How to find the Fourier series of the function $\frac{1}{1+x^2}$from $-\pi$ to $0$ and $\frac{-1}{1+x^2}$from$0$ to $\pi$

Question

I'm stuck in calculating the integration $\int\frac{\sin x}{1+x^2}dx$ while finding the Fourier series of the so-called function.I've tried many exchanging variables but none of the worked.any help would be appreciated.

Answer 1

Let $f(x)=\frac{1}{1+x^2}$. The Fourier series for $f(x)$ is $\frac{1}{\pi}[\frac{1}{2}a_0+\sum_{i=1}^\infty(a_i\cos(ix)+b_i\sin(ix))]$, by definition. $f(x)$ is an even function so each $b_i$ will be $0$.

$$\begin{align} a_0 &=\int_{-\pi}^\pi \frac{1}{1+x^2}\,\mathrm{d}x \\ &=\left[\tan^{-1}(x)\right]_{-\pi}^\pi \\ &=2\tan^{-1}(\pi) \\ \\ a_i&=\int_{-\pi}^\pi \frac{1}{1+x^2}\cos(ix)\,\mathrm{d}x \\ &=\int_{-\pi}^\pi \frac{1}{1+x^2}\sum_{j=0}^\infty\frac{(-1)^j(ix)^{2j}}{(2j)!}\,\mathrm{d}x \\ &=\sum_{j=0}^\infty\int_{-\pi}^\pi \frac{(-1)^j(ix)^{2j}}{(1+x^2)(2j)!}\,\mathrm{d}x \\ &=\sum_{j=0}^\infty\left[\frac{(-1)^j(i x)^{2 j}x\cdot {_2}F_1(1, j+1/2, j+3/2, -x^2)}{(2 j+1) (2 j)!}\right]_{-\pi}^\pi\quad * \\ &=\scriptsize\sum_{j=0}^\infty\left(\frac{(-1)^j(i \pi)^{2 j}\pi\cdot {_2}F_1(1, j+1/2, j+3/2, -\pi^2)-(-1)^j(-i \pi)^{2 j}(-\pi)\cdot {_2}F_1(1, j+1/2, j+3/2, -(-\pi)^2)}{(2 j+1) (2 j)!}\right) \\ &=\sum_{j=0}^\infty\frac{(-1)^j{_2}F_1(1, j+1/2, j+3/2, -\pi^2)(i \pi)^{2 j}\pi}{(2 j+1) (2 j)!}\left(1+(-1)^{2j}\right) \end{align}$$

We can simplify $(1+(-1)^{2j})$ to just $2$.

$$\\a_i= 2\pi\sum_{j=0}^\infty\frac{(-1)^j{_2}F_1(1, j+1/2, j+3/2, -\pi^2)(i \pi)^{2 j}}{(2 j+1) (2 j)!}$$

So we have a final answer and can calculate a few coefficients with WolframAlpha.

$$\begin{align} f(x) &=\frac{\tan^{-1}(\pi)}{\pi} +2\sum_{i=1}^\infty \sum_{j=0}^\infty \frac{(-1)^j{_2}F_1(1, j+1/2, j+3/2, -\pi^2)(i \pi)^{2 j}}{(2 j+1) (2 j)!}\cos(ix) \\ &\approx 0.402+2\left[ 0.194\cos(x)+0.064\cos(2x)+0.027\cos(3x)+0.008\cos(4x) \right] \end{align}$$

By plotting our approximate coefficients in Geogebra, we can show that with just $4$ terms of the series, we achieve an absolute error less than $0.01$ over the interval $[-\pi,\pi]$ and the curve is fitted very well.

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* According to WolframAlpha, where ${_2}F_1(a,b;c;x)$ is the hypergeometric function.

Answer 2

An antiderivative for that integrand will involve Fresnel functions (yuck.) But for the interval $-\pi$ to $\pi$, the answer will be $0$ because it's an odd function over a symmetric interval. The integral from $0$ to $\pi$ will be much more of a problem. Either approximate with Taylor series or do something numerical or dive in with the Fresnel functions.