How we can solve this?$\newcommand{\sech}{\operatorname{sech}}$
$$
\int \sech^4 x \, dx.
$$
I know we can solve the simple case
$$
\int \sech \, dx=\int\frac{dx}{\cosh x}=\int\frac{dx\cosh x}{\cosh ^2x}=\int\frac{d(\sinh x)}{1+\sinh^2 x}=\int \frac{du}{1+u^2}=\tan^{-1}\sinh x+C.
$$
I am stuck with the $\sech^4$ though. Thank you
[Math] Integral $\int \operatorname{sech}^4 x \, dx$
calculusindefinite-integralsintegration
Related Solutions
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty} \bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x = {7 \over \pi^{2}}\,\zeta\pars{3}} \approx 0.8526\ \Large ?}$.
\begin{align} &\color{#f00}{\int_{0}^{\infty} \bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x} \\[5mm] = &\ \int_{0}^{\infty}{\tanh\pars{x} - x \over x^{3}}\,\dd x + \int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x \\[5mm] = & -\,\half\int_{x\ =\ 0}^{x\ \to\ \infty}\bracks{\tanh\pars{x} - x} \,\dd\pars{1 \over x^{2}} + \int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x \\[5mm] = &\ \half\int_{0}^{\infty}{\mrm{sech}^{2}\pars{x} - 1 \over x^{2}}\,\dd x + \int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x = \half\int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x \\[5mm] = &\ 32\sum_{k = 0}^{\infty}\,\sum_{n = 0}^{\infty}\,\,\ \underbrace{% \int_{0}^{\infty}{1 \over \bracks{\pars{2k + 1}\pi}^{\, 2} + 4x^{2}}\, {1 \over \bracks{\pars{2n + 1}\pi}^{\, 2} + 4x^{2}}\,\dd x} _{\ds{1 \over 8\pi^{2}\pars{2k + 1}\pars{2n + 1}\pars{k + n + 1}}} \label{1}\tag{1} \\[5mm] = &\ {4 \over \pi^{2}}\ \underbrace{\sum_{k = 0}^{\infty}{H_{k} + 2\ln\pars{2} \over \pars{2k + 1}^{2}}} _{\ds{{7 \over 4}\,\zeta\pars{3}}}\label{2}\tag{2} = \color{#f00}{{7 \over \pi^{2}}\,\zeta\pars{3}} \end{align} Note that
- In \eqref{1}, we use the identity $\ds{{\tanh\pars{x} \over x} = 8\sum_{j = 0}^{\infty}{1 \over \bracks{\pars{2j + 1}\pi}^{\, 2} + 4x^{2}}}$
- The sum over $\ds{n}$, in \eqref{1}, yields a Digamma Function term $\ds{\Psi\pars{1 + k}}$ which explains the appearance of the Harmonic Number $\ds{H_{k} = \Psi\pars{1 + k} + \gamma}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant.
- $\ds{\sum_{k = 0}^{\infty}{H_{k} \over \pars{2k + 1}^{2}} = {1 \over 4}\bracks{7\zeta\pars{3} - \pi^{2}\ln\pars{2}}}$ is a well known result.
- $\ds{\sum_{k = 0}^{\infty}{1 \over \pars{2k + 1}^{2}} = {1 \over 8}\,\pi^{2}}$.
Let $\operatorname{sech}(x) = t$ and integrate by parts to obtain \begin{align} -I &= \int \limits_0^\infty \frac{x \log(\cosh(x))}{\cosh^3(x)} \, \mathrm{d}x = \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) t^2}{\sqrt{1-t^2}} \, \mathrm{d} t \\ &= \int \limits_0^1 \sqrt{1-t^2} \frac{\mathrm{d}}{\mathrm{d} t} \left[-\log(t) t \operatorname{arsech}(t)\right] \mathrm{d} t \\ &= \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) (1-t^2)}{\sqrt{1-t^2}} \, \mathrm{d} t - \int \limits_0^1 \left[\sqrt{1-t^2} \operatorname{arsech}(t) - \log(t)\right] \mathrm{d} t \, . \end{align} Averaging the second and the fourth expression yields $$- I = \frac{1}{2} \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t)}{\sqrt{1-t^2}} \, \mathrm{d} t - \frac{1}{2}\int \limits_0^1 \sqrt{1-t^2} \operatorname{arsech}(t)\, \mathrm{d} t - \frac{1}{2} \equiv J - K - \frac{1}{2}\, .$$ $K$ can be computed by reversing the previous substitution: $$ K = \frac{1}{2} \int \limits_0^\infty \frac{x \sinh^2(x)}{\cosh^3(x)} \, \mathrm{d} x = \frac{1}{4} \int\limits_0^\infty \frac{\sinh(x) + x \cosh(x)}{\cosh^2(x)} \, \mathrm{d} x = \frac{1}{4}(1+2 \mathrm{G}) = \frac{1}{4} + \frac{\mathrm{G}}{2} \, . $$ For $J$ we can use $t = \frac{2u}{1+u^2}$ to find \begin{align} J &= \int \limits_0^1 \frac{\log(u) \log\left(\frac{2u}{1+u^2}\right)}{1+u^2} \, \mathrm{d} u \\ &= \int \limits_0^1 \frac{-\log(u) \log(1+u^2)}{1+u^2} \, \mathrm{d} u + \int \limits_0^1 \frac{\log^2(u)}{1+u^2} \, \mathrm{d} u - \log(2) \int \limits_0^1 \frac{-\log(u)}{1+u^2} \, \mathrm{d} u \\ &= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) + \mathrm{G} \log(2) - \frac{\pi}{8} \log^2(2) - \frac{3 \pi^3}{32} + \frac{\pi^3}{16} - \mathrm{G} \log(2) \\ &= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} \, . \end{align} The first integral has been calculated here and the others are well-known special values of the Dirichlet beta function. Therefore, $$ -I = J - K - \frac{1}{2} = 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} - \frac{\mathrm{G}}{2} - \frac{3}{4} \, . $$
The problem with the original approach is that for $b=3$ the series only converges for $a > 1$, which leads to the two divergent series in the final answer. This can be avoided by computing the result for sufficiently large values of $a$ first and then taking the limit $a \to 1^+$, which can be justified by analytic continuation. $\Omega_3, \dots, \Omega_8$ are calculated as before after taking the limit inside the series, but the divergent terms are replaced by the regularised versions $$ \Omega_1 = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} = \lim_{a \to 1^+} [1 - \beta(a-1)] = 1 - \beta (0) = \frac{1}{2}$$ and \begin{align} \Omega_2 &= \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1} H_n}{(2n+1)^{a-1}} = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} \int \limits_0^1 \frac{1 - x^n}{1-x} \, \mathrm{d} x \\ &= \lim_{a \to 1^+} \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{a-1}(\sqrt{x})}{\sqrt{x}} - \beta(a-1)}{1-x} \, \mathrm{d} x = \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{0}(\sqrt{x})}{\sqrt{x}} - \beta(0)}{1-x} \, \mathrm{d} x \\ &= \int \limits_0^1 \frac{\frac{1}{1+x} - \frac{1}{2}}{1-x} \, \mathrm{d} x = \frac{1}{2}\int \limits_0^1 \frac{\mathrm{d} x}{1+x} = \frac{1}{2} \log(2) \, . \end{align}
Best Answer
Note that $$ \int \DeclareMathOperator{sech}{sech}{\sech}^4x\,dx=\int{\sech}^2{x}\cdot(1-\tanh^2x)\,dx $$ Letting $u=\tanh x$ gives $du={\sech}^2x$ so $$ \int{\sech}^4x\,dx=\int(1-u^2)\,du=u-\frac{u^3}{3}+C=\tanh x-\frac{1}{3}\tanh^3x+C $$