Leibniz formula for π using a Fourier sine series expansion

In the math.stackexchange.com question How to show this formula using a Fourier sine series expansion? \sum _{k=1}^{\infty }\frac {\left(-1\right)^{k-1}}{2k-1}=\frac {\pi }{4}FMath asked how to prove

\displaystyle\sum _{k=1}^{\infty }\dfrac{\left(-1\right)^{k-1}} {2k-1}=\dfrac {\pi }{4}

using the Fourier sine expansion for the function f(x)=\dfrac {1} {2}\left( \pi -x\right); (range: \left[0,\pi \right]).

Here is my answer.

To obtain the Fourier sine series expansion for f(x) the coefficients a_n must vanish (see below). So let g(x) be the odd function extending f(x) to the interval [-\pi,0[ defined by

g(x)=\left\{\begin{array}{l}\phantom{-}f(x)=\phantom{-}\dfrac{\pi-x}{2},\qquad\quad 0\leq x\leq\pi\\\\-f(-x)=-\dfrac{\pi+x}{2},\qquad -\pi\leq x <0\end{array}\right.

whose graph in [\pi,\pi] is shown in the following figure

g(x)=f(x)=\dfrac{\pi -x}{2}, x\in[0,\pi[ ; \; g(x)=-f(-x)=-\dfrac{\pi +x}{2},x\in[-\pi,0]

We know that the trigonometric Fourier series expansion for g(x) in the interval \left[ -\pi ,\pi \right] is given by

\dfrac{a_{0}}{2}+\displaystyle\sum_{n=1}^{\infty}\left(a_{n}\cos (nx)+b_{n}\sin (nx)\right),

where the coefficients are the integrals

a_{n}=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }g(x)\cos (nx)\,dx,\qquad n=0,1,2,\ldots,

b_{n}=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }g(x)\sin (nx)\,dx,\qquad n=1,2,3,\ldots,

The integrand g(x)\cos (nx) is an odd function, while g(x)\sin (nx) is even. So a_{n}=0 and

b_{n}=\dfrac{2}{\pi }\displaystyle\int_{0}^{\pi }g(x)\sin (nx)\,dx=\dfrac{2}{\pi }\displaystyle\int_{0}^{\pi }\dfrac{\pi -x}{2}\sin (nx)\,dx.

Evaluating this last integral we obtain b_{n}=\frac{1}{n}, which proves that

f(x)=\dfrac{\pi -x}{2}=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n}\sin (nx),\qquad 0\leq x\leq\pi,

(the Fourier sine series for the function f(x) in the interval \left[0,\pi\right]). We see that f(\pi/2)=\pi/4. We just need to confirm that for x=\pi /2 this last series reduces to your series. Indeed since

\sin\big(\dfrac{n\pi }{2}\big)=\left\{\begin{array}{l}\phantom{-}1,\qquad n=1,5,\ldots ,4k+1,\ldots\\\phantom{-}0,\qquad n=2,4,\ldots ,2k+2,\ldots\quad (k=0,1,2,\ldots)\\-1,\qquad n=3,5,\ldots,4k+3,\ldots,\end{array}\right.

we have

\dfrac{\pi }{4}=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n}\sin \big(\dfrac{n\pi }{2}\big)=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\ldots = \displaystyle\sum_{k=1}^{\infty }\dfrac {\left( -1\right)^{k-1}} {2k-1}.

Remark. For a Fourier cosine series we would need to extend f(x) to an even function instead, because then b_n would vanish.

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Análise de Fourier, Cálculo, Matemática, Math, Mathematics Stack Exchange, Séries com as etiquetas , , . ligação permanente.

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