Problem 8

Exercise 1.8 (Gaussian integral with a shift)  

1. Prove that

   $$\int_{\mathbb{R}} e^{-(x+ia)^2}dx = \int_{\mathbb{R}}e^{-x^2}dx.$$ (1.56)

2. Use part (a) to prove

   $$\int e^{-ix\xi-\frac{x^2}{2\sigma^2}}= e^{-\frac{\xi^2\sigma^2}{2}} \sigma\sqrt{2\pi}.$$ (1.57)

Proof. To attack part (a), for $a=0$, we already have the result, so suppose $a>0$. The case $a<0$ is handled similarly. Integrating around the counterclockwise rectangle $\{-R, R, R+ia, -R + ia\}$ captures the nonexisting poles of the function, so we should have

$$\int_{-R}^R e^{-x^2}dx + \int_0^a e^{-(R+iy)^2}dy + \int_R^{-R} e^{-(x+ia)^2}dx + \int_a^0 e^{-(R+iy)^2}dy=0$$ (1.58)

Two of these integrals vanish as $R\to\infty$:

$$\left\vert\int_0^a e^{-(R+iy)^2}dy\right\vert \leq \int_0^a \vert... ...2}\vert dy = \int_0^a \vert e^{-R^2 - 2Ryi +y^2}\vert dy \leq ae^{a^2-R^2}\to 0$$ (1.59)

Similarly

$$\int_a^0 e^{-(R+iy)^2}dy \to 0$$ (1.60)

Therefore,

$$\int_{-\infty}^\infty e^{-x^2}dx+\int_\infty^{-\infty} e^{-(x+ia)^2}dx=0$$ (1.61)

which indicates

$$\int_{-\infty}^\infty e^{-(x+ia)^2}dx=\int_{-\infty}^\infty e^{-x^2}dx$$ (1.62)

For part (b), let $\sigma>0$ be fixed. Show that

$$\int_{\mathbb{R}} e^{-i x\xi}e^{-\frac{x^2}{2\sigma^2}}dx = \sigma\sqrt{2\pi}e^{-\frac{\sigma^2\xi^2}{2}}$$ (1.63)

To apply the previous part, we complete the square:

$$\frac{x^2}{2\sigma^2} +ix\xi =\frac{1}{2\sigma^2}(x^2 + ix\xi 2\sigma^2) = \frac{1}{2\sigma^2}\left[ (x + i\xi\sigma^2)^2-(i\xi\sigma^2)^2 \right]$$ (1.64)

$$= \frac{1}{2\sigma^2}(x + i\xi\sigma^2)^2 -\frac{1}{2\sigma^2}(i\xi\sigma^2)^2$$ (1.65)

$$= \frac{1}{2\sigma^2}(x + i\xi\sigma^2)^2 +\frac{1}{2\sigma^2}\left[\xi^2\sigma^4\right]$$ (1.66)

$$= \frac{1}{2\sigma^2}(x + i\xi\sigma^2)^2 +\frac{1}{2}\xi^2\sigma^2$$ (1.67)

Therefore

$$\int e^{-ix\xi-\frac{x^2}{2\sigma^2}}= e^{-\frac{\xi^2\sigma^2}{2}} \int e^{-\frac{1}{2\sigma^2}(x + i\xi\sigma^2)^2}$$ (1.68)

In this integral, substitute $u=x/\sqrt{2\sigma^2}$ and simplify by applying the previous result

$$\int e^{-\frac{1}{2\sigma^2}(x + i\xi\sigma^2)^2}dx = \int e^{-(x/\sqrt{2\sigma^2}+i\xi\sigma^2/\sqrt{2\sigma^2})^2}d... ...2\sigma^2}\int e^{-(u+i\xi\sigma/\sqrt{2})^2} = \sqrt{2\sigma^2}\int e^{-u^2}du$$ (1.69)

$$= \sqrt{2\sigma^2}\sqrt{\pi}$$ (1.70)

$$= \sigma\sqrt{2\pi}$$ (1.71)

Therefore,

$$\int e^{-ix\xi-\frac{x^2}{2\sigma^2}}= e^{-\frac{\xi^2\sigma^2}{2}} \sigma\sqrt{2\pi}$$ (1.72)

$\qedsymbol$