Closed expression of the following integral? I believe that the following integral has a closed expression, but I haven't been able to check it
$$I(k)=\int_{-\infty}^{\infty}dt\,\text{erf}\left(\frac{t}{b}-i \frac{1}{2}b(k+a)\right) e^{-\frac{t^2}{b^2}-\frac{1}{4}b^2(k+a)^2-i t(k-a)}$$
where $a,\,b$ are real numbers.
Could you give me some advice on how to proceed, or simply the result?
 A: Hint: Let $I(a)=\displaystyle\int_{-\infty}^\infty\frac{\text{erf}\big(t-ia\big)}{\exp\big(A^2+Bi~t+t^2\big)}~dt,~$ and evaluate $I'(a)$, then integrate back 
with regard to a, and let $A=a$ and $B=2\beta$. You'll get $I(a)=-i~\sqrt\pi~\dfrac{\text{erfi}\bigg(\dfrac{a+\beta}{\sqrt2}\bigg)}{\exp\big(a^2+\beta^2\big)}~.$
A: Actually I was working in @Lucian's hint when I realized there is no need to "forget to derive" the exponential part. Let me do it step by step


*

*We derive $I(a)$ w.r.t. $a$
$$I'(a)=\int_{-\infty}^\infty dt\,\left(2 i a e^{-a^2-i b t-t^2} \text{erfi}(a+i t)-\frac{2 i e^{-a^2+(a+i t)^2-i b t-t^2}}{\sqrt{\pi }}\right)\\
=-2a\,I(a)-i \sqrt{2} e^{-\frac{1}{8} (b-2 a)^2}.$$

*We solve the homogenous differential equation and obtain $I_H(a)=Ce^{-a^2}$.

*Using variation of constants we get:
$$C'(a)e^{-a^2}=-i\sqrt{2}e^{-\frac{1}{8}\left(b-2a\right)^2}\\
C(a)=-i \sqrt{\pi } e^{-\frac{b^2}{4}} \text{erfi}\left(\frac{2 a+b}{2 \sqrt{2}}\right)$$
So the final solution is
$$I(a)=-i\sqrt{\pi}e^{-a^2-\frac{b^2}{4}}\text{erfi}\left(\frac{a+\frac{b}{2}}{\sqrt{2}}\right)$$
Which is Lucian's answer.
