Inductively prove that this sequence of integrals is bounded.

Question

EDIT: I have an attempted solution to this in a post below, it is very long, but still incomplete.

EDIT:Alright, I've pretty much almost finished my solution, but my biggest problem is the 2nd integral, rightmost inequality. I cannot find a way to evaluate it and compare it to the n=1 case for the upper integral, which should give me what I need.

EDIT: still stuck!

EDIT: 10 days later, I still need help with this..

Can someone please help me with this? I'm not even sure where to go from here.

Here is the problem:

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Prove that

$$\frac{1}{2}<\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,$$

for any integer $n\ge1$,

And that

$$\frac{1}{2}<\int_{0}^{1}\frac{dx}{\sqrt{4-x+x^{3}}}\le0.52359$$

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Now I can do this for the $n=1$ case, as obviously this simply integrates to $\arcsin{\frac{1}{2}}$ giving $\frac{\pi}{6}$, but I'm completely stumped on any doing any further for this question.

All help very appreciated, thank you for your time.

Answer 1

In the interest of completeness, and at the risk of duplication, I will include my solution of the first part of the question. $$ \frac12\lt\int_0^{1/2}\frac{\mathrm{d}x}{\sqrt{1-x^{2n}}} $$ since $1\lt\dfrac1{\sqrt{1-x^{2n}}}$ on $\left(0,\frac12\right)$.

Furthermore, since $\dfrac1{\sqrt{1-x^{2n}}}\le\dfrac1{\sqrt{1-x^2}}$ on $\left(0,\frac12\right)$ for $n\ge1$, we have $$ \begin{align} \int_0^{1/2}\frac{\mathrm{d}x}{\sqrt{1-x^{2n}}} &\le\int_0^{1/2}\frac{\mathrm{d}x}{\sqrt{1-x^2}}\\ &=\arcsin\left(\tfrac12\right)-\arcsin(0)\\ &=\frac\pi6\\[6pt] &\doteq0.5235987755983 \end{align} $$

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Second Part

$$ \frac12<\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}} $$ since $\frac12\lt\dfrac1{\sqrt{4-x+x^3}}$ on $(0,1)$.

Furthermore, $4-x+x^3$ is convex on $[0,1]$. Looking at its tangents at $0$ and $1$, and its minimum on $[0,1]$,

$\hspace{3cm}$

we get $$ \frac1{\sqrt{4-x+x^3}}\le\min\left(\frac1{\sqrt{4-x}},\frac1{\sqrt{4-2/\sqrt{27}}},\frac1{\sqrt{2+2x}}\right) $$ Therefore, $$ \begin{align} &\int_0^1\frac{\mathrm{d}x}{\sqrt{4-x+x^3}}\\ &\le\int_0^{2/\sqrt{27}}\frac{\mathrm{d}x}{\sqrt{4-x}} +\int_{2/\sqrt{27}}^{1-1/\sqrt{27}}\frac{\mathrm{d}x}{\sqrt{4-2/\sqrt{27}}} +\int_{1-1/\sqrt{27}}^1\frac{\mathrm{d}x}{\sqrt{2+2x}}\\ &=2\left(2-\sqrt{4-2/\sqrt{27}}\right) +\frac{1-3/\sqrt{27}}{\sqrt{4-2/\sqrt{27}}} +\left(2-\sqrt{4-2/\sqrt{27}}\right)\\ &=6-\frac{11-3/\sqrt{27}}{\sqrt{4-2/\sqrt{27}}}\\[6pt] &\doteq0.5182655150624\\[18pt] &\lt0.5235987755983 \end{align} $$

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On Strict Inequality

Suppose we know that $f(x)\le g(x)$ on $[a,b]$ and that $f(x)\lt g(x)$ on $(a,b)$. If $f$ and $g$ are continuous and $a\lt b$, then we can pick any $a'$ and $b'$ so that $a\lt a'\lt b'\lt b$. $g-f$ is continuous on the compact set $[a',b']$ so it attains its minimum on $[a',b']$ that is $$ g(x)-f(x)\ge m\gt0\text{ for }x\in[a',b']\subset(a,b) $$ Therefore, $$ \begin{align} \int_a^b(g(x)-f(x))\,\mathrm{d}x &\ge\int_{a'}^{b'}(g(x)-f(x))\,\mathrm{d}x\\ &\ge\int_{a'}^{b'}m\,\mathrm{d}x\\[6pt] &=m(b'-a')\\[12pt] &\gt0 \end{align} $$ so that we have the strict inequality $$ \int_a^bf(x)\,\mathrm{d}x\lt\int_a^bg(x)\,\mathrm{d}x $$

Answer 2

First note that the leftmost inequality is trivial, since $\frac{1}{\sqrt{1-x^{2n}}}\geq 1$.

Now note that for $x<1$, $x^{2(n+1)} = x^2x^{2n} \leq x^{2n}$. Hence $\frac{1}{\sqrt{1-x^{2(n+1)}}}\leq\frac{1}{\sqrt{1-x^{2n}}}$ for all $x<1$. Thus $\displaystyle\int_0^\frac{1}{2}\frac{\mathrm{d}x}{\sqrt{1-x^{2(n+1)}}} \leq \displaystyle\int_0^\frac{1}{2}\frac{\mathrm{d}x}{\sqrt{1-x^{2n}}}$. The rightmost inequality now follows by induction.

Answer 3

Alright, here is my attempt at a full solution, can anyone tell me if it is correct?

EDIT: Somewhere along the way I realise I have went wrong, which is pointed out in this post.

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For the first integral:

Trivially:

$$\frac{1}{\sqrt{1-x^{2n}}}\ge1, \forall x \in \mathbb{R}, \forall n \in \mathbb{N}$$

From here, we use

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\ge\int_0^\frac{1}{2}dx=\frac{1}{2}$$

$$\implies \frac{1}{2}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}} \forall n\in\mathbb{N}$$

[Note, I have proven the $\le$ relationship here, but I don't know how to also confirm the $<$ relationship between the integral and $\frac{1}{2}$, any help here?]

Now, for $x \in[0,\frac{1}{2}],x^{2(n+1)}\le x^{2n}$

$$\implies \frac{1}{\sqrt{1-x^{2(n+1)}}}\le \frac{1}{\sqrt{1-x^{2n}}}\forall n\in\mathbb{N},x\in[0,\frac{1}{2}]$$

$$\implies\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2(n+1)}}}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\forall n\in\mathbb{N}$$

With this information, we can now prove the statement:

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,\forall n\in\mathbb{N}$$

For $n=1$,

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2(1)}}}=\left[\arcsin{x}\right]_0^\frac{1}{2}=\frac{\pi}{6}\approx0.52359$$

Assume statement true for $n=k$

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2k}}}\le0.52359$$

Now, for $n=k+1$, we know:

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2(k+1)}}}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2k}}}\le0.52359$$

Therefore

$$\frac{1}{2}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,\forall n\in\mathbb{N}$$

Again, I still can't prove the strict inequality on the left, unless I already have and just cannot see it.

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Now for the second integral:

For $x\in[0,1]$,$$4\ge 4-x+x^3\ge3$$

As $x\ge x^3$

So we know that $$\sqrt{4}=2\ge\sqrt{4-x+x^3}$$

$$\implies \frac{1}{2}\le\frac{1}{\sqrt{4-x+x^3}}$$

$$\implies\frac{1}{2}=\int_0^\frac{1}{2}\frac{1}{2}dx\le\int_0^1\frac{dx}{\sqrt{4-x+x^3}}$$

Again, I cannot prove strict inequality here, and would greatly appreciate help for that.

Now for the rightmost inequality.

Statement:

$$\frac{1}{2}\left[\int_0^1\frac{dx}{\sqrt{1-x^2}}-\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}\right]=\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}$$

The proof for this is very trivial. The first integral equates to $\frac{\pi}{2}$, the second to $\frac{\pi}{6}$

Moving on, we know that $$\forall x\in[0,1], \sqrt{4-x+x^3}>\sqrt{1-x^2}, always.$$

$$\implies \frac{1}{\sqrt{4-x+x^3}}<\frac{1}{\sqrt{1-x^2}}$$

In typing this, I have just realised I made a huge mistake, and the rest of my solution to this is now invalid.

I was going to go on to say that $$\int_0^1\frac{dx}{\sqrt{4-x+x^3}}\le\frac{1}{2}\left[\int_0^1\frac{1}{\sqrt{4-x+x^3}}-\int_0^\frac{1}{2}\frac{1}{\sqrt{4-x+x^3}}\right]\le\frac{1}{2}\left[\int_0^1\frac{dx}{\sqrt{1-x^2}}-\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}\right]=\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}$$

But this is clearly incorrect. Any ideas, anyone?

Right now I'm trying to toy with the relationship$$\frac{1}{3}\int_0^1\frac{dx}{\sqrt{1-x^2}}=\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}$$