I'm struggling to solve this problem.
$$ \int_{-\infty}^{\infty} e^{-a(x+di)^2} dx $$
I tried
$$ t = \sqrt{a}(x+di) $$ $$ => dt = \sqrt{a}dx $$ $$ => \int_{-\infty}^{\infty} \frac{e^{-t^2}}{\sqrt{a}} dt $$ since $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$,
$$ => \int_{-\infty}^{\infty} \frac{e^{-t^2}}{\sqrt{a}} dt = \sqrt{\frac{\pi}{a}} $$ $$ => \int_{-\infty}^{\infty} e^{-a(x+di)^2} dx = \sqrt{\frac{\pi}{a}} $$
However, I have found that complex infinity has only magnitude without arguments.
It means I can't regard domain of t as (-$\infty$,$\infty$) when $t = \sqrt{a}(x+di)$ and $x(-\infty, \infty)$.
And I found the correct answer is $\sqrt{\frac{\pi}{a}}e^{ad^2}$. How can I reach to this answer? I lost my way.
Here is the way I like to do it, avoiding contour integration. Fix a real number $a>0$. Define a function of a complex variable $k$ by $$ F(k) := \int_{-\infty}^\infty \exp\left(-a(x+k)^2\right)\;dx $$ When $k$ is real, do the change of variables suggested (the real line maps to the real line for that change of variables) to get $$ F(k) = \sqrt\frac{\pi}{a} \qquad \text{if } k \in \mathbb R. $$ Next, from Morera's theorem, $F$ is an entire function of the complex variable $k$. Therefore $$ F(k) = \sqrt\frac{\pi}{a} \qquad \text{if } k \in \mathbb C. $$
Your assertion about "the correct answer" is wrong. Your answer is actually for $$ \int_{-\infty}^\infty \exp\left(-a(x^2+(id)^2)\right)\;dx = \sqrt\frac{\pi}{a}\;e^{ad^2} $$