Question

For any continuous function $f:(0, \infty) \rightarrow \mathbb{R}$, define

$R_{\alpha}=\{(x, y) \in \mathbb{R}^{2}: 0 \leq y \leq f(x), \quad 1 \leq x \leq \alpha\}$ and $S_{\alpha}=\{(x, y) \in \mathbb{R}^{2}: 0 \leq y \leq f(x), \quad \alpha \leq x \leq \alpha^{2}\}$

where $\mathbb{R}^{2}$ denotes $\mathbb{R} \times \mathbb{R}$. Let $S$ be the set of all continuous functions $f:(0, \infty) \rightarrow \mathbb{R}$, for which area of the region $R_{\alpha}=S_{\alpha}$ for all $\alpha \in[1, \infty)$.

Let $f \in S$ such that $f(1)=1$ and, let $g$ be defined as follows

$g(x)=\lim _{n \rightarrow \infty} \int_{\frac{1}{x}}^{n} f^{2}(t) \cos t \ dt$ for any $x>0, g(0)=0$.

Then the value of $\lim _{h \rightarrow 0} \frac{g\left(\frac{1}{\pi}\right)-g\left(\frac{1}{\pi}+h\right)}{h}$

Answer 1

1

Answer 2

The condition for a function $f:(0, \infty) \rightarrow \mathbb{R}$ to be in the set $S$ is that the area of the region $R_{\alpha}$ is equal to the area of the region $S_{\alpha}$ for all $\alpha \in [1, \infty)$.

The area of $R_{\alpha}$ is given by $\int_1^{\alpha} f(x) dx$. The area of $S_{\alpha}$ is given by $\int_{\alpha}^{\alpha^2} f(x) dx$. So, for $f \in S$, we have $\int_1^{\alpha} f(x) dx = \int_{\alpha}^{\alpha^2} f(x) dx$ for all $\alpha \in [1, \infty)$.

Let $F(x) = \int_1^x f(t) dt$. The condition is $F(\alpha) - F(1) = \int_{\alpha}^{\alpha^2} f(x) dx$. Since $F(1) = \int_1^1 f(t) dt = 0$, we have $F(\alpha) = \int_{\alpha}^{\alpha^2} f(x) dx$ for all $\alpha \in [1, \infty)$.

Differentiating both sides with respect to $\alpha$ using the Leibniz integral rule: $\frac{d}{d\alpha} \int_1^{\alpha} f(x) dx = f(\alpha) \cdot \frac{d\alpha}{d\alpha} - f(1) \cdot \frac{d1}{d\alpha} = f(\alpha)$. $\frac{d}{d\alpha} \int_{\alpha}^{\alpha^2} f(x) dx = f(\alpha^2) \cdot \frac{d(\alpha^2)}{d\alpha} - f(\alpha) \cdot \frac{d\alpha}{d\alpha} = f(\alpha^2) \cdot (2\alpha) - f(\alpha) \cdot 1 = 2\alpha f(\alpha^2) - f(\alpha)$.

Equating the derivatives, we get $f(\alpha) = 2\alpha f(\alpha^2) - f(\alpha)$, which simplifies to $2f(\alpha) = 2\alpha f(\alpha^2)$, or $f(\alpha) = \alpha f(\alpha^2)$ for all $\alpha \in [1, \infty)$.

We are given that $f \in S$ and $f(1)=1$. We need to find a function $f$ that satisfies $f(\alpha) = \alpha f(\alpha^2)$ for $\alpha \in [1, \infty)$ and $f(1)=1$. Let's try a power function $f(x) = x^p$. Substituting into the functional equation: $\alpha^p = \alpha (\alpha^2)^p = \alpha \alpha^{2p} = \alpha^{1+2p}$. For this to hold for all $\alpha \in [1, \infty)$, the exponents must be equal: $p = 1+2p$, which gives $p = -1$. So, $f(x) = x^{-1} = \frac{1}{x}$ is a potential solution. Let's check if $f(1)=1$: $f(1) = \frac{1}{1} = 1$. This condition is satisfied. Let's verify if $f(x) = \frac{1}{x}$ satisfies the area condition for $\alpha \in [1, \infty)$: $\int_1^{\alpha} \frac{1}{x} dx = [\ln|x|]_1^{\alpha} = \ln \alpha - \ln 1 = \ln \alpha$. $\int_{\alpha}^{\alpha^2} \frac{1}{x} dx = [\ln|x|]_{\alpha}^{\alpha^2} = \ln(\alpha^2) - \ln \alpha = 2\ln \alpha - \ln \alpha = \ln \alpha$. Since $\ln \alpha = \ln \alpha$ for $\alpha \in [1, \infty)$, the area condition is satisfied. Thus, the function $f \in S$ with $f(1)=1$ is $f(x) = \frac{1}{x}$.

Now, consider the function $g(x)=\lim _{n \rightarrow \infty} \int_{\frac{1}{x}}^{n} f^{2}(t) \cos t \ dt$ for $x>0$. This is an improper integral $g(x) = \int_{\frac{1}{x}}^{\infty} f^{2}(t) \cos t \ dt$. Substitute $f(t) = \frac{1}{t}$: $g(x) = \int_{\frac{1}{x}}^{\infty} \frac{\cos t}{t^2} dt$. The integral $\int_a^{\infty} \frac{\cos t}{t^2} dt$ converges absolutely for any $a > 0$, since $|\frac{\cos t}{t^2}| \leq \frac{1}{t^2}$ and $\int_a^{\infty} \frac{1}{t^2} dt = [-\frac{1}{t}]_a^{\infty} = \frac{1}{a}$, which is finite. Thus, $g(x)$ is well-defined for $x > 0$.

We need to evaluate $\lim _{h \rightarrow 0} \frac{g\left(\frac{1}{\pi}\right)-g\left(\frac{1}{\pi}+h\right)}{h}$. This expression is the negative of the derivative of $g(x)$ evaluated at $x = \frac{1}{\pi}$. Let $x_0 = \frac{1}{\pi}$. The limit is $-\left. \frac{dg}{dx} \right|_{x=x_0}$.

To find $\frac{dg}{dx}$, we use the Leibniz integral rule for differentiating with respect to the lower limit. Let $H(u) = \int_u^{\infty} k(t) dt$. Then $\frac{dH}{du} = -k(u)$. In our case, $k(t) = f^2(t) \cos t = \frac{\cos t}{t^2}$, and the lower limit is $u = \frac{1}{x}$. So $g(x) = \int_{\frac{1}{x}}^{\infty} \frac{\cos t}{t^2} dt$. Using the chain rule, $\frac{dg}{dx} = \frac{dg}{du} \frac{du}{dx}$, where $u = \frac{1}{x}$. $\frac{du}{dx} = -\frac{1}{x^2}$. $\frac{dg}{du} = -\frac{\cos u}{u^2}$. So, $\frac{dg}{dx} = \left(-\frac{\cos u}{u^2}\right) \left(-\frac{1}{x^2}\right) = \frac{\cos(1/x)}{(1/x)^2} \frac{1}{x^2} = \frac{\cos(1/x)}{1/x^2} \frac{1}{x^2} = x^2 \cos(1/x) \frac{1}{x^2} = \cos(1/x)$.

Wait, let's recheck the differentiation using Leibniz rule for $\int_{a(x)}^{b(x)} k(t) dt$. $\frac{d}{dx} \int_{a(x)}^{b(x)} k(t) dt = k(b(x)) b'(x) - k(a(x)) a'(x)$. Here, $a(x) = \frac{1}{x}$, $b(x) = \infty$. We treat the upper limit as a constant for differentiation purposes, assuming the convergence allows this. $g(x) = \int_{\frac{1}{x}}^{\infty} f^2(t) \cos t dt$. Let the antiderivative of $f^2(t) \cos t$ be $\Phi(t)$. Then $g(x) = [\Phi(t)]_{\frac{1}{x}}^{\infty} = \lim_{N \rightarrow \infty} \Phi(N) - \Phi(\frac{1}{x})$. $\frac{dg}{dx} = \frac{d}{dx} \left( \lim_{N \rightarrow \infty} \Phi(N) \right) - \frac{d}{dx} \Phi(\frac{1}{x})$. Assuming the limit term's derivative is 0 (which is true if the integral converges to a constant value with respect to $x$), we have: $\frac{dg}{dx} = - \Phi'(\frac{1}{x}) \cdot \frac{d}{dx}(\frac{1}{x})$. Since $\Phi'(t) = f^2(t) \cos t$, we have $\Phi'(\frac{1}{x}) = f^2(\frac{1}{x}) \cos(\frac{1}{x})$. $\frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2}$. So, $\frac{dg}{dx} = - f^2(\frac{1}{x}) \cos(\frac{1}{x}) \cdot (-\frac{1}{x^2}) = f^2(\frac{1}{x}) \cos(\frac{1}{x}) \frac{1}{x^2}$.

This matches the previous calculation. We need to evaluate this derivative at $x = \frac{1}{\pi}$. $\left. \frac{dg}{dx} \right|_{x=\frac{1}{\pi}} = f^2\left(\frac{1}{1/\pi}\right) \cos\left(\frac{1}{1/\pi}\right) \frac{1}{(1/\pi)^2} = f^2(\pi) \cos(\pi) \pi^2$. Since $f(x) = \frac{1}{x}$, $f(\pi) = \frac{1}{\pi}$. $\left. \frac{dg}{dx} \right|_{x=\frac{1}{\pi}} = \left(\frac{1}{\pi}\right)^2 \cos(\pi) \pi^2 = \frac{1}{\pi^2} (-1) \pi^2 = -1$.

The required limit is $\lim _{h \rightarrow 0} \frac{g\left(\frac{1}{\pi}\right)-g\left(\frac{1}{\pi}+h\right)}{h}$. This is equal to $-\left. \frac{dg}{dx} \right|_{x=\frac{1}{\pi}}$. So, the value is $-(-1) = 1$.

Therefore, the value of the limit is 1.