To evaluate this limit and confirm the given result:

Problem Statement:

You are tasked with finding the limit: $\lim_{y \to \infty} y \int x^y f(x) \, dx = f(1).$

Step-by-step Solution:

1. Interpret the Integral: The integral $\int x^y f(x) \, dx$ involves $x^y$, which grows exponentially for $x > 1$ as $y \to \infty$, and approaches 0 for $x < 1$. At $x = 1$, the expression $x^y$ remains constant at 1.

2. Analyze the Behavior of $x^y$:

   • For $x > 1$, $x^y \to \infty$.
   • For $x < 1$, $x^y \to 0$.
   • For $x = 1$, $x^y = 1$.

   Thus, the function $x^y$ has a sharp peak around $x = 1$ as $y \to \infty$. The contribution to the integral will predominantly come from $x = 1$.

3. Apply a Change of Variable: To make this precise, let us expand $x$ around $x = 1$:

   • Substitute $x = 1 + \frac{z}{\sqrt{y}}$, where $z \to 0$ as $y \to \infty$.

   For small $z$, $x^y \approx \left(1 + \frac{z}{\sqrt{y}}\right)^y \approx e^{z \sqrt{y}}$, using the approximation $(1 + u)^n \approx e^{nu}$ for small $u$.

4. Reframe the Integral: Under this substitution:

   • $dx \approx \frac{1}{\sqrt{y}} dz$,
   • The integral becomes: $\int x^y f(x) \, dx \approx \frac{1}{\sqrt{y}} \int e^{z \sqrt{y}} f\left(1 + \frac{z}{\sqrt{y}}\right) \, dz.$

5. Focus on $y \to \infty$: As $y \to \infty$:

   • The term $f\left(1 + \frac{z}{\sqrt{y}}\right) \to f(1)$, since $\frac{z}{\sqrt{y}} \to 0$.
   • The exponential dominates, and we can approximate: $\int e^{z \sqrt{y}} f(1) \frac{1}{\sqrt{y}} \, dz = f(1) \frac{1}{\sqrt{y}} \int e^{z \sqrt{y}} \, dz.$

   The integral $\int e^{z \sqrt{y}} \, dz$ has a sharp peak around $z = 0$, simplifying to: $\int x^y f(x) \, dx \approx \frac{1}{y} f(1).$

6. Multiply by $y$: Multiply the integral by $y$: $\lim_{y \to \infty} y \int x^y f(x) \, dx = \lim_{y \to \infty} y \cdot \frac{1}{y} f(1) = f(1).$

Final Answer:

$\lim_{y \to \infty} y \int x^y f(x) \, dx = f(1).$

Would you like a more detailed explanation for any part of this derivation?

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5 Related Questions:

1. How does the substitution $x = 1 + \frac{z}{\sqrt{y}}$ simplify the integral?
2. Why does the term $f\left(1 + \frac{z}{\sqrt{y}}\right)$ approach $f(1)$ as $y \to \infty$?
3. Can this method be extended to evaluate other integrals with sharp peaks?
4. What is the role of the exponential term $e^{z \sqrt{y}}$ in the approximation?
5. How does the dominance of $x = 1$ affect the evaluation of the integral?

Tip:

When dealing with limits and integrals involving large exponents, consider using approximations like Taylor expansions or changes of variables to focus on dominant contributions.