Evaluate the integral $\int_{-\infty}^{\infty} e^{-x^2} e^{ix} dx$ I tried common techniques like integration by parts but did not succeed. The $e^{ix}$ also is causing some trouble. I'm guessing this is a Fourier transform in disguise but I'm not sure how the evaluate this. Any help will be appreciated! 
 A: Here’s another approach to this problem using differentiation under the integral. Our integral can be simplified into$$\int\limits_{-\infty}^{\infty}dx\, e^{-x^2}\cos x+i\int\limits_{-\infty}^{\infty}dx\, e^{-x^2}\sin x=\int\limits_{-\infty}^{\infty}dx\, e^{-x^2}\cos x$$Now consider the general case $I(a)$$$I(a)=\int\limits_{-\infty}^{\infty}dx\ e^{-x^2}\cos ax$$Integration by parts gives us$$a\cdot I(a)=2\int\limits_{-\infty}^{\infty}dx\ xe^{-x^2}\sin ax=-2\frac {\partial}{\partial a}\int\limits_{-\infty}^{\infty}dx\ e^{-x^2}\cos ax=-2I’(a)$$Solving the basic differential equation however you like gives us$$I(a)=C_1e^{-a^2/4}$$To find our constant $C_1$, we set $a=0$ so $I(a)$ reduces to the Gaussian Integral. Hence$$\int\limits_{-\infty}^{\infty}dx\, e^{-x^2}\cos ax\color{blue}{=\sqrt{\pi}e^{-a^2/4}}$$
A: Yes, it's a Fourier transform. Set
$$F(t)=\int_{-\infty}^\infty e^{-x^2}e^{itx}\,dx.$$
Then
$$F'(t)=i\int_{-\infty}^\infty xe^{-x^2}e^{itx}\,dx
=-\frac i2\left[e^{-x^2}e^{itx}\right]_{x=-\infty}^\infty
-\frac{t}2\int_{-\infty}^\infty e^{-x^2}e^{itx}\,dx
=-\frac t2F(t).$$
Solving this differential equation gives
$$F(t)=F(0)e^{-t^2/4}=\sqrt\pi e^{-t^2/4}.$$
In particular,
$$F(1)=\sqrt\pi e^{-1/4}.$$
A: Completing the square works. For any $z \in\mathbb{C}$, we have
$$ \int_{-\infty}^{\infty} e^{-x^2+zx} \, dx
= e^{z^2/4} \int_{-\infty}^{\infty} e^{-\left( x - (z/2)\right)^2} \, dx
= e^{z^2/4} \int_{-\infty}^{\infty} e^{-x^2} \, dx
= \sqrt{\pi} e^{z^2/4}. $$
In the second equality, we substituted $x-\frac{z}{2} \mapsto x$.1) In this case, $z = i$ and hence the answer is
$$\sqrt{\pi}e^{-1/4}.$$

1) In fact, this computation is only heuristic and we need some justification. An easy way is to consider the function
$$f(z) = \int_{-\infty}^{\infty} e^{-\left( x - (z/2)\right)^2} \, dx$$
and check that $f$ is a holomorphic function on $\mathbb{C}$ whose value is identically $\sqrt{\pi}$ on $\mathbb{R}$. Then by the identity theorem, $f \equiv \sqrt{\pi}$.
Alternatively, one may adopt the contour-shifting technique. Indeed, the correct computation tells that the substitution $w = x-\frac{z}{2}$ yields
$$ \int_{-\infty}^{\infty} e^{-\left( x - (z/2)\right)^2} \, dx
= \int_{-\infty-\frac{i}{2}\operatorname{Im}(z)}^{\infty-\frac{i}{2}\operatorname{Im}(z)} e^{-w^2} \, dw. $$
In order to bring the contour back to the real line $\mathbb{R}$, we may utilize the fact that
$$ \oint_{C_R} e^{-w^2} \, dz = 0 $$
where $C_R$ is a rectangular contour joining $\pm R$ and $\pm R - \frac{i}{2}\operatorname{Im}(z)$ and then let $R\to\infty$.
