Show that \[ f(x)=\sum_{n=0}^\infty \frac{e^{i\sqrt n\,x}}{n^2+1} \] converges uniformly for real \(x\). Prove that \[ \lim_{R\to+\infty}\frac1{2R}\int_{-R}^{R}f(x)\,dx \] exists and calculate it.
Solution
For \(N\geq0\), set \[ S_N(x) \coloneqq \sum_{n=0}^N \frac{e^{i\sqrt n\,x}}{n^2+1}. \]
1. The series defining \(f\) converges uniformly on \(\RR\).
For every real \(x\) and every \(n\geq0\), \[ \abs{ \frac{e^{i\sqrt n\,x}}{n^2+1} } = \frac{1}{n^2+1}. \] The numerical series \[ \sum_{n=0}^{\infty}\frac1{n^2+1} \] converges. The Weierstrass M-test therefore gives uniform convergence of \(S_N\) to \(f\) on all of \(\RR\).
2. For \(n=0\), the mean of the \(n\)th summand over \([-R,R]\) is \(1\).For every \(n\geq1\), it is\[ \frac{1}{n^2+1} \frac{\sin(\sqrt n\,R)}{\sqrt n\,R}. \]
The \(n=0\) summand is identically \(1\). If \(n\geq1\), then \[ \begin{aligned} \frac1{2R} \int_{-R}^{R} \frac{e^{i\sqrt n\,x}}{n^2+1}\,dx &= \frac{1}{n^2+1} \frac{1}{2R} \left[ \frac{e^{i\sqrt n\,x}}{i\sqrt n} \right]_{-R}^{R}\\ &= \frac{1}{n^2+1} \frac{\sin(\sqrt n\,R)}{\sqrt n\,R}. \end{aligned} \]
3. For each fixed \(N\),\[ \lim_{R\to\infty} \frac1{2R}\int_{-R}^{R}S_N(x)\,dx = 1. \]
By step 2, the mean of \(S_N\) is \[ 1 + \sum_{n=1}^N \frac{1}{n^2+1} \frac{\sin(\sqrt n\,R)}{\sqrt n\,R}. \] For each fixed \(n\geq1\), \[ \left| \frac{\sin(\sqrt n\,R)}{\sqrt n\,R} \right| \leq \frac{1}{\sqrt n\,R} \longrightarrow0. \] The sum is finite, so its nonconstant terms all tend to \(0\).
4. For every \(N\) and every \(R>0\),\[ \abs{ \frac1{2R} \int_{-R}^{R} \bigl(f(x)-S_N(x)\bigr)\,dx } \leq \sum_{n=N+1}^{\infty}\frac1{n^2+1}. \]
Uniform convergence permits termwise subtraction, and for every \(x\), \[ \abs{f(x)-S_N(x)} \leq \sum_{n=N+1}^{\infty}\frac1{n^2+1}. \] Hence \[ \begin{aligned} \abs{ \frac1{2R} \int_{-R}^{R} (f-S_N)(x)\,dx } &\leq \frac1{2R} \int_{-R}^{R} \abs{f(x)-S_N(x)}\,dx\\ &\leq \sum_{n=N+1}^{\infty}\frac1{n^2+1}. \end{aligned} \] The bound is independent of \(R\).
5. The requested limit exists and equals\[ \boxed{1}. \]
Let \(\varepsilon>0\). Choose \(N\) so large that \[ \sum_{n=N+1}^{\infty}\frac1{n^2+1} < \frac{\varepsilon}{2}. \] By step 3, for all sufficiently large \(R\), \[ \abs{ \frac1{2R}\int_{-R}^{R}S_N(x)\,dx-1 } < \frac{\varepsilon}{2}. \] Step 4 then gives \[ \abs{ \frac1{2R}\int_{-R}^{R}f(x)\,dx-1 } < \varepsilon. \] This proves both existence of the limit and its value.
6. Q.E.D.
Step 1 proves uniform convergence, and step 5 computes the requested mean.