Evaluating definite integral of $e^{i t^2}$ In passing Sakurai's QM book mentions that 
$$\int_{-\infty}^\infty e^{i t^2} dt = \sqrt{i \pi}$$
This is consistent with 7.4.4 in Abramowitz and Stegun which claims for $\Re a > 0, n = 0, 1, 2, ...$
$$\int_{0}^\infty t^{2n} e^{-a t^2} dt = \frac{\Gamma(n+1/2)}{2 a^{n+1/2}},$$
however that restriction $\Re a > 0$ isn't satisfied in this case (this is equality with zero not a greater than condition).  
I noticed that the Sakurai integral can also be written
$$\int_{-\infty}^\infty e^{i t^2} dt = \frac{1}{2}\sqrt{\pi i}(\textrm{erf}(\infty + i\infty)- \textrm{erf}(-\infty - i \infty)),$$
where
$$\textrm{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{t^2} dt,$$
It seems plausible that $\textrm{erf}(\infty + i\infty) = -\textrm{erf}(-\infty - i \infty)) = 1$, but I'm not sure how one would show this (such limits aren't obvious from any of the other relations I seem in A&S).
How would an definite integral like this be evaluated, and what is the origin of this $\Re a > 0$ restriction?
 A: I'd rather claim that
$$
\int_{-\infty}^{+\infty}e^{it^2}\,dt=(1+i)\sqrt{\frac{\pi}{2}}.
$$
Note that
$$
e^{it^2}=\cos(t^2)+i\sin(t^2),
$$
so this really follows from the Fresnel integrals
$$
\int_{-\infty}^{+\infty}\cos(t^2)\,dt = \sqrt{\frac{\pi}{2}}
\quad\text{and}\quad
\int_{-\infty}^{+\infty}\sin(t^2)\,dt = \sqrt{\frac{\pi}{2}}.
$$
Edit
Since
$$
(1+i)\sqrt{\frac{\pi}{2}}=\sqrt{(\pi i)},
$$
I agree with Sakurai. I first thought that the $i$ was outside the square root in your question.
A: Let $f(z)=e^{iz^2}$.  Then, $f$ is entire and by the residue theorem we have
$$\oint_C f(z)\,dz=0 \tag 1$$
for any sufficiently smooth closed contour $C$ in the complex plane. Now, suppose $C$ is comprised of the three segments; $C_1$, $C_2$, and $C_3$, where
$(i)$ $C_1$ is the line segment on the real axis from $(0,0)$ to $(R,0)$.
$(ii)$ $C_2$ is the circular arc with radius $R$, centered at the origin, from $(R,0$ to $(R/\sqrt{2},R/\sqrt{2})$.
$(iii)$ $C_3$ is the line segment extending from $(R/\sqrt{2},R/\sqrt{2}$ to $(0,0)$.
Then, from $(1)$ we have
$$\int_0^R e^{ix^2}dx+\int_0^{\pi/4}e^{iR^2e^{i2\phi}}iRe^{i\phi}\,d\phi+\int_R^0e^{i\left(\frac{1+i}{\sqrt{2}}t\right)^2}\,\left(\frac{1+i}{\sqrt{2}}\right)dt=0 \tag 2$$
If we let $R\to \infty$, the first integral becomes $1/2$ the integral of interest, the second integral in $(2)$ vanishes, while the third integral becomes
$$-\frac{1+i}{\sqrt{2}}\int_0^\infty e^{-t^2}\,dt=-\frac{1+i}{\sqrt{2}}\frac12 \sqrt{\pi}$$
Putting it all together, we find
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty e^{ix^2}\,dx=(1+i)\sqrt{\frac{\pi}{2}}=\sqrt{i\pi}}$$
since $1+i=\sqrt{2}e^{i\pi/4}$
