# Sum of an Alternating Series

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Find, with proof, the sum of the alternating series $$\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}$$

julian, Greg Bernhardt and topsquark
First write:

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\end{align*}

The method of evaluation employs the fact that the function

\begin{align*}
\frac{\sin \dfrac{\pi z}{2}}{\sin \pi z}
\end{align*}

has simple poles at all integer values except when ##\sin \dfrac{\pi z}{2} = 0##. This allows us to write

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{2i} \oint_C \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\end{align*}

where the contour ##C## is defined in Fig 1. Note

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{2} \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3} - \frac{1}{2} \sum_{n=-1}^{-\infty} \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{4i} \oint_{C+C'} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\end{align*}

where the contour ##C'## is defined in Fig 1.

Fig 1.

Consider the square contour shown in Fig 2 with corners at ##(N+\frac{1}{2}) (1+i)##, ##(N+\frac{1}{2}) (-1+i)##, ##(N+\frac{1}{2}) (-1-i)## and ##(N+\frac{1}{2}) (1-i)##. This contour encloses all the points at ##-N, - (N-1) , \dots, -1,0,1, \dots , N-1,N##. We will prove that the integral

\begin{align*}
\oint_{C_N} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz \qquad (*)
\end{align*}

vanishes ##N \rightarrow \infty##. This will imply

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{4i} \oint_{C_0} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\nonumber \\
& = - \frac{1}{8i} \oint_{C_0} \dfrac{1}{z^3 \cos \dfrac{\pi z}{2}} dz \qquad (**)
\end{align*}

where ##C_0## is an infinitesimal clockwise contour around the origin.

Fig 2.

We start by showing that the value of ##\left| \sin \dfrac{\pi z}{2} \csc (\pi z) \right|## around the square ##C_N## is bounded by a constant that is independent of ##N##.

We write ##z=x+iy##

Case 1: ##y > \frac{1}{2}##, we have

\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & = \left| \dfrac{e^{i \pi z/2} - e^{-i \pi z/2}}{e^{i \pi z} - e^{-i \pi z}} \right|
\nonumber \\
& \leq \dfrac{|e^{i \pi z/2}| + |e^{-i \pi z/2}|}{|e^{-i \pi z}| - |e^{i \pi z}|}
\nonumber \\
& = \dfrac{|e^{i \pi x/2 - \pi y/2}| + |e^{-i \pi x + \pi y}|}{|e^{-i \pi x + \pi y}| - |e^{i \pi x - \pi y}|}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{-\pi y/2}}{e^{\pi y} - e^{-\pi y}}
\nonumber \\
& = \dfrac{e^{-\pi y/2} + e^{-3\pi y/2}}{1 - e^{- 2 \pi y}}
\nonumber \\
& \leq \dfrac{e^{-\pi y/4} + e^{-3\pi y/4}}{1 - e^{- \pi}} \qquad \text{ as we are taking } y > \frac{1}{2}
\nonumber \\
& =: A_1
\end{align*}

Case 2: ##y < - \frac{1}{2}##, we have

\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & \leq \dfrac{|e^{i \pi x/2 - \pi y/2}| + |e^{-i \pi x + \pi y}|}{|e^{i \pi x - \pi y}| - |e^{-i \pi x + \pi y}|}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{-\pi y/2}}{e^{-\pi y} - e^{\pi y}}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{3\pi y/2}}{1 - e^{2 \pi y}}
\nonumber \\
& \leq \dfrac{e^{-\pi y/4} + e^{-3\pi y/4}}{1 - e^{- \pi}} \qquad \text{ as we are taking } y < - \frac{1}{2}
\nonumber \\
& =: A_1
\end{align*}

Case 3: ##-\frac{1}{2} \leq y \leq \frac{1}{2}##. We consider ##z = N +\frac{1}{2} + iy##. We make repeated use of ##\sin (\alpha + \beta) = \cos \alpha \sin \beta + \sin \alpha \cos \beta## and ##\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta##. In particular we will use:

\begin{align*}
\sin (\pi (N + \frac{1}{2} + i y)) = \cos (\pi N) \sin (\pi/2 + i \pi y) = (-1)^N \cosh (\pi y)
\end{align*}

We have:

\begin{align*}
& \; \left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right|
\nonumber \\
& = \left| \dfrac{\sin (\pi (N + \frac{1}{2} + i y)/2)}{\sin (\pi (N + \frac{1}{2} + i y))} \right|
\nonumber \\
& = \dfrac{| \cos (\pi N /2) \sin (\pi/4+iy/2) + \sin (\pi N /2) \cos (\pi/4+iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq \dfrac{| \cos (\pi N/2) \sin (\pi/4+iy/2)}{\cosh (\pi y)} + \dfrac{| \sin (\pi N/2) \cos (\pi /4+iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq \dfrac{| \sin (\pi/4+iy/2) |}{\cosh (\pi y) |} + \dfrac{| \cos (\pi/4+iy/2) |}{\cosh (\pi y) |}
\nonumber \\
& \leq \dfrac{| \cos (\pi/4) \sin (iy/2) + \sin (\pi/4) \cos (iy/2) |}{\cosh (\pi y)} + \dfrac{| \cos (\pi/4) \cos (iy/2) - \sin (\pi/4) \sin (iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq 2 \dfrac{| \sinh (y/2) |}{\cosh (\pi y)} + 2 \dfrac{\cosh (y/2)}{\cosh (\pi y)}
\nonumber \\
& = 2 \dfrac{| e^{y/2} - e^{-y/2} |}{e^{\pi y} + e^{-\pi y}} + 2 \dfrac{e^{y/2} + e^{-y/2}}{e^{\pi y} + e^{-\pi y}}
\nonumber \\
& \leq 4 \dfrac{e^{y/2} + e^{-y/2}}{e^{\pi y} + e^{-\pi y}}
\nonumber \\
& = 4 \dfrac{e^{y (1/2 - \pi)} + e^{-y (1/2 + \pi)}}{1 + e^{-\pi y}}
\nonumber \\
& \leq \dfrac{8 (e^{1/4} + e^{-1/4}) e^{\pi/2}}{1 + e^{-\pi / 2}} \qquad \text{ as we are taking } -\frac{1}{2} \leq y \leq \frac{1}{2}
\nonumber \\
& =: A_2
\end{align*}

When ##z = -N - \frac{1}{2} + iy##, we have similarly:

\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & = \left| \dfrac{\sin ((-N - \frac{1}{2} + i y)/2)}{\sin (\pi (-N - \frac{1}{2} + i y))} \right|
\nonumber \\
& \leq A_2
\end{align*}

So choose ##A## such that ##A > \max \{ A_1 , A_2 \}##. Then we have ##\left| \sin \dfrac{\pi z}{2} \csc (\pi z) \right| < A## on ##C_N## with an ##A## independent of ##N##. Then

\begin{align*}
\left| \oint_{C_N} \csc (\pi z) \sin \dfrac{\pi z}{2} \frac{1}{z^3} dz \right| \leq \frac{A}{N^3} (8N+4)
\end{align*}

as ##(8N+4)## is the length of the curve ##C_N##. Letting ##N \rightarrow \infty## we get that the integral, ##(*)##, vanishes. This establishes ##(**)##. We now use ##(**)## to evaluate the sum. The residue is obtained from

\begin{align*}
\dfrac{1}{z^3 \cos \dfrac{\pi z}{2}} & = \dfrac{1}{z^3 \left( 1 - \dfrac{1}{2!} \dfrac{\pi^2 z^2}{4} \right) + \cdots}
\nonumber \\
& = \frac{1}{z^3} \left( 1 + \dfrac{\pi^2 z^2}{8} \right) + \cdots
\nonumber \\
& = \frac{1}{z^3} + \dfrac{\pi^2}{8 z} + \cdots
\end{align*}

So that

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = (-2 \pi i) \times - \frac{1}{8i} \times \dfrac{\pi^2}{8}
\nonumber \\
& = \frac{\pi^3}{32}
\end{align*}

Last edited:
milkism, docnet and topsquark
julian said:
First write:

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\end{align*}

The method of evaluation employs the fact that the function

\begin{align*}
\frac{\sin \dfrac{\pi z}{2}}{\sin \pi z}
\end{align*}

has simple poles at all integer values except when ##\sin \dfrac{\pi z}{2} = 0##. This allows us to write

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{2i} \oint_C \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\end{align*}

where the contour ##C## is defined in Fig 1. Note

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{2} \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3} - \frac{1}{2} \sum_{n=-1}^{-\infty} \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{4i} \oint_{C+C'} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\end{align*}

where the contour ##C'## is defined in Fig 1.

View attachment 344267

Fig 1.

Consider the square contour shown in Fig 2 with corners at ##(N+\frac{1}{2}) (1+i)##, ##(N+\frac{1}{2}) (-1+i)##, ##(N+\frac{1}{2}) (-1-i)## and ##(N+\frac{1}{2}) (1-i)##. This contour encloses all the poles at ##-N, - (N-1) , \dots, -1,0,1, \dots , N-1,N##. We will prove that the integral

\begin{align*}
\oint_{C_N} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz \qquad (*)
\end{align*}

vanishes ##N \rightarrow \infty##. This will imply

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = - \sum_{n=1}^\infty \dfrac{(-1)^n \sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = - \frac{1}{4i} \oint_{C_0} \dfrac{\sin \dfrac{\pi z}{2}}{z^3 \sin \pi z} dz
\nonumber \\
& = - \frac{1}{8i} \oint_{C_0} \dfrac{1}{z^3 \cos \dfrac{\pi z}{2}} dz \qquad (**)
\end{align*}

where ##C_0## is an infinitesimal clockwise contour around the origin.

View attachment 344279
Fig 2.

We start by showing that the value of ##\left| \sin \dfrac{\pi z}{2} \csc (\pi z) \right|## around the square ##C_N## is bounded by a constant that is independent of ##N##.

We write ##z=x+iy##

Case 1: ##y > \frac{1}{2}##, we have

\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & = \left| \dfrac{e^{i \pi z/2} - e^{-i \pi z/2}}{e^{i \pi z} - e^{-i \pi z}} \right|
\nonumber \\
& \leq \dfrac{|e^{i \pi z/2}| + |e^{-i \pi z/2}|}{|e^{-i \pi z}| - |e^{i \pi z}|}
\nonumber \\
& = \dfrac{|e^{i \pi x/2 - \pi y/2}| + |e^{-i \pi x + \pi y}|}{|e^{-i \pi x + \pi y}| - |e^{i \pi x - \pi y}|}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{-\pi y/2}}{e^{\pi y} - e^{-\pi y}}
\nonumber \\
& = \dfrac{e^{-\pi y/2} + e^{-3\pi y/2}}{1 - e^{- 2 \pi y}}
\nonumber \\
& \leq \dfrac{e^{-\pi y/4} + e^{-3\pi y/4}}{1 - e^{- \pi}} \qquad \text{ as we are taking } y > \frac{1}{2}
\nonumber \\
& =: A_1
\end{align*}

Case 2: ##y < - \frac{1}{2}##, we have

\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & \leq \dfrac{|e^{i \pi x/2 - \pi y/2}| + |e^{-i \pi x + \pi y}|}{|e^{i \pi x - \pi y}| - |e^{-i \pi x + \pi y}|}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{-\pi y/2}}{e^{-\pi y} - e^{\pi y}}
\nonumber \\
& = \dfrac{e^{\pi y/2} + e^{3\pi y/2}}{1 - e^{2 \pi y}}
\nonumber \\
& \leq \dfrac{e^{-\pi y/4} + e^{-3\pi y/4}}{1 - e^{- \pi}} \qquad \text{ as we are taking } y < - \frac{1}{2}
\nonumber \\
& =: A_1
\end{align*}

Case 3: ##-\frac{1}{2} \leq y \leq \frac{1}{2}##. We consider ##z = N +\frac{1}{2} + iy##. We make repeated use of ##\sin (\alpha + \beta) = \cos \alpha \sin \beta + \sin \alpha \cos \beta## and ##\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta##. In particular we will use:

\begin{align*}
\sin (\pi (N + \frac{1}{2} + i y)) = \cos (\pi N) \sin (\pi/2 + i \pi y) = (-1)^N \cosh (\pi y)
\end{align*}

We have:

\begin{align*}
& \; \left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right|
\nonumber \\
& = \left| \dfrac{\sin (\pi (N + \frac{1}{2} + i y)/2)}{\sin (\pi (N + \frac{1}{2} + i y))} \right|
\nonumber \\
& = \dfrac{| \cos (\pi N /2) \sin (\pi/4+iy/2) + \sin (\pi N /2) \cos (\pi/4+iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq \dfrac{| \cos (\pi N/2) \sin (\pi/4+iy/2)}{\cosh (\pi y)} + \dfrac{| \sin (\pi N/2) \cos (\pi /4+iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq \dfrac{| \sin (\pi/4+iy/2) |}{\cosh (\pi y) |} + \dfrac{| \cos (\pi/4+iy/2) |}{\cosh (\pi y) |}
\nonumber \\
& \leq \dfrac{| \cos (\pi/4) \sin (iy/2) + \sin (\pi/4) \cos (iy/2) |}{\cosh (\pi y)} + \dfrac{| \cos (\pi/4) \cos (iy/2) - \sin (\pi/4) \sin (iy/2) |}{\cosh (\pi y)}
\nonumber \\
& \leq 2 \dfrac{| \sinh (y/2) |}{\cosh (\pi y)} + 2 \dfrac{\cosh (y/2)}{\cosh (\pi y)}
\nonumber \\
& = 2 \dfrac{| e^{y/2} - e^{-y/2} |}{e^{\pi y} + e^{-\pi y}} + 2 \dfrac{e^{y/2} + e^{-y/2}}{e^{\pi y} + e^{-\pi y}}
\nonumber \\
& \leq 4 \dfrac{e^{y/2} + e^{-y/2}}{e^{\pi y} + e^{-\pi y}}
\nonumber \\
& = 4 \dfrac{e^{y (1/2 - \pi)} + e^{-y (1/2 + \pi)}}{1 + e^{-\pi y}}
\nonumber \\
& \leq \dfrac{8 (e^{1/4} + e^{-1/4}) e^{\pi/2}}{1 + e^{-\pi / 2}} \qquad \text{ as we are taking } -\frac{1}{2} \leq y \leq \frac{1}{2}
\nonumber \\
& =: A_2
\end{align*}

When ##z = -N - \frac{1}{2} + iy##, we have similarly:

\begin{align*}
\left| \dfrac{\sin \dfrac{\pi z}{2}}{\sin \pi z} \right| & = \left| \dfrac{\sin ((-N - \frac{1}{2} + i y)/2)}{\sin (\pi (-N - \frac{1}{2} + i y))} \right|
\nonumber \\
& \leq A_2
\end{align*}

So choose ##A## such that ##A > \max \{ A_1 , A_2 \}##. Then we have ##\left| \sin \dfrac{\pi z}{2} \csc (\pi z) \right| < A## on ##C_N## with an ##A## independent of ##N##. Then

\begin{align*}
\left| \oint_{C_N} \csc (\pi z) \sin \dfrac{\pi z}{2} \frac{1}{z^3} dz \right| \leq \frac{\pi A}{N^3} (8N+4)
\end{align*}

as ##(8N+4)## is the length of the curve ##C_N##. Letting ##N \rightarrow \infty## we get that the integral, ##(*)##, vanishes. This establishes ##(**)##. We now use ##(**)## to evaluate the sum. The residue is obtained from

\begin{align*}
\dfrac{1}{z^3 \cos \dfrac{\pi z}{2}} & = \dfrac{1}{z^3 \left( 1 - \dfrac{1}{2!} \dfrac{\pi^2 z^2}{4} \right) + \cdots}
\nonumber \\
& = \frac{1}{z^3} \left( 1 + \dfrac{\pi^2 z^2}{8} \right) + \cdots
\nonumber \\
& = \frac{1}{z^3} + \dfrac{\pi^2}{8 z} + \cdots
\end{align*}

So that

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = (-2 \pi i) \times - \frac{1}{8i} \times \dfrac{\pi^2}{8}
\nonumber \\
& = \frac{\pi^3}{32}
\end{align*}
That's quite a lengthy solution! It's also the correct solution according to Wolfram :)

Inspired by answer of julian, let me show my sketch using Fourier transform and distribution.
Let us say the given sum S
$$S=Im \sum_{n=1}^\infty \frac{e^\frac{n\pi i}{2}}{n^3}$$
$$\frac{8}{\pi^3}S=Im \sum_{n=1}^\infty \int_{-\infty}^{\infty} dx \ x^{-3}e^{xi} \delta(x-\frac{n\pi}{2})$$
$$= Im \frac{1}{2 \pi } \sum_{n=1}^\infty \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dp \ x^{-3}e^{xi} e^{pxi}e^ {-ip\frac{n\pi}{2}}$$
$$= Im \sum_{n=1}^\infty \frac{1}{2 \pi }\int_{-\infty}^{\infty} dp \ e^ {-i(p-1)\frac{n\pi}{2}} ( \int_{-\infty}^{\infty} dx \ x^{-3} e^{pxi})$$
$$= Im \sum_{n=1}^\infty \frac{1}{2 \pi }\int_{-\infty}^{\infty} dp \ e^ {-i(p-1)\frac{n\pi}{2}} i \frac{\pi}{2} p^2 sgn(p)$$
$$=\frac{1}{4}Im\ i\int_{-\infty}^{\infty} dp \ (\sum_{n=1}^\infty e^ {-i(p-1)\frac{n\pi} {2}} ) p^2 sgn(p) = \frac{1}{4}$$
I am sure that the sum plays a delta function role but not sure it is correct coefficient. If it is OK
$$S=\frac{\pi^3}{32}$$

Last edited:
topsquark
We have

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} & = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n^3}
\nonumber \\
& = \frac{1}{2} \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n^3} \int_0^\infty e^{-y} y^2 dy
\nonumber \\
& = \frac{1}{2} \sum_{n=1}^\infty \sin \dfrac{\pi n}{2} \int_0^\infty e^{-nx} x^2 dx
\nonumber \\
& = \frac{1}{4i} \sum_{n=1}^\infty \int_0^\infty (e^{-nx + \frac{i\pi n}{2}} - e^{-nx + \frac{-i\pi n}{2}}) x^2 dx
\nonumber \\
& = \frac{1}{4i} \int_0^\infty \left( \dfrac{1}{1-e^{-x + \frac{i\pi}{2}}} - \dfrac{1}{1-e^{-x + \frac{-i\pi}{2}}} \right) x^2 dx
\nonumber \\
& = \frac{1}{2} \int_0^\infty \dfrac{x^2} {e^x + e^{-x}} dx
\nonumber \\
& = \frac{1}{4} \int_{-\infty}^\infty \dfrac{x^2} {e^x + e^{-x}} dx
\end{align*}

We can write

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2} {e^x + e^{-x}} dx & = \frac{\partial^2}{\partial \alpha^2} \left. \int_{-\infty}^\infty \dfrac{e^{\alpha x}} {e^x + e^{-x}} dx \right|_{\alpha =0}
\end{align*}

We will evaluate

\begin{align*}
\int_{-\infty}^\infty \dfrac{e^{\alpha x}} {e^x + e^{-x}} dx
\end{align*}

for ##-\frac{1}{2} \leq \alpha \leq \frac{1}{2}## by considering the rectangular contour integral (see figure) of

\begin{align*}
\oint_C \dfrac{e^{\alpha z}} {e^z + e^{-z}} dz
\end{align*}

The integral along the vertical edges vanishes as:

\begin{align*}
f(z) = \dfrac{e^{\alpha (x+iy)}}{e^{x+iy} + e^{-x-iy}} =
\begin{cases}
e^{(\alpha - 1) (x+iy)} & x \rightarrow \infty \\
e^{(\alpha + 1) (x+iy)} & x \rightarrow - \infty \\
\end{cases}
\end{align*}

So that

\begin{align*}
\oint_C \dfrac{e^{\alpha z}}{e^z + e^{-z}} dz & = \int_{-\infty}^\infty \dfrac{e^{\alpha x}}{e^x + e^{-x}} dx + e^{\alpha \pi i} \int_{-\infty+ i \pi}^{\infty + i \pi} \dfrac{e^{\alpha x}}{e^x + e^{-x}} dx
\nonumber \\
& = (1 + e^{\alpha \pi i}) \int_{-\infty}^\infty \dfrac{e^{\alpha x}} {e^x + e^{-x}} dx
\end{align*}

and so

\begin{align*}
\frac{1}{4} \int_{-\infty}^\infty \dfrac{e^{\alpha x}} {e^x + e^{-x}} dx & = \frac{i \pi}{2(1 + e^{\alpha \pi i}) } \frac{1}{2 \pi i} \oint_C \dfrac{e^{\alpha z}}{e^z + e^{-z}} dz
\nonumber \\
& = \frac{i \pi}{2(1 + e^{\alpha \pi i})} \lim_{z \rightarrow \frac{i \pi}{2}} (z - \frac{i \pi}{2}) \dfrac{e^{\alpha z}}{e^z + e^{-z}}
\nonumber \\
& = \frac{i \pi}{2(1 + e^{\alpha \pi i})} \lim_{z \rightarrow \frac{i \pi}{2}} \dfrac{e^{\alpha z}}{e^z - e^{-z}}
\nonumber \\
& = \frac{\pi}{4(1 + e^{\alpha \pi i})} e^{\alpha \frac{i \pi}{2}}
\nonumber \\
& = \frac{\pi}{8} \frac{1} {\cos (\dfrac{\alpha \pi}{2})} .
\end{align*}

We then have

\begin{align*}
\frac{1}{4} \int_{-\infty}^\infty \dfrac{x^2} {e^x + e^{-x}} dx & = \frac{\pi}{8} \frac{\partial^2}{\partial \alpha^2} \left. \frac{1}{\cos (\dfrac{\alpha \pi}{2})} \right|_{\alpha =0}
\nonumber \\
& = \frac{\pi^2}{16} \frac{\partial}{\partial \alpha} \left. \frac{\sin (\dfrac{\alpha \pi}{2})}{\cos^2 (\dfrac{\alpha \pi}{2})} \right|_{\alpha =0}
\nonumber \\
& = \frac{\pi^3}{32} \left. \frac{\cos^3 (\dfrac{\alpha \pi}{2}) + 2 \sin^2 (\dfrac{\alpha \pi}{2}) \cos (\dfrac{\alpha \pi}{2})}{\cos^4 (\dfrac{\alpha \pi}{2})} \right|_{\alpha =0}
\nonumber \\
& = \frac{\pi^3}{32} .
\end{align*}

So finally,

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{32} .
\end{align*}

Last edited:
anuttarasammyak and topsquark
@anuttarasammyak. Is the sum ##Re \sum_{n=1}^\infty e^{-i(p-1) \frac{n \pi}{2}}## related to the periodic delta function though? With period 4? And so the integral ##\int_{-\infty}^\infty dp \cdots## wouldn't pick out a single value of ##p##?

EDIT: Thinking about some more. Put ##p'=p-1## and plug in you are dealing with the periodic delta function results in: ##\int_{-\infty}^\infty dp' \sum_{m=-\infty}^\infty \delta (p'+4m) (p'+1)^2 sgn(p'+1)##. So terms negative in ##p'## don't cancel with terms positive in ##p'##.

Last edited:
julian said:
@anuttarasammyak. Is the sum Re∑n=1∞e−i(p−1)nπ2 related to the periodic delta function though? With period 4? And so the integral ∫−∞∞dp⋯ wouldn't pick out a single value of p?
Thanks for pointing out the difficulty. From my last line
$$\frac{8}{\pi^3}S=\frac{1}{2}\int_{0}^{\infty} p^2 dp \ (\sum_{n=1}^\infty \sin \frac{n\pi} {2}\sin \frac{pn\pi} {2} )$$
In here
$$\sum_{n=1}^\infty \sin \frac{n\pi} {2}\sin \frac{pn\pi} {2}=\frac{1}{2}\sum_{n=1}^\infty [\cos \frac{(p-1)n\pi} {2}-\cos \frac{(p+1)n\pi}{2}]$$
As you pointed out not only p=1 but p=4m+1 is picked up. Performing integral by parts now I observe that my attempt was not solving but just transforming the appearence of the problem.

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milkism
I would try another transformation
$$S=Re(-i\sum_{n=1}^\infty \frac{e^{\frac{\pi n i}{2}}}{n^3})$$
$$=\lim_{x\rightarrow 1} Re(-i (\frac{\pi i}{2})^3 \int dx \int dx \int dx \sum_{n=1}^\infty e^{\frac{\pi nx i}{2}})$$
with convention that all the integral constants are zero.
$$S=\frac{\pi^3}{8}Re( - \lim_{x\rightarrow 1} \int dx \int dx \int dx \frac{1}{e^{\frac{-i \pi x}{2}-1}})$$
$$=\frac{\pi^3}{8}Re( - \lim_{x\rightarrow 1} \int dx \int dx \frac{2i}{\pi} \log(1-e^{\frac{i \pi x}{2}}))$$With help of wolfram introducing polylogarithm function
$$=\frac{\pi^3}{8}Re( - \lim_{x\rightarrow 1} \int dx \frac{-4}{\pi^2} Li_2(e^{\frac{i \pi x}{2}}))$$
$$=\frac{\pi^3}{8}Re( - \lim_{x\rightarrow 1} \frac{8i}{\pi^3} Li_3(e^{\frac{i \pi x}{2}}))$$
$$=Re(-iLi_3(i))$$
it seems coming back to the original problem again. According to wolfram
$$=Re(\frac{\pi^3}{32}-\frac{3}{32}\zeta(3)i)=\frac{\pi^3}{32}$$

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It so happens that the general sum

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^{2k+1}}
\end{align*}

for ##k=0,1,2,\dots## can be easily performed using the above method of post #5, the answer expressed in terms of Euler numbers. Which I do in this post. In post #5, however, I didn't justify interchanging a summation and integration. I do this in this post.

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^{2k+1}} & = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n^{2k+1}}
\nonumber \\
& = \frac{1}{(2k)!} \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n^{2k+1}} \int_0^\infty e^{-y} y^{2k} dy
\nonumber \\
& = \frac{1}{(2k)!} \sum_{n=1}^\infty \sin \dfrac{\pi n}{2} \int_0^\infty e^{-nx} x^{2k} dx
\nonumber \\
& = \frac{1}{(2k)! 2i} \sum_{n=1}^\infty \int_0^\infty (e^{-nx + \frac{i\pi n}{2}} - e^{-nx + \frac{-i\pi n}{2}}) x^2 dx
\nonumber \\
& = \frac{1}{(2k)! 2i} \int_0^\infty \left( \dfrac{1}{1-e^{-x + \frac{i\pi}{2}}} - \dfrac{1}{1-e^{-x + \frac{-i\pi}{2}}} \right) x^{2k} dx
\nonumber \\
& = \frac{1}{(2k)!} \int_0^\infty \dfrac{x^{2k}} {e^x + e^{-x}} dx
\nonumber \\
& = \frac{1}{(2k)! 2} \int_{-\infty}^\infty \dfrac{x^{2k}} {e^x + e^{-x}} dx
\end{align*}

In post #5 I proved

\begin{align*}
\frac{1}{2} \int_{-\infty}^\infty \dfrac{e^{\alpha x}} {e^x + e^{-x}} dx & = \frac{\pi}{4} \frac{1}{\cos (\dfrac{\alpha \pi}{2})} .
\end{align*}

Using this, we have

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^{2k+1}} & = \frac{1}{(2k)! 2} \int_{-\infty}^\infty \dfrac{x^{2k}} {e^x + e^{-x}} dx
\nonumber \\
& = \frac{\pi}{(2k)! 4} \frac{\partial^{2k}}{\partial \alpha^{2k}} \left. \frac{1}{\cos (\dfrac{\alpha \pi}{2})} \right|_{\alpha =0} \qquad (*)
\end{align*}

Euler numbers are defined by

\begin{align*}
\frac{2}{e^t+e^{-t}} = \sum_{n=0}^\infty \dfrac{E_n}{n!} t^n .
\end{align*}

So that

\begin{align*}
\sec t = \sum_{n=0}^\infty \dfrac{(-1)^n E_{2n}}{(2n)!} t^{2n}
\end{align*}

and

\begin{align*}
\sec \dfrac{\pi t}{2} & = \sum_{n=0}^\infty \dfrac{(-1)^n \pi^{2n} E_{2n}}{4^k (2n)!} t^{2n}
\end{align*}

Using this in ##(*)##, results in:

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^{2k+1}} & = \frac{\pi}{(2k)! 4} \frac{\partial^{2k}}{\partial \alpha^{2k}} \left. \frac{1}{\cos (\dfrac{\alpha \pi}{2})} \right|_{\alpha =0}
\nonumber \\
& = \pi^{2k+1} \dfrac{(-1)^k E_{2k}}{(2k)! 4^{k+1}}
\end{align*}

The first few Euler numbers are ##E_0=1##, ##E_2=-1##, and ##E_4=5##.

Interchanging summation and integration

Note in the above, I made an interchange of summation and integration. This needs to be justified.

Case ##k>0##:

As

\begin{align*}
\frac{1}{(2k)!} \sum_{n=1}^\infty \int_0^\infty \left| \sin \dfrac{\pi n}{2} x^{2k} e^{-nx} \right| dx & = \sum_{n=0}^\infty \frac{1}{(2n+1)^{2k+1}}
\nonumber \\
& < \sum_{n=0}^\infty \frac{1}{(2n+1)^2}
\nonumber \\
& = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{(2n)^2}
\nonumber \\
& = (1 - 2^{-2}) \sum_{n=1}^\infty \frac{1}{n^2}
\nonumber \\
& < (1 - 2^{-2}) (1 + \lim_{N \rightarrow \infty} \sum_{n=2}^N \frac{1}{n (n-1)} )
\nonumber \\
& < (1 - 2^{-2}) (1 + \lim_{N \rightarrow \infty} \sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) )
\nonumber \\
& = (1 - 2^{-2}) ( 1 + 1 - \lim_{N \rightarrow \infty} \frac{1}{N}) = (1 - 2^{-2}) 2 < \infty
\end{align*}

you can use Fubini to justify interchanging summation and integration:

\begin{align*}
\sum_{n=1}^\infty \int_0^\infty \sin \dfrac{\pi n}{2} x^{2k} e^{-nx} dx = \int_0^\infty \sum_{n=1}^\infty \sin \dfrac{\pi n}{2} x^{2k} e^{-nx} dx .
\end{align*}

Case ##k=0## (Leibniz formula for ##\pi##):

As

\begin{align*}
\sum_{n=1}^\infty \int_0^\infty \left| \sin \dfrac{\pi n}{2} e^{-nx} \right| dx = \sum_{n=0}^\infty \frac{1}{2n+1} = \infty
\end{align*}

you can't use Fubini to justify this interchange. An equivalent check:

\begin{align*}
\int_0^\infty \sum_{n=1}^\infty \left| \sin \dfrac{\pi n}{2} e^{-nx} \right| dx & = \int_0^\infty (e^{-x} + e^{-3} + e^{-5x} + \cdots) dx
\nonumber \\
& = \int_0^\infty \dfrac{1}{e^x-e^{-x}} dx
\nonumber \\
& = \left[ \frac{1}{2} \ln \left( \dfrac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}} \right) \right]_0^\infty = \infty
\end{align*}

You can, however, use the dominated convergence theorem to prove the interchange is legitimate. The proof for ##k \geq 0## is not much more difficult that the proof for ##k=0##, so I give the proof for ##k \geq 0##.

Using the dominated convergence theorem ##k \geq 0##:

I now justify interchanging summation and integration using the dominated convergence theorem. Define ##f_N (x) = \sum_{n=1}^N \sin \dfrac{\pi n}{2} x^{2k} e^{-nx}##. Then, for appropriate integers ##\ell,\ell'## taken from ##1,2, \dots##, we have

\begin{align*}
f_N (x) & =
\begin{cases}
(e^{-x} - e^{-3x} + e^{-5x} - e^{-7x} + \cdots + e^{-(4\ell-3)x}) x^{2k} & \text{ odd number of terms} \\
(e^{-x} - e^{-3x} + e^{-5x} - e^{-7x} + \cdots - e^{-(4\ell'-1)x}) x^{2k} & \text{ even number of terms} \\
\end{cases}
\nonumber \\
& =
\begin{cases}
e^{-x} \dfrac{1+e^{-(4\ell-1)x}}{1+e^{-2x}} x^{2k} & \text{ odd number of terms} \\
e^{-x} \dfrac{1-e^{-(4\ell'+1)x}}{1+e^{-2x}} x^{2k} & \text{ even number of terms} \\
\end{cases}
\end{align*}

So the limit function for ##x > 0## is

\begin{align*}
\lim_{N \rightarrow \infty} f_N(x) = f(x) = \dfrac{x^{2k}}{e^x+e^{-x}}
\end{align*}

So ##f_N(x)## converges to the function ##f(x)## except at ##x=0##:

\begin{align*}
\lim_{N \rightarrow \infty} |f(x) - f_N (x)| & =
\begin{cases}
\lim_{\ell \rightarrow \infty} e^{-x} \dfrac{e^{-(4\ell-1)x}}{1+e^{-2x}} x^{2k} = 0 & \text{ odd number of terms} \\
\lim_{\ell' \rightarrow \infty} e^{-x} \dfrac{e^{-(4\ell'+1)x}}{1+e^{-2x}} x^{2k} = 0 & \text{ even number of terms} \\
\end{cases}
\end{align*}

It is not a problem that you don't have convergence at the single point ##x=0##?

Also,

\begin{align*}
0 \leq f_N (x)\leq \dfrac{1+e^{-2x}}{1+e^{-2x}} e^{-x} x^{2k} = x^{2k} e^{-x}
\end{align*}

So

\begin{align*}
|f_N (x)| \leq x^{2k} e^{-x} = g(x)
\end{align*}

The function ##g(x)## is integrable. By the dominated convergence theorem

\begin{align*}
\lim_{N \rightarrow \infty} \int_0^\infty \sum_{n=1}^N \sin \dfrac{\pi n}{2} e^{-nx} dx = \int_0^\infty \lim_{N \rightarrow \infty} \sum_{n=1}^N \sin \dfrac{\pi n}{2} e^{-nx} dx
\end{align*}

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This is a different solution method to the ones I've already given. The general sum

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^{2k+1}}
\end{align*}

for ##k = 0,1,2, \dots## can be obtained from the partial fraction expansion for ##\pi \sec \pi x##.

The generating function for Euler numbers is

\begin{align*}
\frac{2}{e^x+e^{-x}} = \sum_{k=0}^\infty \dfrac{E_{2k}}{(2k)!} x^{2k} \qquad (*)
\end{align*}

We have the partial fraction expansion (derived below)

\begin{align*}
\pi \sec \pi x & = \sum_{n=0}^\infty (-1)^n \dfrac{2n+1}{(n+\frac{1}{2})^2 - x^2}
\nonumber \\
& = 4 \sum_{n=0}^\infty (-1)^n \dfrac{2n+1}{(2n+1)^2 - 4x^2}
\end{align*}

From which we have

\begin{align*}
\text{sech } x & = \frac{4}{\pi} \sum_{n=0}^\infty (-1)^n \dfrac{2n+1}{(n+\frac{1}{2})^2 + \frac{x^2}{\pi^2}}
\nonumber \\
& = 4 \pi \sum_{n=0}^\infty (-1)^n \dfrac{2n+1}{(2n+1)^2 \pi^2 + x^2}
\end{align*}

The ##n##th term in this infinite sum is just the geometric expansion:

\begin{align*}
\frac{4}{\pi} (-1)^n \times \frac{1}{2n+1} \sum_{k=0}^\infty \left( \dfrac{-4 x^2}{(2n+1)^2 \pi^2} \right)^k & = \frac{4}{\pi} (-1)^n \times \frac{1}{2n+1} \dfrac{1}{ 1+ \dfrac{4x^2}{(2n+1)^2 \pi^2} }
\nonumber \\
& = 4 \pi (-1)^n \dfrac{2n+1}{(2n+1)^2 \pi^2 + 4x^2}
\end{align*}

Substituting this into the expression for ##\text{sech } x##,

\begin{align*}
\text{sech } x & = \frac{4}{\pi} \sum_{n=0}^\infty (-1)^n \frac{1}{2n+1} \sum_{k=0}^\infty \left( \dfrac{-4 x^2}{(2n+1)^2 \pi^2} \right)^k
\nonumber \\
& = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty (-1)^k \dfrac{4^{k+1}}{(2n+1)^{2k+1} \pi^{2k+1}} x^{2k}
\nonumber \\
& = \sum_{k=0}^\infty (-1)^k \sum_{n=0}^\infty (-1)^n \dfrac{4^{k+1}}{(2n+1)^{2k+1} \pi^{2k+1}} x^{2k}
\end{align*}

Comparing this to ##(*)##,

\begin{align*}
\sum_{k=0}^\infty (-1)^k \sum_{n=0}^\infty (-1)^n \dfrac{4^{k+1}}{(2n+1)^{2k} \pi^{2k+1}} x^{2k} = \sum_{k=0}^\infty \dfrac{E_{2k}}{(2k)!} x^{2k}
\end{align*}

and equating coefficients,

\begin{align*}
(-1)^k \sum_{n=0}^\infty (-1)^n \dfrac{4^{k+1}}{(2n+1)^{2k+1} \pi^{2k+1}} = \dfrac{E_{2k}}{(2k)!}
\end{align*}

or

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^{2k+1}} = \pi^{2k+1} \dfrac{(-1)^k E_{2k}}{4^{k+1} (2k)!}
\end{align*}

For ##k=1##, and using ##E_2=-1##, we have

\begin{align*}
\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^3} = \dfrac{\pi^3}{32} .
\end{align*}

Proving partial fraction expansion:

You can derive the partial fraction expansion for trigonometric functions using the complex contour integration technique I used in post #2 (see my notes here for details). Instead I derive the partial fraction expansion for ##\pi \sec \pi x## another way.

I know of the following proof for the partial fraction expansion for ##\pi \cot \pi x##: Write ##g(y) = \cos (xy)## as the Fourier series

\begin{align*}
g(y) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos (ny)
\end{align*}

for ##-\pi \leq y \leq \pi##. Doing the calculation of the Fourier coefficients using trig identities ##\cos \alpha \cos \beta = \frac{1}{2} [\cos (\alpha - \beta) + \cos (\alpha - \beta)]## and ##\sin (\alpha + \beta) = \cos \alpha \sin \beta + \sin \alpha \cos \beta##, you obtain

\begin{align*}
a_n & = \frac{1}{\pi} \dfrac{2x \sin (x \pi) (-1)^n}{x^2-n^2} , \qquad a_0 = \frac{1}{\pi} \dfrac{2}{x} \sin (x \pi) .
\end{align*}

So that

\begin{align*}
\cos (xy) = \dfrac{\sin (x \pi)}{x \pi} + \sum_{n=1}^\infty \frac{1}{\pi} \dfrac{2x \sin (x \pi) (-1)^n}{x^2-n^2} \cos (ny)
\end{align*}

Putting ##y=\pi##, results in

\begin{align*}
\pi \cot \pi x = \frac{1}{x} + 2x \sum_{n=1}^\infty \frac{1}{x^2-n^2}
\end{align*}

We have

\begin{align*}
2x \sum_{n=0}^\infty \frac{1}{(n+\frac{1}{2})^2 - x^2} & = 8x \sum_{n=0}^\infty \frac{1}{(2n+1)^2 - (2x)^2}
\nonumber \\
& = 8x \sum_{n=1}^\infty \frac{1}{n^2 - (2x)^2} - 8x \sum_{n=1}^\infty \frac{1}{(2n)^2 - (2x)^2}
\nonumber \\
& = 8x \sum_{n=1}^\infty \frac{1}{n^2 - (2x)^2} - 8x \frac{1}{4} \sum_{n=1}^\infty \frac{1}{n^2 - x^2}
\nonumber \\
& = 2 \left( \frac{1}{2x} - \pi \cot 2 \pi x \right) - \left( \frac{1}{x} - \pi \cot \pi x \right)
\nonumber \\
& = \pi \left( \cot \pi x - 2 \cot 2 \pi x \right)
\nonumber \\
& = \pi \left( \cot \pi x - 2 \dfrac{\cos^2 \pi x - \sin^2 \pi x}{2 \sin \pi x \cos \pi x} \right)
\nonumber \\
& = \pi \tan \pi x .
\end{align*}

So that

\begin{align*}
\pi \tan \frac{1}{2} \pi x = 4x \sum_{n=0}^\infty \frac{1}{(2n+1)^2 - x^2}
\end{align*}

We have the identities:

\begin{align*}
\frac{\pi}{\sin \pi x} & = \frac{\pi (\sin^2 \dfrac{\pi x}{2} + \cos^2 \dfrac{\pi x}{2})}{2 \sin \dfrac{\pi x}{2} \cos \dfrac{\pi x}{2}} = \frac{\pi \tan \dfrac{\pi x}{2}}{2} + \frac{\pi \cot \dfrac{\pi x}{2}}{2}
\nonumber \\
\pi \cot \pi x & = \frac{\pi (\cos^2 \dfrac{\pi x}{2} - \sin^2 \dfrac{\pi x}{2})}{2 \sin \dfrac{\pi x}{2} \cos \dfrac{\pi x}{2}} = \frac{\pi \cot \dfrac{\pi x}{2}}{2} - \frac{\pi \tan \dfrac{\pi x}{2}}{2}
\end{align*}

The first minus the second gives

\begin{align*}
\frac{\pi}{\sin \pi x} & = \pi \cot \pi x + \pi \tan \dfrac{\pi x}{2}
\end{align*}

from which we have

\begin{align*}
\frac{\pi}{\sin \pi x} & = \pi \cot \frac{1}{2} \pi x + \pi \tan \dfrac{\pi x}{2}
\nonumber \\
& = \frac{1}{x} + 2x \sum_{n=1}^\infty \frac{1}{x^2-n^2} - 4x \sum_{n=0}^\infty \frac{1}{x^2 - (2n+1)^2}
\nonumber \\
& = \frac{1}{x} + 2x \sum_{n=1}^\infty \frac{(-1)^n}{x^2 - n^2}
\nonumber \\
& = \sum_{n=-\infty}^\infty \frac{(-1)^n}{x + n}
\end{align*}

We use the identity ##\cos \pi x = \sin \pi (x + \frac{1}{2})## to obtain

\begin{align*}
\frac{\pi}{\cos \pi x} & = \sum_{n=-\infty}^\infty \frac{(-1)^n}{x + (n+\frac{1}{2})}
\nonumber \\
& = 2 \sum_{n=0}^\infty (-1)^n \dfrac{n+\frac{1}{2}}{(n+\frac{1}{2})^2 - x^2}
\end{align*}

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As I have noted in an Insight, just create the Fourier serien from the function $\pi ^{2}\cdot x - x^{3}$. This ends up in $12\cdot \sum_{n=1}^{\infty}\frac{\sin(n(\pi - x))}{n^{3}}$. Evaluate both expressions at π/2. and you have the desired proof.

anuttarasammyak
Svein said:
As I have noted in an Insight, just create the Fourier serien from the function $\pi ^{2}\cdot x - x^{3}$. This ends up in $12\cdot \sum_{n=1}^{\infty}\frac{\sin(n(\pi - x))}{n^{3}}$. Evaluate both expressions at π/2. and you have the desired proof.
You mentioned your Insight here. Reading your Insight is where I got the idea to rewrite an alternating sum over odd integers using ##\sin \dfrac{n \pi}{2}##, an idea which has opened up fruitful avenues.

What's intriguing about @Euge's question is its versatility; it invites a range of perspectives and solution methods. Maybe there are still more ways of proving it?

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Svein said:
As I have noted in an Insight, just create the Fourier serien from the function $\pi ^{2}\cdot x - x^{3}$. This ends up in $12\cdot \sum_{n=1}^{\infty}\frac{\sin(n(\pi - x))}{n^{3}}$. Evaluate both expressions at π/2. and you have the desired proof.
$$T(x):=\sum_{n=1}^{\infty}\frac{\sin(n(\pi-x))}{n^3}$$ for ##-\pi < x < \pi##. Regarding this as Fourier series, ##n^{3}## in denominator shows that T(x) is a tertial function of x thinking of integration by parts for the calculation of Fourier components. We can write T(x) with constant a as
$$T(x)=ax(x-\pi)(x+\pi)$$ because $$T(0)=T(\pi)=T(-\pi)=0$$ To determine a, let us compare T'(0) of the both forms
$$1-2^{-2}+3^{-2}-4^{-2}+...=\frac{1}{2}\zeta(2)=-\pi^2 a$$
$$a=-\frac{1}{12}$$
Thus the value we want is
$$T(\frac{\pi}{2})=-\frac{1}{12}(\frac{\pi}{2})(-\frac{\pi}{2})\frac{3\pi}{2}=\frac{\pi^3}{32}$$

With wolfram.

It shows $$\frac{\pi^3}{ 32}=0.9674...$$

Similary say
$$T_+ (x):=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}$$ for ##-\pi < x < \pi##.
$$T_+ (x)=-\frac{1}{6}x(x-\pi)(x+\pi)=2T(x)$$ But from the summation
$$T_+ (\frac{\pi}{2})=T (\frac{\pi}{2})=1-3^{-3}+5^{-3}-...$$ which is confirmed as below shown.

Where I went wrong ? We may have to choose the basic area for repetition where Y of sin Y for n=1 does not change sign, e.g. ##0 < x < 2\pi## for
$$T_+ (x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}$$

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As those know who learned precalculus from Euler, (Intro. to Analysis of the Infinite, paragraphs ##174-175##, one can deduce from his complex series for the trig functions, that in general,
$$1/m^3 - 1/(2n-m)^3 + 1/(2n+m)^3 - 1/(4n-m)^3 + 1/(4n+m)^3 - .... = (k^2+1)π^3/(8n^3k^3),$$ where ##k = \tan(mπ/2n)##.

In particular, when ##m=1, n=2##, then ##k = \tan(π/4) = 1##, so
$$1 - 1/3^3 + 1/5^3 -1/7^3 + 1/9^3 -\cdots = π^3/32.$$

cf. p.146:

Last edited by a moderator:
dextercioby and julian
anuttarasammyak said:
$$T(x):=\sum_{n=1}^{\infty}\frac{\sin(n(\pi-x))}{n^3}$$ for ##-\pi < x < \pi##. Regarding this as Fourier series, ##n^{3}## in denominator shows that T(x) is a tertial function of x thinking of integration by parts for the calculation of Fourier components. We can write T(x) with constant a as
$$T(x)=ax(x-\pi)(x+\pi)$$ because $$T(0)=T(\pi)=T(-\pi)=0$$ To determine a, let us compare T'(0) of the both forms
$$1-2^{-2}+3^{-2}-4^{-2}+...=\frac{1}{2}\zeta(2)=-\pi^2 a$$
$$a=-\frac{1}{12}$$
Thus the value we want is
$$T(\frac{\pi}{2})=-\frac{1}{12}(\frac{\pi}{2})(-\frac{\pi}{2})\frac{3\pi}{2}=\frac{\pi^3}{32}$$

With wolfram.

View attachment 344990

It shows $$\frac{\pi^3}{ 32}=0.9674...$$View attachment 344995
Similary say
$$T_+ (x):=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}$$ for ##-\pi < x < \pi##.
$$T_+ (x)=-\frac{1}{6}x(x-\pi)(x+\pi)=2T(x)$$ But from the summation
$$T_+ (\frac{\pi}{2})=T (\frac{\pi}{2})=1-3^{-3}+5^{-3}-...$$ which is confirmed as below shown.
View attachment 345002
View attachment 345031

Where I went wrong ? We may have to choose the basic area for repetition where Y of sin Y for n=1 does not change sign, e.g. ##0 < x < 2\pi## for
$$T_+ (x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}$$
The correct formula is

\begin{align*}
\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3} = - \frac{1}{12} x (x-\pi)(2\pi-x)
\end{align*}

and it is on the interval ##0 < x < 2 \pi##.

We demonstrate that for the interval ##0 < x < 2 \pi## that the Fourier series of ##\frac{(x-\pi)^3 - \pi^2 (x-\pi)}{12}## is ##\sum_{n=1}^\infty \frac{\sin (nx)}{n^3}##. For the interval ##0 < x < 2 \pi##,

\begin{align*}
s(x) = \sum_{n=1}^\infty b_n \sin (nx) \qquad \text{where} \qquad b_n = \frac{1}{\pi} \int_0^{2 \pi} s (x) \sin (nx)
\end{align*}

So, in our case

\begin{align*}
b_n = \frac{1}{\pi} \int_0^{2 \pi} \frac{(x-\pi)^3 - \pi^2 (x-\pi)}{12} \sin (nx) dx = (-1)^n \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{y^3 - \pi^2 y}{12} \sin (ny) dy
\end{align*}

where we have put ##y = x-\pi##. So

\begin{align*}
b_n & = (-1)^n \frac{1}{12 \pi} \int_{-\pi}^{\pi} (y^3 - \pi^2 y) \sin (ny) dy
\nonumber \\
& = (-1)^n \frac{1}{6 \pi} \left. \left( \left[ \frac{d^3}{dk^3} + \pi^2 \frac{d}{dk} \right] \int_0^\pi \cos (ky) dy \right) \right|_{k=n}
\nonumber \\
& = (-1)^n \frac{1}{6 \pi} \left. \left( \left[ \frac{d^3}{dk^3} + \pi^2 \frac{d}{dk} \right] \frac{1}{k} \sin (k \pi) \right) \right|_{k=n}
\nonumber \\
& = \frac{1}{n^3}
\end{align*}

Last edited:
anuttarasammyak
Hi @anuttarasammyak, I'm looking at your solution method from post #8 in more detail. For simplicity reasons I consider ##e^{inx}## instead of ##e^{\frac{\pi n x i}{2}}## and put ##x=\frac{\pi}{2}## at the end.

At a certain step you basically proceed to integrate this:

\begin{align*}
\ln (1- e^{ix}) = - \sum_{n=1}^\infty \frac{e^{inx}}{n}
\end{align*}

twice with respect to ##x## and then take the imaginary part. You can split ##\ln (1- e^{ix})## into its real and imaginary parts:

\begin{align*}
\ln (1- e^{ix}) & = \ln |1- e^{inx}| + i \theta
\nonumber \\
& = \ln \left| 2 \sin \frac{x}{2} \right| + i \theta
\end{align*}

where

\begin{align*}
\tan \theta = \frac{\sin x}{\cos x - 1} = - \frac{\cos (x/2)}{\sin (x/2)} = - \tan \left( \frac{\pi}{2} - \frac{x}{2} \right)
\end{align*}

So

\begin{align*}
\sum_{n=1}^\infty \frac{e^{inx}}{n} & = - \ln \left| 2 \sin \frac{x}{2} \right| + i \left( \frac{\pi}{2} - \frac{x}{2} \right)
\end{align*}

You didn't split ##\ln (1- e^{ix})## into its real and imaginary part, and proceeded to integrate. This made things overly complicated, we only need the imaginary part:

\begin{align*}
\sum_{n=1}^\infty \frac{\sin (nx)}{n} = \frac{\pi}{2} - \frac{x}{2} \qquad (*)
\end{align*}

Integrating this gives

\begin{align*}
\sum_{n=1}^\infty \frac{\cos (nx)}{n^2} - \sum_{n=1}^\infty \frac{1}{n^2} & = - \int_0^x \left( \frac{\pi}{2} - \frac{x}{2} \right) dx
\nonumber \\
& = - \frac{\pi}{2} x + \frac{x^2}{4}
\end{align*}

Rearranging, and integrating again

\begin{align*}
\sum_{n=1}^\infty \frac{\sin (nx)}{n^3} & = - \frac{\pi}{4} x^2 + \frac{x^3}{12} + \sum_{n=1}^\infty \frac{1}{n^2} x
\end{align*}

Putting ##x = \frac{\pi}{2}## we get:

\begin{align*}
\sum_{n=1}^\infty \frac{\sin (\frac{n \pi}{2})}{n^3} & = - \pi^3 \frac{5}{3 \cdot 32} + \frac{\pi}{2} \sum_{n=1}^\infty \frac{1}{n^2} \qquad (**)
\end{align*}

We can find the sum ##\sum_{n=1}^\infty \frac{1}{n^2}## by integrating ##(*)## from ##0## to ##\pi##. Integrating the LHS of ##(*)## results in:

\begin{align*}
- \sum_{n=1}^\infty \frac{\cos (n\pi)}{n^2} + \sum_{n=1}^\infty \frac{1}{n^2} & = - \sum_{n=1}^\infty \frac{(-1)^n}{n^2} + \sum_{n=1}^\infty \frac{1}{n^2}
\nonumber \\
& = 2 (\sum_{n=0}^\infty \frac{1}{(2n-1)^2})
\nonumber \\
& = 2 (\sum_{n=0}^\infty \frac{1}{n^2} - \sum_{n=0}^\infty \frac{1}{(2n)^2})
\nonumber \\
& = \frac{3}{2} \sum_{n=0}^\infty \frac{1}{n^2}
\end{align*}

Integrating the RHS of ##(*)## results in

\begin{align*}
\int_0^\pi (\frac{\pi}{2} - \frac{x}{2}) dx = \frac{\pi^2}{4} .
\end{align*}

Equating the last two results, we obtain:

\begin{align*}
\sum_{n=0}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
\end{align*}

Substituting this into ##(**)##, finally we get:

\begin{align*}
\sum_{n=1}^\infty \frac{\sin (\frac{n \pi}{2})}{n^3} & = - \pi^3 \frac{5}{3 \cdot 32} + \frac{\pi^3}{12} = \frac{\pi^3}{32} .
\end{align*}

Applying the geometric sum formula and then Abel's theorem

In post #8 you wrote ##\sum_{n=1}^\infty e^{\frac{\pi nxi}{2}} = \frac{1}{e^{-\frac{\pi xi}{2}} - 1}##. This is not legitimate as this

\begin{align*}
(1 - e^{\frac{\pi xi}{2}}) \sum_{n=1}^N e^{\frac{\pi nxi}{2}} = 1 - e^{\frac{\pi (N+1)xi}{2}} = 1 - \cos (\frac{\pi (N+1)x}{2}) - i \sin (\frac{\pi (N+1)x}{2})
\end{align*}

is ill-defined in the limit ##N \rightarrow \infty##.

If you consider the sum ##\sum_{n=1}^\infty r^n e^{inx}## where ##|r| < 1##, then the limit ##\lim_{N \rightarrow \infty} (1 - e^{inx}) \sum_{n=1}^N e^{\frac{inx}{2}}## is well defined. You can then write

\begin{align*}
\sum_{n=1}^\infty r^n e^{inx} = \frac{re^{ix}}{1 - r e^{ix}}
\end{align*}

By integrating the uniformly convergent geometric power series term by term (and multiply by ##i##) you get:

\begin{align*}
\sum_{n=1}^\infty r^n \frac{e^{inx}}{n} & = i \int_0^x \frac{re^{ix}}{1 - r e^{ix}} dx
\nonumber \\
& = - \ln |1 - r e^{ix}| + i \tan^{-1} \frac{r \sin x}{r \cos x-1}
\end{align*}

If you prove that the series ##\sum_{n=1}^\infty \frac{e^{inx}}{n}## converges you would be able to apply Abel's theorem to conclude:

\begin{align*}
\lim_{r \rightarrow 1^-} \sum_{n=1}^\infty r^n \frac{e^{inx}}{n} = - \ln |1 - e^{ix}| + i \tan^{-1} \frac{\sin x}{\cos x-1} .
\end{align*}

You can prove ##\ln (1-z)## converges everywhere on the unit circle except at ##z=1##. Fix ##z## in the unit circle, i.e. ##|z|=1##. We apply Dirichlet's test:

If ##\{ a_n \}## are real numbers and ##\{ b_n \}## complex numbers such that: (i) ##a_1 \geq a_2 \geq \cdots## (ii) ##\lim_{n \rightarrow \infty} a_n = 0## (iii) There exists ##M## such that ##\left| \sum_{n=1}^N b_n \right| \leq M## for all ##N \in \mathbb{N}##; then ##\sum_{n=1}^\infty a_n b_n## converges.

Choose ##a_n = 1/n## and ##b_n = z^n##, then the first two conditions obviously hold. For the third condition:

\begin{align*}
\left| \sum_{n=1}^N z^n \right| = \left| \dfrac{z-z^{N+1}}{1-z} \right| \leq \frac{2}{| 1-z |}
\end{align*}

for all ##N \in \mathbb{N}##. This shows the series converges for ##|z|=1## where ##z \not= 1##.

Last edited:
anuttarasammyak

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