I am tasked with the following problem for a homework assignment:
Evaluate the following integrals by hand and check your results on Mathematica.
- Gaussian integral: $\int_{0}^\infty e^{-(x-\mu)^2 / (2\sigma^2)} \, dx$, assume $\mu$ and $\sigma^2$ are real scalars and $\sigma^2 > 0$.
My question pertains to the end result of this integration. Mathematica lists the result as \begin{equation} \sigma\sqrt{\dfrac{\pi}{2}} \left( 1 + Erf\left(\dfrac{\mu}{\sigma \sqrt{2}}\right) \right) \end{equation}
I'm having some trouble getting this same result, and I believe it's because I don't fully understand the error function. Below I lay out my work and show my final answer, which is similar to the above, but there are some glaring differences.
The below text is my completed solution:
Firstly, take $a = 1/(2\sigma^2)$. Take
\begin{align}
I
&=
\int_{0}^\infty e^{-a(x-\mu)^2} \, dx
\nonumber
\\[.5em]
I^2
&=
\left[ \int_{0}^\infty e^{-a(x-\mu)^2 } \, dx \right]^2
\nonumber
\\[.5em]
&=
\left( \int_{0}^\infty e^{-a(x-\mu)^2} \, dx \right) \left( \int_{0}^\infty e^{-a(y-\mu)^2} \, dy \right)
\nonumber
\end{align}
In the above, $y$ is simply a dummy variable for the sole purpose of easing the difficulty of integration. Now, perform a change of variables such that $\beta = x - \mu$, $\gamma = y - \mu$, $dx = d\beta$, $dy = d\gamma$. Note that with this substitution, our limits of integration change. When $x=0$, $\beta = \mu$ (similarly for $y$). So,
\begin{align}
I^2
&=
\int_{0}^\infty \int_{0}^\infty
e^{-a(x-\mu)^2} e^{-a(y-\mu)^2 }
\, dx dy
\nonumber
\\[.5em]
&=
\int_{-\mu}^\infty \int_{-\mu}^\infty
e^{-a(\beta^2 + \gamma^2)}
\, d\beta d\gamma
\nonumber
\end{align}
Changing to polar coordinates, we have $\beta = r\cos\theta$, $\gamma = r\sin\theta$, and $d\beta d \gamma = r dr d\theta$ such that,
\begin{align}
I^2
&=
\int_0^{2\pi} \int_{\mu}^\infty
r e^{-a r^2}
dr d\theta
\nonumber
\\[.5em]
& \text{Take } u = r^2 \text{ such that } du = 2r dr
\nonumber
\\[.5em]
&=
2\pi \int_{\mu^2}^\infty
\dfrac{e^{-a u}}{2}
du
\nonumber
\\[.5em]
&=
\pi \left[- \dfrac{e^{-u / (2\sigma^2)}}{a} \right]_{\mu^2}^\infty
\nonumber
\\[.5em]
&=
\dfrac{\pi}{a} \, e^{-a\mu^2}
\nonumber
\end{align}
Therefore,
\begin{equation}
\boxed{
I
=
\int_{0}^\infty e^{-(x-\mu)^2 / (2\sigma^2)} \, dx
=
\sigma \sqrt{2\pi} \, e^{-\mu^2 / (4\sigma^2)}
}
\end{equation}
I know how the error function is defined as an integral of an exponential, and the $\mu/(\sigma\sqrt{2})$ seen in the Mathematica solution seems to almost match the term in the exponential of my solution, but I'm still very confused on how to get the error function in the solution. Any help is appreciated, thank you.
