Solving a complex integration I want to solve this integration
$$\int_{-\infty}^{\infty} \mathrm{d}t \ \mathrm{erf} \Big(\frac{t-ic}{T}\Big) \ \mathrm{e}^{-\frac{(t-ib)^2}{T^2}}$$
one can open it by using integration by parts
$$u = \mathrm{erf}\Big(\frac{t-ic}{T}\Big), \ \ \ \ \ \ \ \  \mathrm{d}u = \frac{2}{\sqrt{\pi}T} \ \mathrm{e}^{-\frac{(t-ib)^2}{T^2}} \mathrm{d}t$$
$$v = \int \mathrm{d}t \ \mathrm{e}^{-\frac{(t-ib)^2}{T^2}} = \frac{\sqrt{\pi}T}{2} \mathrm{erf}\Big(\frac{t-ib}{T}\Big)$$
therefor
$$\int_{-\infty}^{\infty} \mathrm{d}t \ \mathrm{erf} \Big(\frac{t-ic}{T}\Big) \ \mathrm{e}^{-\frac{(t-ib)^2}{T^2}} = \frac{\sqrt{\pi}T}{2} \mathrm{erf}\Big(\frac{t-ib}{T}\Big)\mathrm{erf}\Big(\frac{t-ic}{T}\Big)\\
-\int_{-\infty}^{\infty} \mathrm{d}t \ \mathrm{erf} \Big(\frac{t-ib}{T}\Big) \ \mathrm{e}^{-\frac{(t-ic)^2}{T^2}}$$
using integration by parts again
$$\int_{-\infty}^{\infty} \mathrm{d}t \ \mathrm{erf} \Big(\frac{t-ic}{T}\Big) \ \mathrm{e}^{-\frac{(t-ib)^2}{T^2}} = \int_{-\infty}^{\infty} \mathrm{d}t \ \mathrm{erf} \Big(\frac{t-ic}{T}\Big) \ \mathrm{e}^{-\frac{(t-ib)^2}{T^2}}$$
you can see that I get exactly the same integration.
I also use the python to solve it numerically
import numpy as np
from scipy.special import erf
t = np.arange(-300.0,300.0,0.01)
T = 80.0; c = 0.1; b = 0.01;
f = erf((t-1j*c)/T)*np.exp(-(t-1j*b)**2/T**2)
ans = np.trapz(f,t)
ans = (-7.81481625665068e-09, 0.12727909116069697j)

Thus, one can see that the answer to this integration is an imaginary number. But I need the analytical version of the integration.
I would highly appreciate it if someone could help me to solve this complex integration
 A: Looks like: 
$$\int_{-\infty }^{\infty } \text{erf}\left(\frac{t-i c}{T}\right) \exp \left(-\frac{(t-i b)^2}{T^2}\right) \, dt=i \sqrt{\pi } T \text{erfi}\left(\frac{b-c}{\sqrt{2} T}\right)$$
for: $T>0$,$b>0$,$c>0$.
Using CAS for help:
$$\int_{-\infty }^{\infty } \text{erf}\left(\frac{t-i c}{T}\right) \exp \left(-\frac{(t-i b)^2}{T^2}\right) \, dt=\\\int \left(\int_{-\infty }^{\infty } \frac{\partial }{\partial c}\text{erf}\left(\frac{t-i
   c}{T}\right) \exp \left(-\frac{(t-i b)^2}{T^2}\right) \, dt\right) \, dc=\\\int \left(\int_{-\infty }^{\infty } -\frac{\left(2 i e^{-\frac{(-i c+t)^2}{T^2}}\right) \exp \left(-\frac{(t-i
   b)^2}{T^2}\right)}{\sqrt{\pi } T} \, dt\right) \, dc=\\\int -i \sqrt{2} e^{\frac{(b-c)^2}{2 T^2}} \, dc=\\i \sqrt{\pi } T \text{erfi}\left(\frac{b-c}{\sqrt{2} T}\right)$$
In this case integrating constant is zero.
With Mathematica 12.1 numericall check:
N[I Sqrt[\[Pi]] T Erfi[(b - c)/(Sqrt[2] T)] /. T -> 80 /. c -> 1/10 /. b -> 1/100, 30](*-0.127279247461544249487166063223 I*)
