Let f(t) = \frac{1}{2} + \Sigma_{n = 1}^\infty \frac{1}{n^3} \cos (nt). a) Find the...

Question:

Let

{eq}f(t) = \frac{1}{2} + \Sigma_{n = 1}^\infty \frac{1}{n^3} \cos (nt). {/eq}

a) Find the antiderivative.

b) Is the antiderivative periodic?

Derivatives and Anti-Derivatives:

{eq}\\ {/eq}

The derivative of a function represents the slope of the tangent line to a given curve represented by {eq}y=f(x) {/eq}. The derivative is represented by {eq}\dfrac{dy}{dx} {/eq}.

We can evaluate the anti-derivatives by integrating a function and adding an arbitrary constant.

Given: {eq}f(t) = \dfrac{1}{2} + \displaystyle\Sigma_{n = 1}^\infty \dfrac{1}{n^3} \cos (nt). {/eq}

a) Finding the anti-derivative:

{eq}\Rightarrow \displaystyle \int f(t) \ dt =\dfrac{t}{2}+\Sigma_{n=1}^{\infty}\dfrac{\sin nt}{n^4}+C {/eq}

b) Since in the above expression, {eq}\dfrac{t}{2} {/eq} is non-periodic and {eq}\Sigma_{n=1}^{\infty}\dfrac{\sin nt}{n^4} {/eq} is periodic and as we know that the sum of a non-periodic and a periodic function is a non-periodic one. Hence, the antiderivative computed above is not periodic.