Calculating values of integrals using Fourier series and uniform convergence

Question

I have a problem that I don't know how to begin solving. I have f(t) $$ f(t) = \sum_{k=1}^\infty\frac{1}{k^2+1}\sin{kt} $$ First I had to show that this series converges uniformly, I've done that using Weierstrass M-test comparing the series to $$ \sum_{k=1}^\infty\frac{1}{k^2} $$ Then using this result (and possibly Fourier series), I have to solve these 2 integrals $$ \int_{-\pi}^\pi f(t)\sin{3t}dt \qquad \text and \qquad \int_{-\pi}^\pi f(t)\cos{3t}dt $$ The problem is that I have no idea how to begin solving this. I tried just sticking $f(t)$ inside these integrals, and then using uniform convergence to swap the integral and sum, but I end up with an integral that I don't know how to solve.

Could i get some hints?

Answer 1

To continue the answer posted by @JackD'Aurizio, recall that $\sin(nx)\sin(mx)=\frac12(\cos((n-m)x)-\cos((n+m)x))$. Then, we have

$$\int_{-\pi}^{\pi}\sin(nx)\sin(mx)\,dx=\frac12 \int_{-\pi}^{\pi}\cos((n-m)x)\,dx-\frac12 \int_{-\pi}^{\pi}\cos((n+m)x)\,dx$$

For $n=m$

$$\int_{-\pi}^{\pi}\cos((n-m)x)\,dx=2\pi$$

while for $n\ne m$ we have

$$\begin{align} \int_{-\pi}^{\pi}\cos((n-m)x)\,dx&=\left.\frac{\sin((n-m)x)}{n-m}\right|_{-\pi}^{\pi}\\\\ &=\frac{\sin((n-m)\pi)}{n-m}-\frac{\sin(-(n-m)\pi)}{n-m}\\\\ &=0-0\\\\ &=0 \end{align}$$

Similarly, for $n=-m$

$$\int_{-\pi}^{\pi}\cos((n+m)x)\,dx=2\pi$$

while for $n\ne -m$ we have

$$\begin{align} \int_{-\pi}^{\pi}\cos((n+m)x)\,dx&=\left.\frac{\sin((n+m)x)}{n+m}\right|_{-\pi}^{\pi}\\\\ &=\frac{\sin((n+m)\pi)}{n+m}-\frac{\sin(-(n+m)\pi)}{n+m}\\\\ &=0-0\\\\ &=0 \end{align}$$

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Analogously, recall that $\sin(nx)\cos(mx)=\frac12(\sin((n+m)x)+\sin((n-m)x))$. Then,

$$\int_{-\pi}^{\pi}\sin(nx)\cos(mx)\,dx=\frac12 \int_{-\pi}^{\pi}\sin((n+m)x)\,dx-\frac12 \int_{-\pi}^{\pi}\sin((n-m)x))\,dx$$

For $n=-m$

$$\int_{-\pi}^{\pi}\sin((n+m)x)\,dx=0$$

and for $n\ne -m$ we have

$$\begin{align} \int_{-\pi}^{\pi}\sin((n+m)x)\,dx&=\left.\frac{-\cos((n+m)x)}{n+m}\right|_{-\pi}^{\pi}\\\\ &=\frac{-\cos((n+m)\pi)}{n+m}-\frac{-\cos(-(n+m)\pi)}{n+m}\\\\ &=-(-1)^{n+m}+(-1)^{n+m}\\\\ &=0 \end{align}$$

Similarly, for $n=m$

$$\int_{-\pi}^{\pi}\sin((n-m)x)\,dx=0$$

and for $n\ne m$ we have

$$\begin{align} \int_{-\pi}^{\pi}\sin((n-m)x)\,dx&=\left.\frac{-\cos((n-m)x)}{n-m}\right|_{-\pi}^{\pi}\\\\ &=\frac{-\cos((n-m)\pi)}{n-m}-\frac{-\cos(-(n-m)\pi)}{n-m}\\\\ &=-(-1)^{n-m}+(-1)^{n-m}\\\\ &=0 \end{align}$$

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Putting everything together gives

$$ \int_{-\pi}^{\pi}\sin(nx)\sin(mx)\,dx= \begin{cases} \pi,&n=m\\\\ 0,&n\ne m \end{cases} $$

and

$$\int_{-\pi}^{\pi}\sin(nx)\cos(mx)\,dx=0$$

Answer 2

You just have to use the main property (orthogonality) of the Fourier base. Provided that $a,b\in\mathbb{N}$,

$$ \int_{-\pi}^{\pi}\sin(ax)\cos(bx)\,dx=0\tag{1}$$ as well as: $$\int_{-\pi}^{\pi}\cos(ax)\cos(bx)\,dx=\int_{-\pi}^{\pi}\sin(ax)\sin(bx)\,dx= \pi\cdot\delta(a,b)\tag{2} $$ hence your integrals equal $\frac{\pi}{10}$ and $0$ by termwise integration.