Question

`\rm \int_(-\infty)^\infty {e}^{ - {x}^(2) } \cos(2 {x}^(2) )dx \\`​

Answer 1

A rather lengthy solution using a neat method I just learned relying on complex analysis.

First observe that

`e^(-x^2) \cos(2x^2) = \mathrm{Re}\left[e^(-x^2) e^(i\,2x^2)\right] = \mathrm{Re}\left[e^(a x^2)\right]`

where
`a=-1+2i`.

Normally we would consider the integrand as a function of complex numbers and swapping out
`x` for
`z\in\Bbb C`, but since it's entire and has no poles, we cannot use the residue theorem right away. Instead, we introduce a new function
`g(z)` such that

`f(z) = (e^(a z^2))/(g(z))`

has at least one pole we can work with, along with the property (1) that
`g(z)` has period
`w` so
`g(z)=g(z+w)`.

Now in the complex plane, we integrate
`f(z)` along a rectangular contour
`\Gamma` with vertices at
`-R`,
`R`,
`R+ib`, and
`-R+ib` with positive orientation, and where
`b=\mathrm{Im}(w)`. It's easy to show the integrals along the vertical sides will vanish as
`R\to\infty`, which leaves us with

`\displaystyle \int_\Gamma f(z) \, dz = \int_(-R)^R f(z) \, dz + \int_(R+ib)^(-R+ib) f(z) \, dz = \int_(-R)^R f(z) - f(z+w) \, dz`

Suppose further that our cooked up function has the property (2) that, in the limit, this integral converges to the one we want to evaluate, so

`f(z) - f(z+w) = e^(a z^2)`

Use (2) to solve for
`g(z)`.

`\displaystyle f(z) - f(z+w) = (e^(a z^2) - e^(a(z+w)^2))/(g(z)) = e^(a z^2) \\\\ ~~~~ \implies g(z) = 1 - e^(2azw) e^(aw^2)`

Use (1) to solve for the period
`w`.

`\displaystyle g(z) = g(z+w) \iff 1 - e^(2azw) e^(aw^2) = 1 - e^(2a(z+w)w) e^(aw^2) \\\\ ~~~~ \implies e^(2aw^2) = 1 \\\\ ~~~~ \implies 2aw^2 = i\,2\pi k \\\\ ~~~~ \implies w^2 = \frac{i\pi}a k`

Note that
`aw^2 = i\pi`, so in fact

`g(z) = 1 + e^(2azw)`

Take the simplest non-zero pole and let
`k=1`, so
`w=\sqrt{\frac{i\pi}a}`. Of the two possible square roots, let's take the one with the positive imaginary part, which we can write as

`w = \displaystyle -\sqrt{\frac\pi{\sqrt5}} e^{-i\,\frac12 \tan^(-1)\left(\frac12\right)}`

and note that the rectangle has height

`b = \mathrm{Im}(w) = \sqrt{\frac\pi{\sqrt5}} \sin\left(\frac12 \tan^(-1)\left(\frac12\right)\right) = \sqrt{(\sqrt5-2)/(10)\,\pi}`

Find the poles of
`g(z)` that lie inside
`\Gamma`.

`g(z_p) = 1 + e^(2azw) = 0 \implies z_p = \frac{(2k+1)\pi}2 e^{i\,\frac14 \tan^(-1)\left(\frac43\right)}`

We only need the pole with
`k=0`, since it's the only one with imaginary part between 0 and
`b`. You'll find the residue here is

`\displaystyle r = \mathrm{Res}\left((e^(az^2))/(g(z)), z=z_p\right) = \frac12 \sqrt{-\frac{5a}\pi}`

Then by the residue theorem,

`\displaystyle \lim_(R\to\infty) \int_(-R)^R f(z) - f(z+w) \, dz = \int_(-\infty)^\infty e^((-1+2i)z^2) \, dz  = 2\pi i r \\\\ ~~~~ \implies \int_(-\infty)^\infty e^(-x^2) \cos(2x^2) \, dx = \mathrm{Re}\left[2\pi i r\right] = \sqrt{\frac\pi{\sqrt5}} \cos\left(\frac12 \tan^(-1)\left(\frac12\right)\right)`

We can rewrite

`\cos\left(\frac12 \tan^(-1)\left(\frac12\right)\right) = \sqrt{(5+\sqrt5)/(10)}`

so that the result is equivalent to

`\sqrt{\frac\pi{\sqrt5}} \cos\left(\frac12 \tan^(-1)\left(\frac12\right)\right) = \boxed{\sqrt{\frac{\pi\phi}5}}`