# Is $F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos (nt)}{2^n} \text{d} t$ continuous in $\mathbb{R}$?

Is my reasoning for whether $$F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos (nt)}{2^n} \text{d} t$$ is continuous in $\mathbb{R}$ correct?:

Proof

I claim it is continuous in $\mathbb{R}$. $\sum_{0}^{\infty}\frac{\cos(nx)}{2^n}$ is uniformly convergent in $\mathbb{R}$ (and the functions in the series are continuous), therefore $$F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos(nt)}{2^n} \text{d} t = \sum_{0}^{\infty} \int_{0}^{x}\frac{\cos(nt)}{2^n} \text{d} t=x+\sum_{1}^{\infty} \frac{\sin(nt)}{2^n\cdot n}\; .$$

$\sum_{1}^{\infty} \frac{\sin(nt)}{2^n\cdot n}$ is uniformly convergent in $\mathbb{R}$, so $F(x)$ is a uniformly convergent series of continuous functions, which means it is continuous.

Have I correctly solved this exercise? Thank you!

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Your argument is indeed correct. I would just add some justification as why the series are actually uniformly continuous. – azarel Jan 28 '12 at 15:30
Looks good. You could save a step by noting that the integrand is continuous, due to uniform convergence, and then appeal to the fundamental theorem of calculus to argue that $F(x)$ is continuous. – David Mitra Jan 28 '12 at 15:35
Thanks, both of you! David Mitra, I will take note of that in the future. I guess this question is closed, perhaps one of you should submit your comment as an answer? (I can't submit an answer since I am a guest user) – ro44 Jan 28 '12 at 15:40

It looks good, but some additional justifications should be made. At the start (and for the later series) you should state why the series converges uniformly (by the Weierstrass_M-test, e.g.).

Also, to justify that the sum of the series is integrable and that switching the order of summation and integration is valid, you should state that the terms of $G(x)=\sum\limits_{n=0}^\infty {\cos(nt)\over 2^n}$ are integrable over any interval $[0,x]$.

But, can save a few steps in your argument. Since $\sum\limits_{n=0}^\infty {\cos(nt)\over 2^n}$ converges uniformly to $G(x)$, and since the terms of this series are continuous, $G(x)$ is a continuous function. The Fundamental Theorem of Calculus immediately gives you the continuity of $F(x)=\int_0^x G(t)\,dt$.

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You could use the M-test as follows.

$$\displaystyle \sum\limits_{n = 0}^\infty {\left|\frac{{\cos nt}}{{{2^n}}}\right|} < \sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}}}} < 2$$

and do the same for the $\sin tn$ series to make your proof complete.

Otherwise, the proof is correct.

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