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# Decide whether the statement is true or false. Give an explanation for your answer. If f(x) is a positive periodic function, then $\int _ { 0 } ^ { \infty } f ( x ) d x$ diverges.

Solution

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Let us take an example of positive periodic function as

$f(x) = 2 + \sin x$

Now

\begin{align*} \int_{0}^{\infty} (2+ \sin x) \,dx &= \lim\limits_{b \to \infty} \int_{0}^{b} (2+ \sin x) \,dx \\ &= \lim\limits_{b \to \infty} \int_{0}^{b} 2 \,dx + \lim\limits_{b \to \infty} \int_{0}^{b} \sin x \,dx \\ &= \lim\limits_{b \to \infty} \eval{2x}_{0}^{b} + \lim\limits_{b \to \infty} \eval{-\cos x}_{0}^{b} \\ &= \lim\limits_{b \to \infty} \eval{2x}_{0}^{b} + \lim\limits_{b \to \infty} \eval{\cos x}_{b}^{0} \\ &= \lim\limits_{b \to \infty} 2(b-0) + \lim\limits_{b \to \infty} (\cos 0 - \cos b) \\ &= 2(\infty-0) + \cos 0 - \cos \infty \\ &= \infty + 1 - \cos \infty \\ &= \infty \end{align*}

The above integral diverges

Hence If $f(x)$ is a positive periodic function, then $\int_{0}^{\infty} f(x) \,dx$ diverges.

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