Does $ \int _0^1 \frac{\arctan \left(\frac{x}{x+1}\right)}{\arctan \left(\frac{-2x^2+2x+1}{2}\right)}dx $ exactly equal to $ \frac{1}{2} $?

Question

Is the integral exactly equal to $\frac12$ or is it just an extremely good approximation?

Answer 1

Call the integral $I$. Let $x \mapsto 1-x$ then add them together using

$$\arctan{a}+\arctan{b} = \arctan\left(\frac{a+b}{1-ab}\right)$$

Add the two integrals; you'll get $\displaystyle 2I = \int_0^1 \, dx = 1$ hence $\displaystyle I = \frac{1}{2}$.

Details: The transformation $x \mapsto x-1$ preserves the denominator; we get:

$\displaystyle I = \int_0^1 \displaystyle \frac{\arctan \left(\frac{x}{x+1}\right)}{\arctan{\left(-x^2+x+\frac{1}{2}\right)}} \,{dx}= \int_0^1 \displaystyle \frac{\arctan \left(\frac{1-x}{2-x}\right)}{\arctan{\left(-x^2+x+\frac{1}{2}\right)}} \,{dx}$ and adding them together:

$$\begin{aligned} \displaystyle 2I & = \int_0^1 \displaystyle \frac{\arctan \left(\frac{x}{x+1}\right)+\arctan \left(\frac{1-x}{2-x}\right)}{\arctan{\left(-x^2+x+\frac{1}{2}\right)}} \,{dx} \\& =\int_0^1 \frac{\arctan \bigg(\frac{\frac{x}{x+1}+\frac{x-1}{x-2}}{1-\frac{x}{x+1}\cdot \frac{x-1}{x-2}}\bigg)}{\arctan \left(-x^2+x+\frac{1}{2}\right)}\,{dx} \\& = \int_0^1 \frac{\arctan \left(-x^2+x+\frac{1}{2}\right)}{\arctan \left(-x^2+x+\frac{1}{2}\right)}\,{dx} \\& = \int_0^1 \,{dx} \\& = 1\end{aligned} $$

Hence $\displaystyle I = \frac{1}{2}$. I guess the most tedious step is where we simplify:

$\displaystyle \frac{\frac{x}{x+1}+\frac{x-1}{x-2}}{1-\frac{x}{x+1}\cdot \frac{x-1}{x-2}} = \frac{x(x-2)+(x-1)(x+1)}{(x+1)(x-2)-x(x-1)} = -\frac{1}{2} (2x^2-2x-1) = -x^2+x+\frac{1}{2}.$

Answer 2

With $f(x)$ as that integrand, you have $f(0)=0$ and $f(1)=1$. So the curve passes through two opposite corners of the unit square.

Now the claim is that $f$'s graph passes through the unit square symmetrically about $(1/2,1/2)$, which would imply that the given integral is $1/2$ (half the unit square's area). To prove this, you need to prove that $f(x+1/2)-1/2$ is an odd function. That is, that: $$ \begin{align} &&f(x+1/2)-1/2&=-[f(-x+1/2)-1/2]\\ \iff&& f(x+1/2)-1/2&=-f(-x+1/2)+1/2\\ \iff&& f(x+1/2)&=-f(-x+1/2)+1\\ \iff&& \frac{\arctan \left(\frac{x+1/2}{x+3/2}\right)}{\arctan \left(\frac{-2(x+1/2)^2+2(x+1/2)+1}{2}\right)} &=-\frac{\arctan \left(\frac{-x+1/2}{-x+3/2}\right)}{\arctan \left(\frac{-2(-x+1/2)^2+2(-x+1/2)+1}{2}\right)}+1\\ \iff&& \frac{\arctan \left(\frac{x+1/2}{x+3/2}\right)}{\arctan \left(-x^2+3/4\right)} &=-\frac{\arctan \left(\frac{-x+1/2}{-x+3/2}\right)}{\arctan \left(-x^2+3/4\right)}+1\\ \iff&& \arctan \left(\frac{x+1/2}{x+3/2}\right)&=-\arctan \left(\frac{-x+1/2}{-x+3/2}\right)+\arctan \left(-x^2+3/4\right) \end{align} $$

Now we can prove $$\arctan \left(\frac{x+1/2}{x+3/2}\right)+\arctan \left(\frac{-x+1/2}{-x+3/2}\right)=\arctan \left(-x^2+3/4\right)$$ The arctangent addition formula applies to the left side: $$ \begin{align} &&\arctan \left(\frac{\frac{x+1/2}{x+3/2}+\frac{-x+1/2}{-x+3/2}}{1-\frac{x+1/2}{x+3/2}\frac{-x+1/2}{-x+3/2}}\right)&=\arctan \left(-x^2+3/4\right)\\ \iff&&\arctan \left(\frac{(x+1/2)(-x+3/2)+(-x+1/2)(x+3/2)}{(x+3/2)(-x+3/2)-(x+1/2)(-x+1/2)}\right)&=\arctan \left(-x^2+3/4\right)\\ \iff&&\arctan \left(\frac{-2x^2+3/2}{2}\right)&=\arctan \left(-x^2+3/4\right)\\ \end{align} $$

And the claim holds.