Evaluate [latex]\underset{n→∞}{\lim} \int_{0}^{n}{\left(1 − \frac{x}{n}\right)^{n} e^{−2x}} dx.[/latex] (11.29)

To express (11.29) as a Lebesgue integral, we define [latex]f_{n}(x) = χ_{[0,n]}(x) · \left(1 − \frac{x}{n}\right)^{n} e^{−2x},[/latex] so that [latex]\int_{0}^{n}{\left(1 − \frac{x}{n}\right)^{n} e^{−2x}} dx = \int_{[0,∞)}{f_{n}} dm.[/latex] Noting that [latex]|f_{n}(x)| ≤ e^{−2x}[/latex] for all [latex]n ∈ \mathbb{N}, x ≥ 0,[/latex] and that [latex]\int_{[0,∞)}{e^{−2x}}dm[/latex] can be evaluated as an improper Riemann integral, [latex]\int_{[0,∞)}{e^{−2x}}dm = \underset{n→∞}{\lim} \int_{0}^{n}{e^{−2x}} dx = \frac{1}{2},[/latex] we can apply the dominated convergence theorem to obtain [latex]\underset{n→∞}{\lim} \int_{0}^{n}{\left(1 − \frac{x}{n}\right)^{n} e^{−2x}} dx = \int_{[0,∞)}{\underset{n→∞}{\lim}f_{n}} dm[/latex] [latex]= \int_{[0,∞)} e^{−3x} dx[/latex] [latex]= \frac{1}{3}.[/latex]

Question 11.9: Evaluate limn→∞∫0^n (1 − x/n)^n e^−2x dx. (11.29)

Elements of Real Analysis

Evaluate

$\underset{n→∞}{\lim} \int_{0}^{n}{\left(1 − \frac{x}{n}\right)^{n} e^{−2x}} dx.$ (11.29)

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To express (11.29) as a Lebesgue integral, we define

$f_{n}(x) = χ_{[0,n]}(x) · \left(1 − \frac{x}{n}\right)^{n} e^{−2x},$

so that

$\int_{0}^{n}{\left(1 − \frac{x}{n}\right)^{n} e^{−2x}} dx = \int_{[0,∞)}{f_{n}} dm.$

Noting that $|f_{n}(x)| ≤ e^{−2x}$ for all $n ∈ \mathbb{N}, x ≥ 0,$ and that $\int_{[0,∞)}{e^{−2x}}dm$ can be evaluated as an improper Riemann integral,

$\int_{[0,∞)}{e^{−2x}}dm = \underset{n→∞}{\lim} \int_{0}^{n}{e^{−2x}} dx = \frac{1}{2},$

we can apply the dominated convergence theorem to obtain

$\underset{n→∞}{\lim} \int_{0}^{n}{\left(1 − \frac{x}{n}\right)^{n} e^{−2x}} dx = \int_{[0,∞)}{\underset{n→∞}{\lim}f_{n}} dm$

$= \int_{[0,∞)} e^{−3x} dx$

$= \frac{1}{3}.$