6

$\def\d{\mathrm{d}}$If $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(x)=x+\cos x$, find$$\int^{\pi}_{0}f^{-1}(x) \,\d x.$$

Try: put $x=f(t)$ and $\d x=f'(t) \,\d t$, so \begin{align*} \int^{f^{-1}(\pi)}_{f^{-1}(0)}tf'(t) \,\d t &=\int^{f^{-1}(\pi)}_{f^{-1}(0)}t(1-\sin t) \,\d t\\ &= \left.\left(t \cos t-\sin t+\frac{t^2}{2}\right)\right|^{f^{-1}(\pi)}_{f^{-1}(0)}. \end{align*}

Could someone help me how to find $f^{-1}(\pi)$ and $f^{-1}(0)$, thanks.

Ѕᴀᴀᴅ
  • 34,263
DXT
  • 11,241

3 Answers3

7

Since $\cos(\pi-x)=-\cos x$, we have $$f(\pi-x)=\pi-f(x).$$ Applying $f^{-1}$ to both sides and replacing $x$ by $f^{-1}(x)$, this gives $$f^{-1}(\pi-x)=\pi-f^{-1}(x).$$ Then, $$\int^\pi_0f^{-1}(\pi-x)\,dx=\int^\pi_0(\pi-f^{-1}(x))\,dx.$$ The substitution $x\to\pi-x$ on the LHS gives $$\int^\pi_0f^{-1}(x)\,dx=\pi^2-\int^\pi_0f^{-1}(x)\,dx,$$ i.e. $$\int^\pi_0f^{-1}(x)\,dx=\frac{\pi^2}2,$$ as @Patrick Stevens guessed from numerical calculation in a comment, already.
REMARK: originally, I started with $x_0=f^{-1}(0)$, and found that then, we must have $f^{-1}(\pi)=\pi-x_0$. If we plug both values into the expression given by the OP and use $\sin(\pi-x_0)=\sin x_0$, $\cos x_0=-x_0$ and $\cos(\pi-x_0)=x_0$, it miraculously simplifies to $\pi^2/2$. Thinking about "why?", I came up with the above.

7

Using graph and symmetry arguments, we can find the integral. The graph of inverse of $f(x)$ will look like:

enter image description here

It is now possible to show that the inverse function has a point of symmetry $(\pi/2, \pi/2)$.

The red regions actually cancel out, and green region is half of the rectangle namely $\frac{\pi(\pi-2)}{2}$. The blue region is $\pi$.

So the answer is $\dfrac{\pi^2}{2}$.

King Tut
  • 3,074
  • Nice visualisation! –  Feb 12 '18 at 10:06
  • Actually. I like your answer for not using graph at all. – King Tut Feb 12 '18 at 10:41
  • 1
    The popular vote seems to be divided. Don't be afraid of pictures, though! Famous mathematicians like Littlewood didn't condemn pictures, as long as they can be immediately translated into rigorous text. –  Feb 12 '18 at 14:18
0

Denote the single solution of the equation $\cos t=t$ by $\alpha\ (\approx0.739085)$.

One has $f'(x)=1-\sin x\geq0$ with isolated points of equality. It follows that $f$ is strictly increasing. Furthermore $$f(-\alpha)=-\alpha+\cos\alpha=0,\quad f(\pi+\alpha)=\pi+\alpha-\cos\alpha=\pi\ .$$ Let $g:=f^{-1}$ be the inverse function of $f$. Then $$\int_0^\pi g(t)\>dt=\int_{-\alpha}^{\pi+\alpha} g\bigl(f(x)\bigr)\>f'(x)\>dx=\int_{-\alpha}^{\pi+\alpha}x f'(x)\>dx\ .$$ Since $f'$ is even with respect to the midpoint ${\pi\over2}$ of the interval of integration we rewrite the first factor in the last integral as ${\pi\over2}+\bigl(x-{\pi\over2}\bigr)$ and obtain $$\int_{-\alpha}^{\pi+\alpha}x f'(x)\>dx={\pi\over2}\int_{-\alpha}^{\pi+\alpha}f'(x)\>dx={\pi\over2}\bigl(f(\pi+\alpha)-f(-\alpha)\bigr)={\pi^2\over2}\ .$$

Christian Blatter
  • 226,825