spiderifilms6e

2022-01-21

Prove that:

$\sum _{n=0}^{\mathrm{\infty}}\frac{{(-1)}^{n}}{{(2n+1)}^{2m+1}}=\frac{{(-1)}^{m}{E}_{2m}{\pi}^{2m+1}}{{4}^{m+1}\left(2m\right)!}$

nick1337

Expert2022-01-27Added 699 answers

Because it wasnt

star233

Skilled2022-01-27Added 352 answers

The Dirichlet beta function is defined as
$\beta (2m+1)=\sum _{n=0}^{\mathrm{\infty}}\frac{(-1{)}^{n}}{(2n+1{)}^{2m+1}}$
Then $\beta (1)=\frac{\pi}{4}$ and
$\beta (2m+1)=-\sum _{k=1}^{m}\frac{(-{\pi}^{2}/4{)}^{k}}{(2k)!}\beta (2m-2k+1)$ (1)
If we reindex recursion derived below, we get that the even Euler numbers are defined by ${E}_{0}=1$ and
${E}_{2m}=-\sum _{k=1}^{m}((2m),(2k)){E}_{2m-2k}$ (2)
then notice that (1) is the same as (2) if we set
$\beta (2m+1)=\frac{(-1{)}^{m}{E}_{2m}{\pi}^{2m+1}}{{4}^{m+1}(2m)!}$

RizerMix

Expert2022-01-27Added 583 answers

My proof works through the following lines: the LHS is:
$\frac{1}{(2m)!}{\int}_{0}^{1}\frac{)\mathrm{log}x{)}^{2m}}{1+{x}^{2}}dx=\frac{1}{2\cdot (2m)!}{\int}_{0}^{+\mathrm{\infty}}\frac{(\mathrm{log}x{)}^{2m}}{1+{x}^{2}}dx$
so we just need to compute:
$\frac{{d}^{2m}}{d{k}^{2m}}{\int}_{0}^{+\mathrm{\infty}}\frac{{x}^{k}}{1+{x}^{2}}{|}_{k=0}$
but:
${\int}_{0}^{+\mathrm{\infty}}\frac{{x}^{1/r}}{1+{x}^{2}}dx=r{\int}_{0}^{+\mathrm{\infty}}\frac{{y}^{r}}{1+{y}^{2r}}dy$
$=\frac{\pi /2}{\mathrm{cos}(\pi /(2r))}$
by the residue theorem, so
$\sum _{n=0}^{+\mathrm{\infty}}\frac{(-1{)}^{n}}{(2n+1{)}^{2m+1}}=\frac{{E}_{2m}}{2\cdot (2m)!}(\frac{\pi}{2}{)}^{2m+1}$
where ${E}_{2m}$ is just the absolute value of an Euler number, that belongs to $\mathbb{N}$ .

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