Bukti alternatif komputasi $\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}$
Dalam solusi ini kami menunjukkan itu
$$\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag1$$
menggunakan identitas
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
Pertanyaan saya di sini adalah dapatkah kita membuktikannya $(1)$dengan cara yang berbeda menggunakan metode dasar? Namun, jangan biarkan batasan pertanyaan saya membatasi pendekatan Anda, semua pendekatan dihargai.
Inti dari posting ini (tantangan) adalah mempelajari teknik yang berbeda jika memungkinkan dan membuat situs ini lebih menghibur.
Terima kasih.
Jawaban
Sketsa (untuk saat ini). Menggunakan identitas$$\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}\,{\rm d}x=\frac{4^n}{n\binom{2n}n}\,\,\,\text{and}\,\,\,\sum_{n\ge1}\frac{H_n}nx^n=\operatorname{Li}_2(x)+\frac12\log^2(1-x)$$ memberi $$\sum_{n\ge1}\frac{4^nH_n}{n^2\binom{2n}n}=\sum_{n\ge1}\frac{H_n}n\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}\,{\rm d}x=\frac12\int_0^1\frac{2\operatorname{Li}_2(x)+\log^2(1-x)}{x\sqrt{1-x}}\,{\rm d}x$$ Bercermin dan menegakkan $\sqrt x\mapsto x$ kemudian menghasilkan \begin{align*} \frac12\int_0^1\frac{2\operatorname{Li}_2(x)+\log^2(1-x)}{x\sqrt{1-x}}\,{\rm d}x&=\frac12\int_0^1\frac{2\operatorname{Li}_2(1-x)+\log^2(x)}{(1-x)\sqrt{x}}\,{\rm d}x\\ &=\int_0^1\frac{2\operatorname{Li}_2(1-x^2)+4\log^2(x)}{1-x^2}\,{\rm d}x \end{align*} Integral terakhir mengevaluasi sebagai $7\zeta(3)$menggunakan deret geometris. Untuk integral pertama, terapkan IBP dua kali untuk mendapatkan\begin{align*} \int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\,{\rm d}x&=-\left[\frac12\operatorname{Li}_2(1-x^2)\log\left(\frac{1-x}{1+x}\right)\right]_0^1+2\int_0^1\frac{x\log x\log\left(\frac{1-x}{1+x}\right)}{1-x^2}\,{\rm d}x\\ &=-\left[\frac12 x\log x\log^2\left(\frac{1-x}{1+x}\right)\right]_0^1+\int_0^1(1+\log x)\log^2\left(\frac{1-x}{1+x}\right)\,{\rm d}x\\ &=\frac{\pi^2}6+\int_0^1\log x\log^2\left(\frac{1-x}{1+x}\right)\,{\rm d}x \end{align*} Saat ini saya tidak yakin bagaimana mendekati integral yang tersisa dengan cara yang elegan.
Catatan samping: Menggunakan perluasan deret biasa dari logaritma dan representasi integral dari bilangan harmonik mengarah ke evaluasi integral yang tersisa. Namun, metode ini agak tidak elegan dan saya akan melihat apakah saya dapat menemukan sesuatu yang lebih memuaskan.
Kami menggunakan bentuk kuat dari fungsi Beta yang disajikan dalam buku, (Almost) Impossible Integrals, Sums, and Series ,$\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} dx = \operatorname{B}(a,b)$, (lihat halaman $72$-$73$).
Set $a=b=n$ kita punya
$$\int_0^1\frac{2x^{n-1}}{(1+x)^{2n}}dx=\frac{\Gamma^2(n)}{\Gamma(2n)}=\frac{2}{n{2n\choose n}}$$
Begitu $$\frac{1}{n{2n\choose n}}=\int_0^1\frac{x^{n-1}}{(1+x)^{2n}}dx=\int_0^1\frac1x\left(\frac{x}{(1+x)^2}\right)^ndx$$
$$\Longrightarrow \sum_{n=1}^\infty\frac{4^nH_n}{n^2{2n\choose n}}=\int_0^1\frac1x\left(\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{4x}{(1+x)^2}\right)^n\right)dx$$
$$=\int_0^1\frac1x\left(\text{Li}_2\left(\frac{4x}{(1+x)^2}\right)+\frac12\ln\left(1-\frac{4x}{(1+x)^2}\right)\right)dx$$
$$\overset{IBP}{=}\int_0^1\frac{2+2x}{x(1-x)}\ln x\ln\left(\frac{1-x}{1+x}\right)dx$$
$$=\int_0^1\left(\frac2x+\frac{4}{1-x}\right)\ln x\ln\left(\frac{1-x}{1+x}\right)dx$$
$$\small{=2\int_0^1\frac{\ln x\ln(1-x)}{x}dx+4\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{1-x}dx}_{1-x\to x}-2\int_0^1\frac{\ln x\ln(1+x)}{x}dx-4\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx}$$
$$=6\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x}dx}_{\zeta(3)}-2\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{x}dx}_{-\frac34\zeta(3)}-4\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx}_{\zeta(3)-\frac32\ln(2)\zeta(2)}$$
$$=6\ln(2)\zeta(2)+\frac72\zeta(3)$$
Integral terakhir dihitung di sini .
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{4^{n}H_{n} \over n^{2}{2n \choose n}}} = \int_{0}^{4}\sum_{n = 1}^{\infty}{H_{n} \over n{2n \choose n}} \,x^{n - 1}\,\dd x \\[5mm] = &\ \int_{0}^{4}\sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\,{\Gamma\pars{n} \Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}\,\dd x \\[5mm] = &\ \int_{0}^{4}\sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\int_{0}^{1}t^{n - 1} \pars{1 - t}^{n}\,\dd t\,\dd x \\[5mm] = &\ \int_{0}^{4}\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n} \bracks{xt\pars{1 - t}}^{\, n}\,{\dd t\,\dd x \over tx} \\[5mm] = &\ \int_{0}^{4}\int_{0}^{1}\braces{% -\,{\ln\pars{1 - xt\bracks{1-t}} \over 1 - xt\pars{1-t}}} {\dd t\,\dd x \over tx} \\[5mm] = &\ \int_{0}^{1}{2\ln^{2}\pars{\verts{1 - 2t}} + \mrm{Li}_{2}\pars{4\bracks{1 - t})\, t}\over t}\,\dd t \\[5mm] = &\ 2\int_{-1/2}^{1/2}{2\ln^{2}\pars{\verts{2t}} + \mrm{Li}_{2}\pars{1 - 4t^{2}}\over 1 + 2t}\,\dd t \\[5mm] = &\ 4\int_{0}^{1/2}{2\ln^{2}\pars{2t} + \mrm{Li}_{2}\pars{1 - 4t^{2}} \over 1 - 4t^{2}}\,\dd t \\[5mm] = &\ 2\int_{0}^{1}{2\ln^{2}\pars{t} + \mrm{Li}_{2}\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t \\[5mm] = &\ 4\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{t} \over 1 - t^{2}}\,\dd t} _{\ds{\color{red}{\LARGE\S}:\ {7 \over 4}\,\zeta\pars{3}}}\ +\ 2\, \underbrace{\int_{0}^{1}{\mrm{Li}_{2}\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t}_{\ds{\color{red}{\LARGE *}:\ {1 \over 2}\,\pi^{2}\ln\pars{2} - {7 \over 4}\,\zeta\pars{3}}} \\[5mm] = &\ \bbx{6\ln\pars{2}\,\zeta\pars{2} + {7 \over 2}\,\zeta\pars{3}} \\ & \end{align}
$\left\{\begin{array}{lcl} \ds{\color{red}{\LARGE\S}} & \ds{:} & \mbox{First} "Partial\ Fraction\ Split.\ \mbox{Next, integrate}\ twice\ \mbox{by parts.} \\[2mm] \ds{\color{red}{\LARGE *}} & \ds{:} & \mbox{After integration by parts, the final expression seems to be a doable and} \\ && \mbox{known integral.} \end{array}\right.$