To determine whether the improper integral

$\int_0^\infty x e^{-\sqrt{x}} \, dx$

converges or diverges, we can analyze its behavior over the range of integration, particularly at both $x \to 0^+$ and $x \to \infty$.

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Step 1: Split the integral

Since improper integrals are evaluated by splitting the domain, consider: $\int_0^\infty x e^{-\sqrt{x}} \, dx = \int_0^a x e^{-\sqrt{x}} \, dx + \int_a^\infty x e^{-\sqrt{x}} \, dx$ for some $a > 0$. We will examine the convergence at $x \to 0^+$ and $x \to \infty$ separately.

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Step 2: Substitution to simplify $x \to \infty$

For the integral at $x \to \infty$, use the substitution $u = \sqrt{x}$, so that $x = u^2$ and $dx = 2u \, du$. Substituting into the integral gives: $\int_a^\infty x e^{-\sqrt{x}} \, dx = \int_{\sqrt{a}}^\infty u^2 e^{-u} \cdot 2u \, du = 2 \int_{\sqrt{a}}^\infty u^3 e^{-u} \, du.$

Now analyze the term $u^3 e^{-u}$:

• For $u \to \infty$, the exponential $e^{-u}$ dominates $u^3$, causing $u^3 e^{-u} \to 0$.
• The integral $\int_{\sqrt{a}}^\infty u^3 e^{-u} \, du$ converges because $e^{-u}$ decreases faster than any polynomial growth.

Thus, the integral at $x \to \infty$ converges.

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Step 3: Behavior near $x \to 0^+$

For the integral at $x \to 0^+$: $\int_0^a x e^{-\sqrt{x}} \, dx.$ At $x \to 0^+$, $e^{-\sqrt{x}} \to 1$, so the integrand behaves like $x$. Specifically:

• Near $x = 0$, $x e^{-\sqrt{x}} \approx x$, and $\int_0^a x \, dx = \left[\frac{x^2}{2}\right]_0^a$, which converges to a finite value.

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Step 4: Conclude

Both parts of the integral ($x \to 0^+$ and $x \to \infty$) converge. Therefore, the improper integral $\int_0^\infty x e^{-\sqrt{x}} \, dx$ converges.

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Final Result

The improper integral converges. If you'd like, I can compute its exact value using integration techniques.

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Would you like a detailed evaluation of the integral, or do you have further questions? Here are related questions you might consider:

1. How does substitution simplify improper integrals in general?
2. Why does exponential decay dominate polynomial growth for large $x$?
3. Can this integral be solved explicitly, and what methods would you use?
4. How does convergence differ for improper integrals of other forms (e.g., $e^{-x^2}$)?
5. What tests exist for determining convergence of improper integrals?

Tip: Substitution is a powerful tool in improper integral problems, especially when one term grows or decays rapidly.