Generalization of $\int_0^\alpha \sqrt{1+\cos^2\theta}\,d\theta>\sqrt{\alpha^2+\sin^2\alpha}$

Question

I came across a problem that required proving a specific case and then going on to generalize it. While I have no problem with the first part, I need some confidence from someone about the second part. Here we go.

The problem statement:

Show that for $\displaystyle 0<\alpha\leq \frac{\pi}{2}$ $$ \int_0^\alpha \sqrt{1+\cos^2\theta}\,d\theta>\sqrt{\alpha^2+\sin^2\alpha}\tag{1} $$
Generalize the result in part (1).

My Answer Attempt:
Clearly, LHS of $(1)$ represents the length of the curve $f(\theta)=\sin\theta$ from $\theta=0$ to $\theta=\alpha$ and the RHS represents the distance from the origin to the point $(\alpha,\sin\alpha)$, that is the length of the line from origin to the said point.

Curve in red and Line in blue.

Since both the curve and the line pass through the origin, and the shortest distance between two points is a straight line, we have $(1)$ proven.

The generalization that I've come up with:

For any continuous (not necessarily smooth) curve $f(x)$ in $[0,a]$ such that $f(0)=0$ and $a\in\mathbb{R}$, we have $$\left |\int_0^a \sqrt{1+\big[f'(x)\big]^2}dx\right|\geq \sqrt{a^2+\big[f(a)\big]^2} $$ Equality holding if the curve is a straight line or $a=0$.

This is basically saying that the shortest distance between two points is a straight line. The points in our case being the origin and a point on the curve $f(x)$.

The curve has to pass through the origin because otherwise the relation may not hold true. Example: The curve $f(x)=3$ in $[0,\infty)$.

Is this good?

Interesting problem. It works perfect for $f(x)=\tan(x)$ and, for this case, $$\text{lhs - rhs}=\frac{a^5 \left(23 a^2+42\right)}{1890 \sqrt{2}}+\cdots$$ — Claude Leibovici, Oct 14 '21 at 13:41
I took this case to give an example. Unfortunately, I cannot make a proof for the general case but I have the feeling that your conjecture is correct. Cheers :-) — Claude Leibovici, Oct 15 '21 at 02:35
Does the generalization necessarily require taking $sin(x)$ and $cos(x)$ as $f(x)$ and $f'(x)$? Could you recognize $sin^2(x) + cox^2(x) = 1$ so that $\sqrt({sin^2(x)}) = sin(x)$ is the left hand integral? Is that too specific? That is how I see the problem, recognizing sin(x) is a bounded curve. — Gwendolyn Anderson, Oct 15 '21 at 16:31
Of course I meant $\sqrt({2 + sin^2(x)})$. You would then make a generalization for alpha. — Gwendolyn Anderson, Oct 15 '21 at 17:18
I am a bit confused why the reasoning you proposed (shortest path is the linear line) should not apply to the general case directly. Could you clarify that for me? — Thomas, Oct 16 '21 at 08:40
So you just wanted a confirmation of the fact that the reasoning apply to the general case? By the way nice trick :) — Thomas, Oct 16 '21 at 13:34

score 1 · Answer 1 · answered Oct 15 '21 at 14:48

1

My "geometrically" solution.

https://i.stack.imgur.com/wYgjx.png

answered Oct 15 '21 at 14:48

eyal20

86

3

Please typeset your answer using MathJax. – StubbornAtom Oct 15 '21 at 16:24

Thomas Bakx · Accepted Answer · 2021-10-16T07:02:45.227

1

The inequality naturally only makes sense for continuously differentiable $f$. We will use that if $F$ and $G$ are continuously differentiable, and $F(0)=G(0)$ and $F'(y) \geq G'(y)$ for all $y \in [0,z]$ then $F(y) \geq G(y)$ on that interval.

Equality holds for $a=0$. Differentiating both sides w.r.t $a$ gives $$\sqrt{1+(f'(a))^2} \geq \frac{a+f(a)f'(a)}{\sqrt{a^2+f(a)^2}}$$ Which after squaring both sides and rearranging amounts to $$a^2f'(a)^2 + f(a)^2 \geq 2a f(a) f'(a)$$ or $$ (f(a)-a f'(a))^2=0.$$ Insisting that this holds for all $a$ is easily seen to imply linearity of $f$.

edited Oct 16 '21 at 07:02

answered Oct 15 '21 at 19:23

Thomas Bakx

1,113

I think you should mention that $F'(x)\geq G'(x)$ for all $x\in[0,a]$ implies $F(x)\geq G(x)$ – dezdichado Oct 15 '21 at 21:02
@ThomasBakx I've made some edits to the question. – Abhishek A Udupa Oct 16 '21 at 01:35
What edits exactly? – Thomas Bakx Oct 16 '21 at 06:59
The absolute value signs are not necessary. – Thomas Bakx Oct 16 '21 at 07:05
@ThomasBakx I don't think the function should be continuously differentiable. It can have a finite number of non-differentiable points like cusps or sharp corners etc... – Abhishek A Udupa Oct 16 '21 at 13:04
If $f$ is piecewise $C^1$ and continuous, just apply the proof to each interval on which $f$ is $C^1$ and conclude by using the triangle inequality. I'm just saying that if $f$ is merely continuous, you'll need to think about your definition of 'arc length'. The formula involving $f'$ makes no sense in that case. Moreover, I'm pretty sure you can construct continuous functions for which 'arc length' is very hard to make sense of. – Thomas Bakx Oct 16 '21 at 13:12
See https://math.stackexchange.com/questions/3562210/arc-length-of-non-differentiable-functions. – Thomas Bakx Oct 16 '21 at 13:20
@ThomasBakx I see. I'm yet to learn math to that level. – Abhishek A Udupa Oct 17 '21 at 09:42
No problem! Please don't forget to accept my answer, if you're happy with it :) – Thomas Bakx Oct 17 '21 at 12:33

score 0 · Answer 3 · answered Oct 16 '21 at 16:47

This is not an answer but some ideas: The integral involved in the problem is an Elliptic integral and maybe can be analyzed with the theory of this functions

Recall the incomplete elliptic integral of the second kind:

$$E(\varphi,k) = \int_{0}^{\varphi} \sqrt{1-k^2\sin^2 \theta}d\theta$$

where $k^2<1$

Then the left hand side of the inequality is

$$ \int_0^\alpha \sqrt{1+\cos^2\theta}\,d\theta = \int_0^\alpha \sqrt{2-\sin^2\theta}\,d\theta = \sqrt{2} \int_0^\alpha \sqrt{1-\frac{1}{2}\sin^2\theta}\,d\theta = \sqrt{2} E\left(\alpha, \frac{1}{\sqrt{2}}\right) $$

Here is a formula which states that

$$\max \left(\sin \phi, \phi\sqrt{1-k^2\sin^2 \phi} \right)\leq E(\phi,k)\leq \phi$$

Therefore

$$\max \left(\sin \alpha, \alpha\sqrt{1-\frac{1}{2}\sin^2 \alpha} \right)\leq E(\alpha,\frac{1}{\sqrt{2}})$$

Hence

$$\boxed{\int_0^\alpha \sqrt{1+\cos^2\theta}\,d\theta\geq \sqrt{2} \max \left(\sin \alpha, \alpha\sqrt{1-\frac{1}{2}\sin^2 \alpha} \right)}$$ This inequality is far from being the one we look for but I hope it works as food for thought

Generalization of $\int_0^\alpha \sqrt{1+\cos^2\theta}\,d\theta>\sqrt{\alpha^2+\sin^2\alpha}$

3 Answers3