I am trying to take a Fourier transform of a truncated Gaussian function $f(t,T)=e^{-at^2}I(|t|\leq T)$, where $I(B)$ is the indicator function returning one when boolean variable $B$ is true and zero otherwise. I thus must solve the following integral:
$$\tilde{f}(\xi,T)=\int_{-T}^Te^{-at^2}e^{-2\pi i \xi t}dt.$$
Wolfram Mathematica "helpfully" returns this "solution" (which can be derived by completing the square):
$$\tilde{f}(\xi,T)=\frac{\sqrt{\pi } e^{-\frac{\pi ^2 \xi^2}{a}} \left(\text{erf}\left(\frac{a T -i\pi \xi}{\sqrt{a}}\right)+\text{erf}\left(\frac{a T + i\pi \xi}{\sqrt{a}}\right)\right)}{2 \sqrt{a}}.$$
The symmetrical nature of $f(t,T)$ implies that $\tilde{f}(\xi,T)$ is a real function (this is because a Fourier transform of any even real function is real and even), and thus I am wondering if there is a way to write it without imaginary numbers. This reduces this writing the following function without the use of imaginary numbers:
$$g(u,v)=\text{erf}(u+iv)+\text{erf}(u-iv),$$
where arguments $u$ and $v$ are real.
However, I am stuck at evaluating erf for complex variables. I understand that it is possible to do this with complex analysis, but unfortunately, I am not versed in it at all. I thus would be grateful to anyone who could show me how to write the expression for $\tilde{f}(\xi,T)$ without imaginary numbers, and, if possible, carry out the derivation without complex analysis (or stuff simple enough for someone with good background in multivariate calculus and some elementary real analysis).
UPDATE
I updated the question by somewhat simplifying $\tilde{f}(\xi,T)$, as well as indicating that one can get it by the usual method of completing the square (this was also mentioned in a comment by @tired). I also made explicit the reason why $\tilde{f}(\xi,T)$ should be a real function. For clarity, I also adjusted the definition of $g(u,v)$ by using the fact that the error function is odd (i.e., $\operatorname{erf}(x)=-\operatorname{erf}(x)$). The form of $g(u,v)$ given by @tired in the first comment to this question should be valid.
Also, @tired's point about the conjugate of error function being equal to the error function of the conjugate (that is, $\operatorname{erf}^*(x)= \operatorname{erf}(x^*)$) indeed confirms that $g(u,v)$ is real. This can be shown explicitly as follows:
$$\begin{align} g(u,v)&=\text{erf}(u+iv)+\text{erf}(u-iv)\\ &=\mathfrak{Re}[\text{erf}(u+iv)]+i\mathfrak{Im}[\text{erf}(u+iv)]+\mathfrak{Re}[\text{erf}(u-iv)]+i\mathfrak{Im}[\text{erf}(u-iv)]\\ &=\mathfrak{Re}[\text{erf}(u+iv)]+i\mathfrak{Im}[\text{erf}(u+iv)]+\mathfrak{Re}[\text{erf}^*(u+iv)]+i\mathfrak{Im}[\text{erf}^*(u+iv)]\\ &=2\mathfrak{Re}[\text{erf}(u+iv)], \end{align}$$ where $\mathfrak{Re}[a+ib]=a$ and $\mathfrak{Im}[a+ib]=b$.
Unfortunately, it seems that this is as far as the expression for $\tilde{f}(\xi,T)$ can be simplified, since it is very difficult to analytically separate real and imaginary parts of $\text{erf}(u+iv)$ per @RonGordon's answer to this question (however, good approximation exists per same answer).