Find $\int_0^1 \frac{f(x)}{\sqrt{1+x^2}}dx$

Question

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?

Answer 1

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.

Answer 2

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:

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

Answer 3

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 $$ \int_0^1 \frac{f(x)}{\sqrt{1+x^2}} \, dx = \frac 1k \int_0^1 f(x) f'(x) \, dx = \frac 1{2k}( f^2(1) - f^2(0)) = \frac 1{2k} = \frac 12 \ln(1+\sqrt 2) \,. $$