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Let $f(x)$ be continuous on $[0;1]$, with $f(0) = 0; f(1) = 1$ and $$\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx = \dfrac{1}{\ln\left(1+\sqrt{2}\right)}$$ Find ${\displaystyle \int_0^1} \dfrac{f(x)}{\sqrt{1+x^2}} \,dx$

  • Attempt:

I tried to use Cauchy-Scharwz as below:

$$\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \cdot \int_0^1 \dfrac{f(x)}{\sqrt{1+x^2}} \,dx \geq \left(\int_0^1 \sqrt{[f'(x)]^2 \sqrt{1+x^2}} \cdot \sqrt{\dfrac{f(x)}{\sqrt{1+x^2}}} \,dx\right)^2$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left(\int_0^1 f'(x) \sqrt{f(x)} \,dx\right)^2$$

I was able to find ${\displaystyle \int_0^1} f'(x) \sqrt{f(x)} \,dx = \dfrac{2}{3}$, but the problem is I can't show if the equality is happen or not, so my attempt isn't helpful at all.

Is there a better way to approach this?

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  • $\begingroup$ What?! How did you get that @AdityaDwivedi $\endgroup$
    – Babu
    May 7, 2021 at 11:24

3 Answers 3

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Rewriting the given condition as $$I = \color{blue}{\ln(1+\sqrt 2)}\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx = 1$$ we can notice that $$I = \color{blue}{\int_0^1 \frac{1}{\sqrt{1+x^2}}dx}\cdot\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx = 1 \tag{1}$$ However, by CS inequality we have $$\begin{align} I&=\int_0^1 \frac{1}{\sqrt{1+x^2}}dx\cdot\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \\ &\ge \left(\int f'(x) dx\right)^2 \\ &= (f(1)-f(0))^2 \\ &= 1 \tag{2}\end{align}$$ Hence the equality of CS inequality holds for $$g(x) = \frac{1}{\sqrt{1+x^2}} \ \ \text{ and } \ \ h(x) = [f'(x)]^2 \sqrt{1+x^2}$$ As Martin R said, now solve $h(x) = Cg(x)$ for $f(x)$, then find the integral that you want.

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  • $\begingroup$ Which equality holds for those? $\endgroup$
    – Babu
    May 7, 2021 at 11:00
  • $\begingroup$ @Buraian Equality of Cauchy Swarz $\endgroup$
    – VIVID
    May 7, 2021 at 11:03
  • $\begingroup$ But which integral of cauchy schwarz? There are a few here. Where is $h(x)$ to be kept in place of? @VIVID $\endgroup$
    – Babu
    May 7, 2021 at 11:15
  • $\begingroup$ @Buraian $(1)$ in the post says $\int_0^1 g(x)dx \cdot \int_0^1 h(x) dx = 1$ while $(2)$ shows that $\int_0^1 g(x)dx \cdot \int_0^1 h(x) dx \ge 1$. $\endgroup$
    – VIVID
    May 7, 2021 at 11:17
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    $\begingroup$ Equality in the CS inequality holds exactly if the functions are a constant multiple of each other. $\endgroup$
    – Martin R
    May 7, 2021 at 11:31
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From the @VIVID answer, we can then find $f(x)$ and solve for the original integral

As he shows:

$$1 = \int_0^1 \frac{1}{\sqrt{1+x^2}}dx\cdot\int_0^1 [f'(x)]^2 \sqrt{1+x^2} \,dx \geq \left(\int_0^1 f'(x) dx\right)^2 = 1$$

The equality happens when:

$$[f'(x)]^2 \sqrt{1+x^2} = \frac{k}{\sqrt{1+x^2}} \,\,\text{(k is a constant)}$$ $$\implies f'(x) = \sqrt{k} \cdot \frac{1}{\sqrt{1+x^2}} \,\,\,\text{ or }\,\,\, f(x) = \sqrt{k} \,\ln(|\sqrt{x^2 + 1} + x|) + C$$

From the conditions $f(0) = 0; f(1) = 1$, you can find $k$ and $C$. Hence the function will be:

$$f(x) = \frac{1}{\ln(1 + \sqrt{2})} \cdot \ln(|\sqrt{x^2 + 1} + x|)$$

Now solve for the integral:

$$\int_0^1 \frac{f(x)}{\sqrt{1+x^2}} \, dx = \frac{1}{\ln(1 + \sqrt{2})} \cdot \int_0^1 \frac{\ln(|\sqrt{x^2 + 1} + x|)}{\sqrt{1+x^2}} \, dx$$

Let $u = \ln(|\sqrt{x^2 + 1} + x|)$ then the integral become:

$$\frac{1}{\ln(1 + \sqrt{2})} \cdot \int_0^{\ln(1 + \sqrt{2})} u \,du = \color{red}{\frac{1}{2} \cdot \ln(1 + \sqrt{2})}$$

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  • $\begingroup$ Why did we make the equality happen again? $\endgroup$
    – Babu
    May 7, 2021 at 11:44
  • $\begingroup$ @Buraian It's because both sides of the inequality are equal to $1$, hence the equality must occur. $\endgroup$
    – The 2nd
    May 7, 2021 at 11:48
  • $\begingroup$ Oh I was going to write it all after @MartinR 's hint, but got something to do :) $\endgroup$
    – VIVID
    May 7, 2021 at 12:06
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    $\begingroup$ @VIVID Well, at least your post is a hint for me to write this down :)) $\endgroup$
    – The 2nd
    May 7, 2021 at 13:14
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VIVID demonstrated that equality holds in the Cauchy-Schwarz inequality $$ \tag{*} 1 = \int_0^1 \frac{dx}{\sqrt{1+x^2}} \int_0^1 f'(x)^2 \sqrt{1+x^2} \, dx \ge \left( \int_0^1 f'(x) \right)^2 = 1 \, . $$ It follows that $$ f'(x) = \frac{k}{\sqrt{1+x^2}} $$ for some (positive) constant $k$. Substituting this back into $(*)$ gives $$ 1 = k^2 \left(\int_0^1 \frac{dx}{\sqrt{1+x^2}} \right)^2 = k^2 (\ln(1+\sqrt 2))^2 $$ so that $\frac 1k = \ln(1+\sqrt 2)$. Then $$ \int_0^1 \frac{f(x)}{\sqrt{1+x^2}} \, dx = \frac 1k \int_0^1 f(x) f'(x) \, dx = \frac 1{2k}( f^2(1) - f^2(0)) = \frac 1{2k} = \frac 12 \ln(1+\sqrt 2) \,. $$

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    $\begingroup$ This is nice as you don't have to find $f(x)$ explicitly:) $\endgroup$
    – The 2nd
    May 7, 2021 at 13:18

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