February | 2018 | Mathematical Poetry

We will be looking at another Gaussian integral, this time with a trigonometric function included:

$\begin{aligned} I&=\int_{-\infty}^{\infty}e^{-ax^2}\cos bx\,dx \end{aligned}$

We will use the very powerful method of differentiation under the integral sign, getting a differential equation that we can solve with initial conditions. Consider

$\begin{aligned} I(b)&=\int_{\mathbb{R}}e^{-ax^2}\cos bx\,dx \end{aligned}$

Differentiating both sides with respect to $b$ ,

$\begin{aligned} I'(b)&=-\int_{\mathbb{R}}e^{-ax^2}x\sin bx\,dx \end{aligned}$

Using integration by parts,

$\begin{aligned} I'(b)&=\int_{\mathbb{R}}-e^{-ax^2}x\sin bx\,dx \\ I'(b)&=\frac{e^{-ax^2}}{2a}\sin bx \Biggr|_{-\infty}^{\infty}-\frac{b}{2a}\int_{\mathbb{R}} e^{-ax^2}\cos bx\,dx \end{aligned}$

The first term vanishes, while the second term is again our original function $I(b)$ . Our differential equation is thus

$\begin{aligned} I'(b)&=-\frac{b}{2a}I(b) \end{aligned}$

This diff eq. can be solved by separation of variables:

$\begin{aligned} \frac{dI}{db}&=-\frac{b}{2a}I \\ \frac{dI}{I}&=-\frac{b}{2a}\,db \\ \int\frac{dI}{I}&=-\int \frac{b}{2a}\,db \\ \log I&=-\frac{b^2}{4a}+C \\ I&=e^{\frac{-b^2}{4a}}e^C \\I&=Ae^{\frac{-b^2}{4a}} \end{aligned}$