Có một biểu mẫu đóng cho $\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}?$
tôi đã tìm thấy
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx=I\tag1.$$
Mathematica không tìm thấy$I$, vì vậy tôi không chắc liệu có biểu mẫu đóng cho nó hay không. Tôi chỉ đang thử ở đây.
Ý tưởng đầu tiên nảy ra trong đầu tôi là sử dụng chuỗi Fourier của $-\ln(\cos x)=\ln(2)+\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}$ và chúng ta có
$$I=8\ln(2)\underbrace{\int_0^{\pi/2}x^2\cot x\ dx}_{\frac32\ln(2)\zeta(2)-\frac78\zeta(3)}+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x^2 \cot x\cos(2nx)\ dx.$$
Tôi bị mắc kẹt ở đây. Bất kì sự trợ giúp nào đều được đánh giá cao.
Chứng minh rằng $(1)$
từ đây chúng tôi có
$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
thay thế $x$ bởi $\sqrt{x}$ chúng tôi nhận được $$\sum_{n=1}^\infty\frac{2^{2n}x^n}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{x})$$
nhân cả hai bên với $-\frac{\ln(1-x)}{x}$ sau đó $\int_0^1$ Và sử dụng $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ chúng tôi nhận được
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=2\int_0^1\frac{\arcsin^2(\sqrt{x})\ln(1-x)}{x}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx$$
Trả lời
$$S=-8 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^4}{90}-\frac{1}{3} \log ^4(2)+\frac{4}{3} \pi ^2 \log ^2(2)$$
Bằng chứng $1$. Cái này . Bằng chứng$2$. Cái này . Bằng chứng$3$. Cái này . Tặng kem:$$\small \int_0^{\frac{\pi }{2}} x^3 \cot (x) \log (\cos (x)) \, dx=\frac{3}{2} \pi \text{Li}_4\left(\frac{1}{2}\right)+\frac{9}{16} \pi \zeta (3) \log (2)-\frac{\pi ^5}{120}+\frac{1}{16} \pi \log ^4(2)-\frac{1}{8} \pi ^3 \log ^2(2)$$
Cảm ơn @ user97357329 vì gợi ý của anh ấy trong các nhận xét.
Trong cuốn sách Các phép tích phân gần như bất khả thi, Tổng và Chuỗi , trang$247$ Eq $(3.288)$ chúng ta có
$$\cot x\ln(\cos x)=\sum_{n=1}^\infty(-1)^n\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac1n\right)\sin(2nx)$$
$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\sin(2nx),\quad 0<x<\frac{\pi}{2}$$
Vì vậy,
$$\int_0^{\pi/2}x^2\cot x\ln(\cos x)dx=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\int_0^{\pi/2}x^2\sin(2nx)dx\right)$$
$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\frac{\cos(n\pi)}{4n^3}-\frac{3\zeta(2)\cos(n\pi)}{4n}-\frac{1}{4n^3}\right)$$
$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\frac{(-1)^n}{4n^3}-\frac{3\zeta(2)(-1)^n}{4n}-\frac{1}{4n^3}\right)$$
$$=\frac14\int_0^1\frac{1-t}{t(1+t)}\left(\sum_{n=1}^\infty\frac{t^n}{n^3}-\frac{3\zeta(2)t^n}{n}-\frac{(-t)^n}{n^3}\right)dt$$
$$=\frac14\int_0^1\left(\frac1t-\frac2{1+t}\right)\left(\text{Li}_3(t)+3\zeta(2)\ln(1-t)-\text{Li}_3(-t)\right)dt$$
$$=\frac14\underbrace{\int_0^1\frac{\text{Li}_3(t)-\text{Li}_3(-t)}{t}dt}_{\mathcal{I}_1}-\frac12\underbrace{\int_0^1\frac{\text{Li}_3(t)-\text{Li}_3(-t)}{1+t}dt}_{\mathcal{I}_2}$$ $$+\frac34\zeta(2)\underbrace{\int_0^1\frac{\ln(1-t)}{t}dt}_{\mathcal{I}_3}-\frac32\zeta(2)\underbrace{\int_0^1\frac{\ln(1-t)}{1+t}dt}_{\mathcal{I}_4}$$
$$\mathcal{I}_1=\text{Li}_4(1)-\text{Li}_4(-1)=\zeta(4)+\frac78\zeta(4)=\boxed{\frac{15}{8}\zeta(4)}$$
Bằng cách tích hợp theo các bộ phận, chúng tôi có
$$\mathcal{I}_2=\frac74\ln(2)\zeta(3)-\int_0^1\frac{\ln(1+t)\text{Li}_2(t)}{t}dt+\int_0^1\frac{\ln(1+t)\text{Li}_2(-t)}{t}dt$$
$$=\frac74\ln(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1 t^{n-1}\text{Li}_2(t)dt-\frac12\text{Li}_2^2(-t)|_0^1$$
$$=\frac74\ln(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n} \left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)-\frac5{16}\zeta(4)$$ $$=\frac74\ln(2)\zeta(3)-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac5{16}\zeta(4)$$ thay thế
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$
chúng tôi nhận được
$$\mathcal{I}_2=\boxed{-2\operatorname{Li_4}\left(\frac12\right)-\frac{25}{16}\zeta(4)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42}$$
$$\mathcal{I}_3=-\text{Li}_2(1)=\boxed{-\zeta(2)}$$
$$\mathcal{I}_4=\int_0^1\frac{\ln(1-t)}{1+t}dt=\int_0^1\frac{\ln x}{2-x}dx=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln xdx$$ $$=-\sum_{n=1}^\infty\frac{1}{n^22^n}=-\text{Li}_2\left(\frac12\right)=\boxed{\frac12\ln^22-\frac12\zeta(2)}$$
Kết hợp tất cả các kết quả đóng hộp mà chúng tôi cuối cùng nhận được
$$\int_0^{\pi/2}x^2\cot x\ln(\cos x)dx=\text{Li}_4\left(\frac12\right)-\frac18\zeta(4)-\ln^2(2)\zeta(2)+\frac{1}{24}\ln^4(2)$$
cái nào cho chúng ta
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\pi/2}x^{2}\cot\pars{x}\cos\pars{2nx}\,\dd x} \\[5mm] & = \left. \Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\bracks{-\ic\ln\pars{z}}^{\, 2}\, \pars{{z^{2} + 1 \over z^{2} - 1}\,\ic}z^{2n}\,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left. -\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}z^{2n - 1}\ln^{2}\pars{z}\, {z^{2} + 1 \over z^{2} - 1}\,\dd z \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \Re\int_{1}^{0}\pars{-1}^{n + 1}\,\ic\,y^{2n - 1} \bracks{\ln\pars{y} + {\pi \over 2}\,\ic}^{\, 2}\, {-y^{2} + 1 \over -y^{2} - 1}\,\ic\,\dd y \\[2mm] &\ + \int_{0}^{1}x^{2n - 1}\ln^{2}\pars{x}\, {x^{2} + 1 \over x^{2} - 1}\,\dd x \\[5mm] = &\ \pars{-1}^{n}\int_{0}^{1}y^{2n - 1} \bracks{\ln^{2}\pars{y} - {\pi^{2} \over 4}} {1 - y^{2} \over 1 + y^{2}}\,\dd y \\[2mm] &\ - \int_{0}^{1}x^{2n - 1}\ln^{2}\pars{x}\, {1 + x^{2} \over 1 - x^{2}}\,\dd x \\[5mm] & = \pars{-1}^{n}\,\mathcal{I}''\pars{2n - 1} - \pars{-1}^{n}\,{\pi^{2} \over 4}\,\mathcal{I}\pars{2n - 1} - \mathcal{J}''\pars{2n - 1} \\ &\ \mbox{where}\quad \left\{\begin{array}{rcl} \ds{\mathcal{I}\pars{\nu}} & \ds{\equiv} & \ds{\int_{0}^{1}y^{\nu}\,{1 - y^{2} \over 1 + y^{2}}\,\dd y} \\[2mm] \ds{\mathcal{J}\pars{\nu}} & \ds{\equiv} & \ds{\int_{0}^{1}\pars{y^{\nu} - 1}\,{1 + y^{2} \over 1 - y^{2}}\,\dd y} \end{array}\right. \end{align}Hãy đánh giá một số tích phân chúng ta cần để đánh giá kết quả chính của chúng ta:
$\ds{\Large\mathcal{I}\pars{\nu}:\ ?.}$ \begin{align} \mathcal{I}\pars{\nu} & \equiv \int_{0}^{1}y^{\nu}\,{1 - y^{2} \over 1 + y^{2}}\,\dd y = \int_{0}^{1}{y^{\nu} - 2y^{\nu + 2} + y^{\nu + 4} \over 1 - y^{4}}\,\dd y \\[5mm] & = {1 \over 4}\int_{0}^{1}{y^{\nu/4 - 3/4} - 2y^{\nu/4 - 1/4} + y^{\nu/4 + 1/4} \over 1 - y}\,\dd y \\[5mm] & = {1 \over 4}\bracks{% 2\int_{0}^{1}{1 - y^{\nu/4 - 1/4} \over 1 - y}\,\dd y - \int_{0}^{1}{1 - y^{\nu/4 - 3/4} \over 1 - y}\,\dd y - \int_{0}^{1}{1 - y^{\nu/4 + 1/4} \over 1 - y}\,\dd y} \\[5mm] & = {1 \over 4}\bracks{% 2\Psi\pars{{\nu \over 4} + {3 \over 4}} - \Psi\pars{{\nu \over 4} + {1 \over 4}} - \Psi\pars{{\nu \over 4} + {5 \over 4}}} \\[5mm] & = -\,{1 \over \nu + 1} - {1 \over 2}\,\Psi\pars{\nu + 1 \over 4} + {1 \over 2}\,\Psi\pars{\nu + 3 \over 4} \\[1cm] &\ \left\{\begin{array}{rcl} \ds{\mathcal{I}\pars{2n - 1}} & \ds{=} & \ds{-\,{1 \over 2n} - {1 \over 2}\,\Psi\pars{n \over 2} + {1 \over 2}\,\Psi\pars{n + 1 \over 2}} \\[2mm] \ds{\mathcal{I}''\pars{2n - 1}} & \ds{=} & \ds{-\,{1 \over 4n^{3}} - {1 \over 32}\,\Psi\, ''\pars{n \over 2} + {1 \over 32}\,\Psi\, ''\pars{n + 1 \over 2}} \end{array}\right. \end{align}
$\ds{\Large\mathcal{J}\pars{\nu}:\ ?.}$ \begin{align} \mathcal{J}\pars{\nu} & \equiv \int_{0}^{1}\pars{y^{\nu} - 1}\,{1 + y^{2} \over 1 - y^{2}}\,\dd y = {1 \over 2}\int_{0}^{1}{-y^{-1/2} - y^{1/2} + y^{\nu/2 - 1/2} + y^{\nu/2 + 1/2} \over 1 - y}\,\dd y \\[5mm] & = {1 \over 2}\bracks{\Psi\pars{1 \over 2} + \Psi\pars{3 \over 2} - \Psi\pars{{\nu \over 2} + {1 \over 2}} - \Psi\pars{{\nu \over 2} + {3 \over 2}}} \\[5mm] & = 1 - \gamma - 2\ln\pars{2} - {1 \over 2}\,\Psi\pars{\nu + 1 \over 2} - {1 \over 2}\,\Psi\pars{\nu + 3 \over 2} \\[1cm] \mathcal{J}\, ''\pars{2n - 1} & = -\,{1 \over 8}\,\Psi\,''\pars{n} -\,{1 \over 8}\,\Psi\,''\pars{n + 1} = -\,{1 \over 4}\,\Psi\,''\pars{n} - {1 \over 4n^{3}} \end{align}