Can we avoid the use of the geometric interpretation combined with polar coordinates change of variable for proving that

$$\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(x)}{ \cos(a \cos(x) \cos(y))} d x dy =\frac{\pi}{ 2a}\log\left( \frac{\displaystyle 1+ \tan \left(\frac{a}{2}\right)}{ \displaystyle 1-\tan\left(\frac{a}{2}\right)}\right)$$ ?
What if we go further and we also consider the case

$$\int_0^{\pi/2}\int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos(x)}{ \cos(a \cos(x) \cos(y) \cos(z))} d x dy \ dz$$ ? What can we say about the closed form of this one?

  • $\begingroup$ Actually, now you've made me pretty curious as to what geometric interpretation you had in mind in the first place. $\endgroup$ – Lucian Oct 6 '14 at 18:35
  • $\begingroup$ @Lucian if you let $x \mapsto \pi/2-x$ you get an integration over the positive octant of the unit sphere $\endgroup$ – user 1357113 Oct 6 '14 at 18:47

The simplest way I can see doing these integrals is to expand the secant term into a Taylor series, say

$$\sec{q} = \sum_{k=0}^{\infty} s_{2 k} q^{2 k}$$

where $s_{2 k}$ are known but not really important here. By using this expansion, however, we see that the double integral is

$$\sum_{k=0}^{\infty} s_{2 k} a^{2 k} \int_0^{\pi/2} dx \, \cos^{2 k+1}{x} \, \int_0^{\pi/2} dy \, \cos^{2 k}{y} $$

These integrals are easily done and may be found here. (Scroll down to the Addendum.) The binomial coefficient terms cancel and we get, for the integral,

$$\frac{\pi}{2} \sum_{k=0}^{\infty} s_{2 k} \frac{a^{2 k}}{2 k+1}$$

which we may rewrite as

$$\frac{\pi}{2 a} \sum_{k=0}^{\infty} s_{2 k} \frac{a^{2 k+1}}{2 k+1}$$

If we differentiate the sum, we get the series back for $\sec{a}$! So really, the integral is simply $\pi/(2 a)$ times the antiderivative of $\sec{a}$ (zero when $a=0$), so we get that the integral is

$$\frac{\pi}{2 a} \log{(\sec{a}+\tan{a})}$$

which is equivalent to the stated result.

The 3d case, is conceptually simple at this point; again, it is a sum:

$$\frac{\pi^2}{4} \sum_{k=0}^{\infty} s_{2 k} \frac{a^{2 k}}{2 k+1} \frac1{2^{2 k}} \binom{2 k}{k}$$

You would now need to know that $s_{2 k}=|E_{2 k}|/(2 k)!$, where $E_{2 k}$ are the Euler numbers. I am not sure there is an easy way to do this sum, but perhaps there is.


It should be clear that this problem is a special case of the following result: let $f(x)$ be an even-valued function over the reals. Then

$$\int_0^{\pi/2} dx \, \int_0^{\pi/2} dy \, \cos{y} \: f(a \cos{x} \cos{y}) = \frac{\pi}{2 a} F(a)$$

where $F'(x) = f(x)$ and $F(0)=0$.


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