= Solution
The <dominated convergence theorem> says that if measurable $f_n$ converge <almost everywhere> to $f$ and $|f_n|\leq H$ almost everywhere for a single <Lebesgue integrable function> $H\geq0$, then $f$ is integrable and
$$
\int|f_n-f|\longrightarrow0,\qquad \int f_n\longrightarrow\int f.
$$
Here, for $x>0$, put $H_0(x)=x^{-1/2}e^{-x}$ and $f_n(x)=H_0(x)(1-(-1)^ne^{-nx})$. The <exponential function> gives $e^{-nx}\to0$ at every such $x$, so $f_n\to H_0$. Also $|f_n(x)|\leq2H_0(x)$. This dominator is integrable: on $(0,1)$, $H_0(x)\leq x^{-1/2}$, whose integral is two; on $[1,\infty)$, $H_0(x)\leq e^{-x}$, whose integral is finite. DCT therefore proves the required limit.
The limiting value can also be evaluated. Substitute $x=t^2$ and use the <Gaussian integral>:
$$
\boxed{\lim_{n\to\infty}\int_0^\infty f_n(x)\,dx
=\int_0^\infty x^{-1/2}e^{-x}\,dx=\sqrt\pi.}
$$
In fact, scaling the second exponential term gives the exact integral $\sqrt\pi\,[1-(-1)^n/\sqrt{n+1}]$. The integral error decays as $n^{-1/2}$ despite exponential convergence at each fixed positive $x$; the slow part lies near the endpoint zero.
Back to article page
