How to take this integral? It looks like as Gamma but I'm confused.

Question

x and b are real. b is constant.

$$\int _{ -\infty }^{ +\infty }{ { e }^{ { (ix+b) }^{ 2 } } } dx$$

What the answer should be?

Answer 1

By contour shifting,

$$\int_{-\infty}^\infty e^{-(x - bi)^2} \, dx = \int_{-\infty}^\infty e^{-x^2}\, dx \tag{*}$$

To justify this, consider the contour integral

$$\int_{C(R)} e^{-z^2}\, dz$$

where $R > 0$ and $C(R)$ is a positively oriented rectangle parallel to the axes, with vertices at $-R, R, R - bi$, and $-R - bi$. The integrals along the vertical edges of $C(R)$ are $O(e^{-R^2})$, so they are negligible as $R\to \infty$. Furthermore, since $e^{-z^2}$ is entire, Cauchy's theorem gives $\int_{C(R)} e^{-z^2}\, dz = 0$. Hence, the contour shifting is justified and $(*)$ holds. Since $(ix + b)^2 = (i(x - bi))^2 = -(x - bi)^2$, we have

$$\int_{-\infty}^\infty e^{(ix + b)^2}\, dx = \int_{-\infty}^\infty e^{-(x - bi)^2}\, dx = \int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}.$$

Answer 2

To give a slightly longer take than Spencer's answer, $$ \int _{ -\infty }^{ +\infty }{ { e }^{ { (ix+b) }^{ 2 } } } dx = \int_{b - i\infty}^{b + i\infty} e^{z^2}dz, $$ interpreting it now as a vertical contour integral (or alternatively still, as an inverse Mellin Transform). As $e^{z^2}$ is entire and has extremely rapid decay in vertical strips, we can shift the line of integration to be along $\text{Re } z = 0$ instead. [This is equivalent to using the Residue Theorem on rectangles of increasing height and taking the limit].

So your integral is the same as $$ \int_{-i\infty}^{i\infty} e^{z^2} dz = \int_{-\infty}^\infty e^{-x^2} dx = \sqrt \pi, $$ which is the standard Gaussian integral. There is little gained in the explicit shift to complex variables, except perhaps clarity. $\diamondsuit$

As an aside, it also is nice if you know that the Mellin Transform is a sort of change of variables of the typical Fourier transform; then this integral takes on a sort of enhanced meaning.