Question on an infinite summation series

[phymath7]

Is the infinite series ##\sum_{n=1,3,5,...}^\infty \frac {1} {n^6}## somewhat related to the Riemann zeta function?The attached image suggest the value to be inverse of the co-efficient of the series.Is there any integral representation of the series from where the series can be evaluated?

 

[PeroK]

I think these identities can be calculated using an appropriate Fourier series expansion:

https://math.stackexchange.com/ques...y-frac1n6-frac-pi6945-by-fourier-series-of-x2

 

[pasmith]

These series are usually evaluated by applying Parseval's Theorem to in this case [itex]x^3[/itex] on [itex][-\pi,\pi][/itex]. But you should perhaps look at reference 20, and also see how the [itex]c_n[/itex] were defined in the first place.

(If you have questions about an extract from another book or document, please tell us where it came from so we can consult the original in its context, if we have it available.)

When the sum converges absolutely, we have [tex]
\sum_{\mbox{odd $n$}} \frac1{n^z} = \sum_{n=1}^\infty \frac1{n^z} - \sum_{\mbox{even $n$}} \frac1{n^z}.[/tex] Since [tex]
\sum_{\mbox{even $n$}} \frac1{n^z} = \sum_{r=1}^\infty \frac1{(2r)^z} = 2^{-z} \sum_{r=1}^\infty \frac1{r^z}[/tex] we have [tex]
\sum_{\mbox{odd $n$}} \frac1{n^z} = (1 - 2^{-z}) \sum_{n=1}^\infty \frac1{n^z}.[/tex]

 

[julian]

Is the infinite series ##\sum_{n=1,3,5,...}^\infty \frac {1} {n^6}## somewhat related to the Riemann zeta function?The attached image suggest the value to be inverse of the co-efficient of the series.Is there any integral representation of the series from where the series can be evaluated?

It is closely connected to the Riemann zeta function, as noted by @pasmith. Such summations, of the form ##\sum_{\text{odd } n} \frac{1}{n^{2k}}##, can be evaluated directly without needing Fourier series knowledge, using a complex contour integration technique. I have attached some notes detailing this method, where I've derived a number of established results (draft version), including the closed formula for ##\zeta(2k)## in terms of Bernoulli numbers:

\begin{align*}
\zeta(2k) = \sum_{n=1}^\infty \frac{1}{n^{2k}} = (-1)^{k+1} (2 \pi)^{2k} \dfrac{B_{2k}}{2 (2k)!}
\end{align*}

 

[julian]

@phymath7 You asked if there is any integral representation of the series that could facilitate its evaluation. In my notes, I provided a complex integral representation of the sum that can be easily evaluated. There is also real integral representations available, and they can be evaluated.

The sum has the following essentially equivalent integral representations:

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} & = \frac{1}{4 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} - 1)^2} dx + \frac{1}{4 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} + 1)^2} dx
\end{align*}

and

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = (1 - 2^{-6}) \frac{1}{2 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx
\end{align*}

which I derive below. Comparing them you find

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx = \frac{2^5}{2^5-1} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} + 1)^2} dx
\end{align*}

This relation can also be proven directly. Substituting this into the second integral representation you get another integral representation

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = \frac{2^6-1}{2^6-2} \frac{1}{2 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} dx
\end{align*}

This integral can be evaluated with complex contour integration techniques. More on this later.

Deriving integral representations of the sum

I now derive these integral representations. We have

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} & = \frac{1}{2} \sum_{n=1}^\infty \dfrac{1 - \cos \pi n }{n^6}
\nonumber \\
& = \frac{1}{2 \cdot 5!} \sum_{n=1}^\infty \dfrac{1 - \cos \pi n}{n^6} \int_0^\infty e^{-y} y^5 dy
\nonumber \\
& = \frac{1}{2 \cdot 5!} \sum_{n=1}^\infty [1 - \cos \pi n] \int_0^\infty e^{-nx} x^5 dx
\nonumber \\
& = \frac{1}{2 \cdot 5!} \sum_{n=1}^\infty \int_0^\infty [1 - \cos \pi n] e^{-nx} x^5 dx
\nonumber \\
& = \frac{1}{4 \cdot 5!} \sum_{n=1}^\infty \int_0^\infty (2 e^{-nx} - e^{-nx + i \pi n} - e^{-nx + -i \pi n} ) x^5 dx
\nonumber \\
& = \frac{1}{4 \cdot 5!} \int_0^\infty \left( \dfrac{2}{e^{x} - 1} - \dfrac{1}{e^{x - i \pi} - 1} - \dfrac{1}{e^{x + i\pi} - 1} \right) x^5 dx
\nonumber \\
& = \frac{1}{2 \cdot 5!} \int_0^\infty \left( \dfrac{1}{e^x - 1} + \dfrac{1}{e^{x} + 1} \right) x^5 dx
\nonumber \\
& = \frac{1}{2 \cdot 5!} \int_0^\infty \dfrac{x^5}{e^{x} - 1} dx + \frac{1}{2 \cdot 5!} \int_0^\infty \dfrac{x^5}{e^{x} + 1} dx
\end{align*}

Integrating by parts and then extending the range of integration:

\begin{align*}
\int_0^\infty \dfrac{x^5}{e^{x} - 1} dx & = \frac{1}{6} \int_0^\infty \dfrac{x^6 e^x}{(e^{x} - 1)^2} dx = \frac{1}{12} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} - 1)^2} dx
\nonumber \\
\int_0^\infty \dfrac{x^5}{e^{x} + 1} dx & = \frac{1}{6} \int_0^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} = \frac{1}{12} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} dx
\end{align*}

Using these in the previous equation, we obtain the first integral representation of the sum quoted at the beginning.

Recall the identity

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = (1 - 2^{-6}) \sum_{n=0}^\infty \dfrac{1}{n^6}
\end{align*}

We have

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{n^6} & = \frac{1}{5!} \sum_{n=1}^\infty \dfrac{1}{n^6} \int_0^\infty e^{-y} y^5 dy
\nonumber \\
& = \frac{1}{5!} \sum_{n=1}^\infty \int_0^\infty e^{-nx} x^5 dx
\nonumber \\
& = \frac{1}{5!} \int_0^\infty \dfrac{x^5}{e^{x} - 1} dx
\nonumber \\
& = \frac{1}{6 \cdot 5!} \int_0^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx
\nonumber \\
& = \frac{1}{12 \cdot 5!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx
\end{align*}

where in the second to last step we integrated by parts, followed by extending the range of integration. Substituting this into the identity I just mentioned results in the second integral representation of the sum quoted at the beginning.

As noted at the beginning, another representation follows from the first two:

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = \frac{2^6-1}{2^6-2} \frac{1}{2 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} dx
\end{align*}

I will illustrate a method of evaluating integrals of the form ##\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} dx## with the simpler example of the sum ##\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2}##.


Evaluating the integral: with simpler example

We look at the example:

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

and evaluate the integral of the integral representation. This is simpler to do with this example. We have

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac{1}{2} \sum_{n=1}^\infty \dfrac{1 - \cos \pi n}{n^2} \int_0^\infty e^{-y} y dy
\nonumber \\
& = \frac{1}{2} \sum_{n=1}^\infty [1 - \cos \pi n] \int_0^\infty e^{-nx} x dx
\nonumber \\
& = \frac{1}{2} \sum_{n=1}^\infty \int_0^\infty [1 - \cos \pi n] e^{-nx} x dx
\nonumber \\
& = \frac{1}{4} \sum_{n=1}^\infty \int_0^\infty (2 e^{-nx} - e^{-nx + i\pi n} - e^{-nx + -i\pi n} ) x dx
\nonumber \\
& = \frac{1}{4} \int_0^\infty \left( \dfrac{2}{e^{x} - 1} - \dfrac{1}{e^{x - i \pi} - 1} - \dfrac{1}{e^{x + i\pi} - 1} \right) x dx
\nonumber \\
& = \frac{1}{2} \int_0^\infty \dfrac{x}{e^x - 1} dx + \frac{1}{2} \int_0^\infty \dfrac{x}{e^{x} + 1} dx
\nonumber \\
& = \frac{1}{4} \int_0^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{4} \int_0^\infty \dfrac{x^2 x^x}{(e^x + 1)^2} dx
\nonumber \\
& = \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx
\end{align*}

So

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx \qquad (*)
\end{align*}

Alternatively, we can write

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

We have

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

So

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = (1 - 2^{-2}) \sum_{n=0}^\infty \dfrac{1}{n^2}
\nonumber \\
& = \frac{3}{16} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx \qquad (**)
\end{align*}

Comparing ##(*)## and ##(**)##, we have

\begin{align*}
\frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = \frac{3}{16} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\end{align*}

From which we get

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx = 2 \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx
\end{align*}

Substituting this into ##(**)## gives

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} = \frac{3}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx \qquad (***)
\end{align*}

We now evaluate this integral using complex analysis. Consider the rectangular contour, ##C##, in the figure

and the complex contour integral:

\begin{align*}
\oint_C \dfrac{e^z z^3}{(e^z + 1)^2} dz
\end{align*}

The integral along the vertical edges vanishes as:

\begin{align*}
f(z) = \dfrac{(x+iy)^3 e^{(x+iy)}}{(e^{x+iy} + 1)^2} =
\begin{cases}
(x+iy)^3 e^{-(x+iy)} & x \rightarrow \infty \\
(x+iy)^3 e^{(x+iy)} & x \rightarrow - \infty \\
\end{cases}
\end{align*}

So that

\begin{align*}
\oint_C \dfrac{e^z z^3}{(e^z + 1)^2} dz & = \int_{-\infty}^\infty \dfrac{e^x x^3}{(e^x + 1)^2} dx - \int_{-\infty+2 \pi i}^{\infty + 2 \pi i} \dfrac{e^x (x+ 2 \pi i)^3}{(e^x - 1)^2} dx
\nonumber \\
& = - \int_{-\infty}^\infty \dfrac{e^x (3 x^2 (2 \pi i) + 3x (2 \pi i)^2 + (2\pi i)^3)}{(e^x + 1)^2} dx
\nonumber \\
& = - 6 \pi i \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx + 12 \pi^2 \int_{-\infty}^\infty \dfrac{x e^x}{(e^x + 1)^2} dx
\nonumber \\
& i 8 \pi^3 \int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx
\end{align*}

The function ##\dfrac{e^z z^3}{(e^z + 1)^2}## has residue ##3 \pi^2## at ##z=i \pi## (we calculate this in a moment). So that

\begin{align*}
6 \pi^3 = - 6 \pi \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx + 8 \pi^3 \int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx
\end{align*}

Which rearranged is

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = - \pi^2 + \frac{4}{3} \pi^2 \int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx
\end{align*}

The integral on the RHS is straightforward to do:

\begin{align*}
\int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx & = \int_{-\infty}^\infty \left( - \frac{d}{dx} \dfrac{1}{e^x + 1} \right) dx
\nonumber \\
& =\left[ - \dfrac{1}{e^x + 1} \right]_{-\infty}^\infty = 1
\end{align*}

So,

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = \frac{\pi^2}{3}
\end{align*}

Plugging this into ##(***)##, we finally have

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} = \frac{\pi^2}{8} .
\end{align*}

Calculating the residue:

We now calculate

\begin{align*}
\text{Res}_{z = \pi i} \dfrac{e^z z^3}{(e^z + 1)^2}
\end{align*}

We have, where ##z_0## is the position of the pole ##\pi i##:

\begin{align*}
\frac{1}{(e^z + 1)^2} & = \frac{1}{(e^z - e^{z_0})^2}
\nonumber \\
& = \frac{e^{-2 z_0}}{(e^{z-z_0} - 1)^2}
\nonumber \\
& = \frac{e^{-2 z_0}}{[(z-z_0) + \frac{1}{2!} (z-z_0)^2 + \cdots]^2}
\nonumber \\
& = \dfrac{e^{-2 z_0}}{(z-z_0)^2 [1 + \frac{1}{2!} (z-z_0) + \cdots]^2}
\nonumber \\
& = \dfrac{e^{-2z_0}}{(z-z_0)^2} [1 - (z-z_0) + \cdots]
\nonumber \\
& = \dfrac{e^{-2z_0}}{(z-z_0)^2} - \dfrac{e^{-2z_0}}{z-z_0} + \cdots
\end{align*}

We use this in extracting the residue of ##f(z)## at ##z_0##:

\begin{align*}
\frac{e^z z^3}{(e^z + 1)^2} & = \frac{e^z z^3}{(e^z - e^{z_0})^2}
\nonumber \\
& = e^{-2z_0} \frac{e^{z_0} e^{z-z_0} [z_0 +(z-z_0)]^3}{(z-z_0)^2} - e^{-2z_0} \dfrac{z_0^3 e^{z_0}}{z-z_0} + \cdots
\nonumber \\
& = e^{-z_0} \dfrac{[1 + (z-z_0) + \cdots] [z_0^3 + 3z_0^2 (z-z_0) + \cdots]}{(z-z_0)^2} - e^{-z_0} \dfrac{z_0^3}{z-z_0} + \cdots
\nonumber \\
& = e^{-z_0} \dfrac{z_0^3}{z-z_0} + e^{-z_0} \dfrac{3 z_0^2}{z-z_0} + \cdots
\end{align*}

So

\begin{align*}
\text{Res}_{z = \pi i} \dfrac{e^z z^3}{(e^z + 1)^2} & = 3 \pi^2 .
\end{align*}

 

[julian]

General sum, its integral representations, and evaluation of integral

The general sum, ##\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}}##, has the following essentially equivalent integral representations:

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}} & = \frac{1}{4 \cdot (2k)!} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^{x} - 1)^2} dx + \frac{1}{4 \cdot (2k)!} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^{x} + 1)^2} dx
\end{align*}

and

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}} = (1 - 2^{-2k}) \frac{1}{2 \cdot (2k)!} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx
\end{align*}

which I derive below. Comparing these you find

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x - 1)^2} dx = \frac{2^{2k-1}}{2^{2k-1}-1} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^{x} + 1)^2} dx
\end{align*}

This relation can also be proven directly. Substituting this into the second integral representation you get another integral representation

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}} = \frac{2^{2k}-1}{2^{2k}-2} \frac{1}{2 \cdot (2k)!} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} dx
\end{align*}

Below we will evaluate this integral and obtain the expected result:

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}} = (1-2^{-2k}) \times (-1)^{k+1} (2 \pi)^{2k} \frac{B_{2k}}{2 (2k)!} .
\end{align*}

where ##B_{2k}## are the Bernoulli numbers.

Deriving integral representations of sum

I now derive these integral representations. We have

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

where we have integrated by parts and then extended the range of integration. This is first integral representation the sum quoted at the beginning.

Recall the identity

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

We have

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

Substituting this into the identity just mentioned we get the second integral representation of the sum quoted above.

As noted at the beginning, another representation follows from the first two:

\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^{2k}} = \frac{2^{2k}-1}{2^{2k}-2} \frac{1}{2 \cdot (2k)!} \int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} dx .
\end{align*}

We next perform the integral on the RHS.

Performing the integral

Consider the integral:

\begin{align*}
\int_{-\infty}^\infty \dfrac{e^{\alpha x} e^x}{(e^{x} + 1)^2} dx
\end{align*}

where ##-\frac{1}{2} \leq \alpha \leq \frac{1}{2}##. Then

\begin{align*}
\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} dx = \left. \dfrac{\partial^{2k}}{\partial \alpha^{2k}} \int_{-\infty}^\infty \dfrac{e^{\alpha x} e^x}{(e^{x} + 1)^2} dx \right|_{\alpha=0}
\end{align*}

We now evaluate this integral using complex analysis. Consider the rectangular contour, ##C##, in the figure

and the integral

\begin{align*}
\oint_C \dfrac{e^{\alpha z} e^z}{(e^z + 1)^2} dz
\end{align*}

whose integrand has a pole at ##\pi i##. The integral along the vertical edges vanishes as:

\begin{align*}
f(z) = \dfrac{e^{\alpha (x+iy)} e^{(x+iy)}}{(e^{x+iy} + 1)^2} =
\begin{cases}
e^{- (1 - \alpha) (x+iy)} & x \rightarrow \infty \\
e^{(1 + \alpha) (x+iy)} & x \rightarrow - \infty \\
\end{cases}
\end{align*}

So that

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

Which rearranged is

\begin{align*}
\int_{-\infty}^\infty \dfrac{e^{\alpha x} e^x}{(e^x + 1)^2} dx & = \frac{2 \pi i}{1 - e^{i 2 \alpha \pi}} \frac{1}{2 \pi i} \oint_C \dfrac{e^{\alpha z} e^z}{(e^z + 1)^2} dz
\nonumber \\
& = \frac{2 \pi i}{1 - e^{i 2 \alpha \pi}} Res_{z=\pi i} [f(z)] .
\end{align*}

We calculate the residue. Expand about pole, ##z_0 = i \pi##:

\begin{align*}
\frac{1}{(e^z + 1)^2} & = \frac{1}{(e^{z_0 + (z-z_0)} - e^{z_0})^2}
\nonumber \\
& = e^{-2 z_0} \dfrac{1}{[(z-z_0) + \frac{1}{2!} (z-z_0)^2+ \cdots]^2}
\nonumber \\
& = e^{-2 z_0} \dfrac{1}{(z-z_0)^2 [(1 + \frac{1}{2!} (z-z_0)+ \cdots]^2}
\nonumber \\
& = e^{-2 z_0} \dfrac{1}{(z-z_0)^2 [1 + (z-z_0) + \cdots]}
\nonumber \\
& = e^{-2 z_0} \dfrac{1}{(z-z_0)^2} [(1 - (z-z_0) + \cdots]
\nonumber \\
& = e^{-2 z_0} \dfrac{1}{(z-z_0)^2} - e^{-2 z_0} \dfrac{1}{z-z_0} + \cdots
\end{align*}

Using this we can find the residue:

\begin{align*}
\frac{e^{\alpha z} e^z}{(e^z + 1)^2} & = e^{- 2 z_0} \dfrac{ e^{(\alpha + 1) z_0 + (\alpha + 1) (z-z_0)} }{ (z-z_0)^2 } - e^{- 2 z_0} \dfrac{ e^{(\alpha + 1) z_0} }{ z-z_0 } + \cdots
\nonumber \\
& = e^{- 2 z_0} \dfrac{ e^{(\alpha + 1) z_0} [1 + (\alpha + 1) (z-z_0) + \cdots] }{ (z-z_0)^2 } - e^{- 2 z_0} \dfrac{ e^{(\alpha + 1) z_0} }{ z-z_0 } + \cdots
\nonumber \\
& = e^{- 2 z_0} \dfrac{ e^{(\alpha + 1) z_0} }{ (z-z_0)^2 } + e^{- 2 z_0} \dfrac{ \alpha e^{(\alpha + 1) z_0} }{ z-z_0 } + \cdots
\end{align*}

So that

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

Then

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

We have the series expansion of ##x \pi \csc x \pi## in terms of Bernoulli numbers:

\begin{align*}
\frac{ x \pi }{ \sin( x \pi) } = \sum_{n=0}^\infty \dfrac{ 2 (2^{2n-1} - 1) (-1)^{n+1} \pi^{2n} B_{2n} }{ (2n)! } x^{2n} .
\end{align*}

Employing this we obtain

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