Lately I stumbled accros the magnifient paper by Roger Nelsen, which can be found here Symmetry and Integration In this paper it is shown that $$ \int_0^{\pi/2} \frac{\mathrm{d}x}{1 + \tan(x)^{\sqrt{2}}} $$ Takes up exactly half of the area of the rectangle with base $\pi/2$ and height $1$. This idea is supported geometrically by a very convincing figure. The argument was also done mathematically by noting that if $$ f(x) + f(a+b-x)$$ happens to be a constant, then one can proceed as above.
I wanted to find an counter intuitive example where it would be misleading to look at the figure. The integral I found was an elliptic one, namely $$ A = \int_0^{\pi/2} \sqrt{1 + \cos^2 x} \mathrm{d}x \approx 1.9100 $$ Which when drawn looks something like
The image is very suggestive into thinking that the area of the figure should be $$ \tilde{A} = \frac{\pi}{4}\bigl( 1 + \sqrt{2}\bigr) \approx 1.8961 $$ Where only the formula for the trapezoid was used. This idea is wrong since clearly $1.9100 > 1.8961$. Mathematically one can argue by looking at $f(x) + f(a+b-x)$ one sees that $$ f(x) + f(\pi/2-x) = \sqrt{1+\cos^2x} + \sqrt{1+\sin^2x} $$ it is not constant. The function wobbles very slightly between $1+\sqrt{2}$ and $\sqrt{6}$. This proves that the function $\sqrt{1+\cos^2x}$ does not have the symmetric properties one thought it had.
Using the same argument on $\sin^2x$ or $\cos^2x$ on $\pi/2$ shows that $f(x)+f(\pi/2-x)=1$, and hence the function "fills" exacly half of the square with base $\pi/2$ and height $1$.
My question about this problem is two fold:
- What is it in the elliptic function that breaks the symmetry? Clearly it is not the trigonometric part.
- Why is the integral bigger than the "expected" value?
I can use Romberg, Trapezoid or any numerical integration method to evaluate the elliptic integral. But I still do not see intuitively why the area is bigger, nor what breaks the nice symmetry.
The Taylor series for $\sqrt{1+u}$ to the fourth term is $$t(u)=1+\frac{u}{2}-\frac{u^2}{8}+\frac{u^3}{16}-\frac{5u^4}{128}.$$ For $u \le 1$ the full taylor series is, from the second term onwards, alternating in sign and with strictly decreasing absolute values: If $a_n$ is the absolute value of the coefficient of $u^n$ then (for $n\ge 1$) we have $a_{n+1}=\frac{2n-1}{2n+2}\cdot a_n.$ It follows that, if we substitute $u=\cos^2 x$ into $t(u),$ we have $\sqrt{1+\cos^2 x}\ge t(\cos^2 x).$ The latter contains even powers of cosine, and can be integrated explicitly from $0$ to $\pi/2$. The result is $$\int_0^{\pi/2} t(\cos^2 x) dx=\frac{19857\pi}{32768}=1.9037..$$ Note that this exceeds $\pi/4 (1+\sqrt{2})=1.896..$ and is a bit less than the value $1.910098..$ of the integral of $\sqrt{1+\cos^2 x}.$