Let \(f:[0,\infty)\to\mathbb R\) be uniformly continuous. Suppose \[ \lim_{b\to\infty}\int_0^b f(x)\,dx \] exists and is finite. Prove that \[ \lim_{x\to\infty}f(x)=0. \]
Solution
1. Convergence of the improper integral implies the following tailCauchy property: for every \(\eta>0\) there is \(A>0\) such that\[ \left| \int_a^b f(x)\,dx \right|<\eta \]whenever \(b\ge a\ge A\).
Set \[ F(t)\coloneqq\int_0^t f(x)\,dx. \] By hypothesis, \(F(t)\) has a finite limit as \(t\to\infty\). Hence for sufficiently large \(a,b\), \[ \abs{F(b)-F(a)}<\eta. \] Since \[ F(b)-F(a)=\int_a^b f(x)\,dx, \] the stated property follows.
2. If \(f(x)\) does not tend to \(0\), then there are\[ \varepsilon>0 \]and a sequence \(x_n\to\infty\) such that\[ \abs{f(x_n)}\ge\varepsilon \]for every \(n\).
This is exactly the negation of \[ \lim_{x\to\infty}f(x)=0. \]
3. Uniform continuity turns every point \(x_n\) from step 2 into aninterval of fixed width on which \(f\) has the same sign as \(f(x_n)\) andabsolute value at least \(\varepsilon/2\).
Uniform continuity gives \(\delta>0\) such that \[ \abs{x-y}<\delta \quad\Longrightarrow\quad \abs{f(x)-f(y)}<\frac{\varepsilon}{2}. \] For every sufficiently large \(n\), one has \(x_n>\delta/2\). If \[ x\in \left[ x_n-\frac{\delta}{2}, x_n+\frac{\delta}{2} \right], \] then \[ \abs{f(x)-f(x_n)} < \frac{\varepsilon}{2}. \] Since \(\abs{f(x_n)}\ge\varepsilon\), this implies that \(f(x)\) has the same sign as \(f(x_n)\) and \[ \abs{f(x)} \ge \frac{\varepsilon}{2}. \]
4. Steps 1 and 3 are incompatible.
Apply step 1 with \[ \eta\coloneqq\frac{\varepsilon\delta}{4}. \] Choose \(n\) so large that \[ x_n-\frac{\delta}{2}\ge A. \] Set \[ a_n=x_n-\frac{\delta}{2}, \qquad b_n=x_n+\frac{\delta}{2}. \] By step 3, \(f\) has one sign throughout \([a_n,b_n]\) and \(\abs{f(x)}\ge\varepsilon/2\) there. Therefore \[ \left| \int_{a_n}^{b_n}f(x)\,dx \right| = \int_{a_n}^{b_n}\abs{f(x)}\,dx \ge \frac{\varepsilon}{2}\delta = \frac{\varepsilon\delta}{2}. \] But step 1 gives \[ \left| \int_{a_n}^{b_n}f(x)\,dx \right| < \frac{\varepsilon\delta}{4}, \] a contradiction.
5. Hence\[ \boxed{\lim_{x\to\infty}f(x)=0}. \]
Step 4 contradicts the negation in step 2, so the desired limit must hold.
6. Q.E.D.
Step 5 is the claimed conclusion.