx &= \frac{1}{q} \int \mathrm{erf}(y) dy \\
&= \frac{1}{q} \left[y\,\mathrm{erf}(y) – \frac{2}{\sqrt{\pi}} \int y\,\mathrm{e}^{-y^2} dy \right] \\
&= \frac{1}{q} y\,\mathrm{erf}(y) + \frac{1}{q\sqrt{\pi}} \mathrm{e}^{-y^2} \\
&= x\,\mathrm{erf}(qx) + \frac{1}{q\sqrt{\pi}} \mathrm{e}^{-q^{2}x^{2}}
\end{align}
using integration by parts and the substitution \(y=qx\)
\begin{equation}
I = \frac{\sqrt{\pi}}{2q} x \,\mathrm{erf}(qx) \,- \frac{\sqrt{\pi}}{2q} x \,\mathrm{erf}(qx)
\,- \frac{1}{2q^2} \mathrm{e}^{-q^{2}x^{2}} \Big|_{0}^{u}
= \frac{1}{2q^2} \left[1 – \mathrm{e}^{-q^{2}u^{2}} \right]
\end{equation}

3.321.5

\begin{align}
I &= \int\limits_{0}^{u} x^{2} \mathrm{e}^{-q^{2}x^{2}} dx
= \frac{-1}{2q^2} x \mathrm{e}^{-q^{2}x^{2}} + \frac{1}{2q^2} \int\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{1}{2q^3} \left[-qx \mathrm{e}^{-q^{2}x^{2}} + \frac{\sqrt{\pi}}{2} \mathrm{erf}(qx) \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^3} \left[\frac{\sqrt{\pi}}{2} \mathrm{erf}(qu) \,-\, qu \mathrm{e}^{-q^{2}x^{2}} \right]
\end{align}
We used integration by parts, 3.321.4, and 3.321.2.

3.321.6

\begin{align}
I &= \int\limits_{0}^{u} x^{3} \mathrm{e}^{-q^{2}x^{2}} dx
= \frac{\sqrt{\pi}}{4q^3}x\,\mathrm{erf}(qx) \,- \frac{x^2}{2q^2}\mathrm{e}^{-q^{2}x^{2}}
– \frac{\sqrt{\pi}}{4q^3} \int \mathrm{erf}(qx) dx + \frac{1}{2q^2} \int x\,\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{\sqrt{\pi}}{4q^3}\,x\,\mathrm{erf}(qx) \,- \frac{x^2}{2q^2}\mathrm{e}^{-q^{2}x^{2}}
– \frac{\sqrt{\pi}}{4q^3}\,x\,\mathrm{erf}(qx) \,- \frac{1}{4q^4}\mathrm{e}^{-q^{2}x^{2}}
– \frac{1}{4q^4}\mathrm{e}^{-q^{2}x^{2}} \\
&= \frac{1}{2q^4} (-q^{2}x^{2} – 1)\mathrm{e}^{-q^{2}x^{2}}\Big|_{0}^{u} \\
&= \frac{1}{2q^4} \left[1-(1+q^{2}u^{2}) \mathrm{e}^{-q^{2}u^{2}} \right]
\end{align}
using integration by parts, 3.321.5, the work done in the solution of 3.321.4 and the result 3.321.4.

3.321.7

\begin{align}
I &= \int\limits_{0}^{u} x^{4} \mathrm{e}^{-q^{2}x^{2}} dx
= -\frac{1}{2q^4}x\mathrm{e}^{-q^{2}x^{2}} – \frac{1}{2q^4}q^{2}x^{3}\mathrm{e}^{-q^{2}x^{2}}
+ \frac{1}{2q^4}\int \mathrm{e}^{-q^{2}x^{2}} dx + \frac{q^{2}}{2q^4}\int x^{2}\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{1}{2q^5}\left[-qx\,\mathrm{e}^{-q^{2}x^{2}} – q^{3}x^{3}\,\mathrm{e}^{-q^{2}x^{2}}
+ \frac{\sqrt{\pi}}{2} \mathrm{erf}(qx) + \frac{\sqrt{\pi}}{4} \mathrm{erf}(qx)
-\frac{1}{2}qx\,\mathrm{e}^{-q^{2}x^{2}} \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^5} \left[\frac{3\sqrt{\pi}}{4}\mathrm{erf}(qx)
– qx\,\mathrm{e}^{-q^{2}x^{2}}\left(\frac{3}{2} + q^{2}x^{2}\right) \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^5} \left[\frac{3\sqrt{\pi}}{4}\mathrm{erf}(qu)
– qu\,\mathrm{e}^{-q^{2}u^{2}}\left(\frac{3}{2} + q^{2}u^{2}\right) \right]
\end{align}
using integration by parts, 3.321.6, 3.321.2, and 3.321.5.

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