A method of evaluating a double integral that nobody taught you in school

Many times we are given integrals to evaluate. The standard way to evaluate is to find a series of transformations that will render the integral into something we know how to evaluate and then proceed. Examples of such transformations are substitutions, parts, replacement of an integrand with another integral, reversing order of integration, and so on.

However, once in a blue moon, we find an integral that is really something else. The trick that will allow us to locate the closed form is no mere algebraic transformation but is something else entirely. Yes, there is the use of symmetry, but there is something else entirely that just comes with experience.

The problem before us is to evaluate the following double integral:

$$\int_0^{2 \pi} dx \, \int_0^{\pi} dy\, \sin{y} \, e^{\sin{y} (\cos{x}-\sin{x})}$$

A direct evaluation of this integral is at best a mess. No matter what we will have some intermediate step involving a Bessel function or worse. Maybe that is the way we must proceed. However, I know the answer is quite simple and, thus, I do not want to involve Mr. Bessel.

Instead, the first insight I had was to notice the integral element $\sin{y} \,dx \, dy$. This describes an element of solid angle in a unit sphere, and wouldn’t you know, the integration limits imply we are integrating over the full unit sphere.

Notation becomes crucial at a time like this, so if we are going to think unit sphere, let’s notate the integration variables as azimuthal and polar angles, respectively. Thus, we write the integral as

$$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sin{\theta} (\cos{\phi}-\sin{\phi})}$$

All we have done is write the integral suggestively – we have done nothing to actually transform it into something else.

Now, note that we can replace the $\cos{\phi}-\sin{\phi}$ in the exponential with simply $\sqrt{2} \cos{\phi}$, as we are integrating over the whole azimuth anyway. Thus, the integral is

$$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sqrt{2} \sin{\theta} \cos{\phi}}$$

Now, note that on the unit sphere, $\sin{\theta} \cos{\phi} = x$, the $x$-coordinate. Thus, by simple geometry, we can replace this double integral with a single integral over $x$ by noting that the function we are integrating over only depends on $x$. The contribution to the integral from a ring a distance $x$ from the plane $x=0$ is $2 \pi \sqrt{1-x^2} e^{\sqrt{2} x} ds$, where $ds$ is an element of arc length. Of course, over a cross-section of the sphere, $y=\pm \sqrt{1-x^2}$ and

$$ds = \sqrt{1+\left ( \frac{dy}{dx} \right )^2} dx = \frac{dx}{\sqrt{1-x^2}}$$

Thus the integral is simply

$$2 \pi \int_{-1}^1 dx \, e^{\sqrt{2} x} = 2 \sqrt{2} \pi \sinh{\sqrt{2}}$$

I have independently verified the numerical value.