Find $\int_0^\infty f(x)dx$ given that $f(0)=1$ and for all $a>0$, the arc length from $0$ to $a$ equals the $x$-intercept of the tangent at $a$.

Question

I made up the following problem.

A curve $y=f(x)$ passes through $(0,1)$, and for all $a>0$, the arc length of $y=f(x)$ from $x=0$ to $x=a$ equals the $x$-intercept of the tangent to $y=f(x)$ at $x=a$.

That is, in the graph below, the red arc and the red line segment have the same length.

What is the area under the curve, $A=\int_0^\infty f(x)dx$ ?

My attempt

We have

$$\int_0^a \sqrt{1+(f'(x))^2}dx=a-\frac{f(a)}{f'(a)}$$

Differentiating both sides with respect to $a$ gives

$$\sqrt{1+(f'(a))^2}=1-\frac{(f'(a))^2-f(a)f''(a)}{(f'(a))^2}$$

$$(y')^6+(y')^4-(yy'')^2=0$$

But I don't know how to solve this differential equation.

Then I approximated the curve with a sequence of triangles as shown below.

Each of the colored line segments has length $\epsilon$, which approaches $0$.

Let $h_n$ be the height, and let $\theta_n$ be the lower-right angle, of the $n$th triangle from the left. Let $\theta_0=\pi/2$.

$h_n=1-\epsilon\sum\limits_{k=0}^{n-1} \sin{\theta_k}$

$\theta_n = \arctan{\left(\dfrac{h_n}{n\epsilon -\epsilon\sum\limits_{k=1}^{n-1}\cos{\theta_k}}\right)}$

I made an Excel simulation using these two equations, and it suggests that $A=1/3$. (The length of the longest side of the triangles seems to approach $1/2$.)

I tried to rearrange the triangles in some clever way to get a total area of $1/3$, but I haven't found a way.

Answer 1

Hint Regard $x$ as a function of $y$, so that the tangent line to the graph at $(x(b), b)$ is $T(y) = x'(b) (y - b) + x(b)$; then our condition is: $$\int_b^1 \sqrt{1 + x'(y)^2} \,dy = T(0) = x(b) - b x'(b) .$$ Differentiating with respect to $b$, renaming $b$ as $y$, and rearranging gives a more manageable differential equation, in fact a separable equation of first order in $x'(y)$: $$y^2 x''(y)^2 - x'(y)^2 - 1 = 0 .$$

Writing $u(y) := x'(y)$ reduces the equation to $y^2 u'(y)^2 - u^2 - 1 = 0$, and separating and integrating yields $$u(y) = \pm \sinh (\log y + C).$$ The tangent to the original curve through $(0, 1)$ passes through $(0, 0)$, so $u(1) = x'(1) = 0$, and substituting gives $C = 0$, so $$u(y) = \pm\sinh(\log y) = \pm\left(\frac{y}{2} - \frac{1}{2 y}\right) .$$ Since $x(y)$ is decreasing on $(0, 1)$, $u(y) = x'(y)$ is negative on that interval, hence we need the solution with $\pm = +$: $$u(y) = \frac{y}{2} - \frac{1}{2 y}.$$ Integrating again yields $$x(y) = \frac{1}{4} y^2 - \frac{1}{2} \log y + D .$$ Since our curve passes through $(0, 1)$, we find that $D = -\frac{1}{4}$, so our curve is $$x(y) = \frac{1}{4}(y^2 - 1) - \frac{1}{2} \log y .$$ Finally, the area under the curve is $$\int_0^\infty y(x) \,dx = \int_0^1 x(y) \,dy = \int_0^1 \left[\frac{1}{4}(y^2 - 1) - \frac{1}{2} \log y\right] \,dy = \boxed{\frac{1}{3}} ,$$ in agreement with your numerical computation.

Remark Inverting gives $y$ as a function of $x$ in terms of the Lambert $W$ function: $$y(x) = \sqrt{-W(-e^{-4 x - 1})}.$$