How to compute the value of the analytic continuation of the following integral?

Question

Let $F_1(z)=\int_0^\infty e^{-(z+1)^2t}dt$

How can I find the value of the analytic continuation of $F_1(z)$ corresponding to $z=2-4i$ ? (1)

I tried first: $(3-4i)^2=-(7+24i)$ so then:

$\int_0^\infty e^{-(3-4i)^2t} dt=\int_0^\infty e^{(7+24i)t} dt$

but the integral $\int_0^\infty e^{(7+24i)t} dt$ does not converge (I confirmed that the integral does not converge with Wolfram Alpha).

I wondered if I am doing something wrong or I am not understanding the question of the book (1) because it says that the answer is $\cfrac{-7+24i}{625}$

Answer 1

Hint: Since the books says "analytic continuation", it should occur to you that the value of $F_1(z)$ at $z=2-4i$ cannot be defined by the integral. So first integrate it while assuming $z\in\mathbb R$, so that it does converge, and you get an elementary function in $z$. What is the analytic continuation of this function?