Integración incorrecta de $\log x \operatorname{sech} x$

Question

¿Cómo demostrar lo siguiente? $$ \int_0^\infty \log x \operatorname{sech}x\,dx = \frac{\pi}{2} \log\left( \frac{4\pi^3}{\Gamma(1/4)^4} \right) $$ Obtuve el lado derecho con CAS. Parece que esta función tiene muchos polos en el eje imaginario, por lo que no se puede utilizar una integral de contorno simple. También he probado lo siguiente $$\begin{align*} \int_0^\infty \log x \operatorname{sech} x\, dx &= \int_0^\infty \operatorname{sech} x \left. \frac{\partial}{\partial s} x^s \right|_{s=0} dx \\ &= \left. \frac{\partial}{\partial s} \int_0^\infty x^s \operatorname{sech} x\,dx \right|_{s=0} \end{align*}$$ Sin embargo, la última integral es muy difícil de evaluar y contiene términos de $\zeta$ funciones.

Answer 1

Sugerencia . Se puede escribir $$\begin{align} \int_0^\infty \log x \operatorname{sech}x\,dx &= 2\int_{0}^{\infty} \frac{\partial_s \left.\left(x^s\right)\right|_{s=0} }{1 + e^{-2x}} \:e^{-x}\: dx \\\\ &=2\: \left.\partial_s \left(\int_{0}^{\infty} \frac{x^s e^{-x}}{1 + e^{-2x}} \; dx \right)\right|_{s=0}\\\\ &= 2\:\left.\partial_s \left(\int_{0}^{\infty} \sum_{n=0}^{\infty} (-1)^n x^s e^{-(2n+1)x} \; dx \right)\right|_{s=0}\\\\ &=2\: \left.\partial_s \left(\sum_{n=0}^{\infty} (-1)^n \, \frac{\Gamma(s+1)}{(2n+1)^{s+1}} \right)\right|_{s=0}\\\\ &=2\: \left.\partial_s \left( \Gamma(s+1)\beta(s+1)\right)\right|_{s=0}\\\\ &= -2\:\gamma \cdot\beta(1)+2\:\beta'(1) \end{align}$$ donde $\gamma $ es la constante de Euler-Mascheroni, $\beta(\cdot)$ es el Función beta de Dirichlet $$ \beta(s)=\sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)^{s}} $$ se pueden utilizar los valores especiales $$ \beta(1)=\frac{\pi}4,\qquad \beta'(1) = \frac{\pi}{4} \left[ \gamma + 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$ probado aquí para obtener

$$\int_0^\infty \frac{\log x}{\cosh x}\,dx = \frac{\pi}{2} \:\log\left( \frac{4\pi^3}{\Gamma(1/4)^4} \right) $$

como se anunció.

Answer 2

Esta prueba directa fue descubierta por Abdulhafeez Ayinde Abdulsalam (véase https://arxiv.org/abs/2203.02675 ).

Sea $\displaystyle \Delta\left(a\right) = \int_0^{\infty} \frac{\log\left(x^2 + a^2\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x$ .

Entonces \begin{align*} \Delta\left(a\right) &= \int_0^{\infty} \frac{\log\left(\left|a\right| - ix\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x + \int_0^{\infty} \frac{\log\left(\left|a\right| + ix\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x \end{align*} donde $i=\sqrt{-1}$ . \begin{align*} \Delta\left(a\right) &= 2\int_0^{\infty} \frac{\log\left(\left|a\right| - ix\right)}{e^{-2 \pi x} + 1} e^{-\pi x} \, \mathrm{d}x + 2\int_0^{\infty} \frac{\log\left(\left|a\right| + ix\right)}{e^{-2 \pi x} + 1} e^{-\pi x} \, \mathrm{d}x \\&= -\frac{2}{\pi}\int_0^{\infty} \log\left(\left|a\right| - ix\right) \,\, \mathrm{d}\left(\arctan\left(e^{-\pi x}\right)\right) - \frac{2}{\pi}\int_0^{\infty} \log\left(\left|a\right| + ix\right) \,\, \mathrm{d}\left(\arctan\left(e^{-\pi x}\right)\right) \\&=\frac{-2i}{\pi}\int_0^{\infty} \frac{\arctan\left(e^{-\pi x}\right)}{\left|a\right| - ix}\, \mathrm{d}x + \frac{2i}{\pi}\int_0^{\infty} \frac{\arctan\left(e^{-\pi x}\right)}{\left|a\right| + ix} \, \mathrm{d}x + \ln{a} \end{align*} Por lo tanto \begin{align*} \Delta\left(a\right) - \ln{a} &= \frac{-2i}{\pi}\int_0^{\infty} \arctan\left(e^{-\pi x}\right) \int_0^{\infty} e^{-t\left(\left|a\right| - ix\right)} \,\, \mathrm{d}t \, \mathrm{d}x \\&\quad+ \frac{2i}{\pi}\int_0^{\infty} \arctan\left(e^{-\pi x}\right) \int_0^{\infty} e^{-t\left(\left|a\right| + ix\right)} \,\, \mathrm{d}t \, \mathrm{d}x \\&= \frac{-2i}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} e^{itx} \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t \\&\quad+\frac{2i}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} e^{-itx} \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t \\&= \frac{-2i}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} \left(e^{itx} - e^{-itx}\right) \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t \end{align*} Por la fórmula de Euler, $$e^{itx} - e^{-itx} = 2i\sin{\left(tx\right)}.$$ Por lo tanto \begin{align*} \Delta\left(a\right) - \ln{a} &= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} \sin{\left(tx\right)} \arctan\left(e^{-\pi x}\right)\mathrm{d}x \,\, \mathrm{d}t \\&= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\int_0^{\infty} \mathrm{d}\left(\frac{-\cos{\left(tx\right)}}{t}\right) \arctan\left(e^{-\pi x}\right) \,\, \mathrm{d}t \\&= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\left( \frac{\pi}{4t} - \frac{\pi}{2t}\int_0^{\infty}\frac{\cos{\left(tx\right)}}{\cosh\left(\pi x\right)} \, \mathrm{d}x\right) \,\, \mathrm{d}t \\&= \frac{4}{\pi}\int_0^{\infty}e^{-\left|a\right|t}\left( \frac{\pi}{4t} - \frac{1}{2t}\int_0^{\infty}\frac{\cos{\left(\frac{tx}{\pi}\right)}}{\cosh\left(x\right)}\, \mathrm{d}x\right) \,\, \mathrm{d}t \\&= \frac{4}{\pi}\int_0^{\infty} e^{-\left|a\right|t} \left(\frac{\pi}{4t} - \frac{\pi}{4t}\mathrm{sech}\left(\frac{t}{2}\right)\right) \,\, \mathrm{d}t \\&= \int_0^{\infty} e^{-\left|a\right|t}\left(\frac{1}{t} - \frac{1}{t}\mathrm{sech}\left(\frac{t}{2}\right)\right) \,\, \mathrm{d}t \\&= \int_0^{\infty} \left(\frac{e^{-\left|a\right|t}}{t} - \frac{2e^{-\left(\left|a\right| + \frac{1}{2}\right)t}}{t\left(1 + e^{-t}\right)} \right)\,\, \mathrm{d}t {\stackrel{\,\,t \rightarrow 2t}{=}} \int_0^{\infty} \left(\frac{e^{-2\left|a\right|t}}{t} - \frac{2e^{-\left(2\left|a\right| + 1\right)t}}{t\left(1 + e^{-2t}\right)} \right)\,\, \mathrm{d}t \\&{\stackrel{z \rightarrow e^{-t}}{=}} -\int_0^{1} \left(z^{2\left|a\right|} - \frac{2z^{2\left|a\right|+1}}{1 + z^2} \right)\,\, \frac{\mathrm{d}z}{z\ln{z}} = -\int_0^{1} \frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{1}{z} - \frac{2}{1 + z^2}\right)\,\, \mathrm{d}z \\&= -\int_0^{1} \frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{1 + z^2 - 2z}{z\left(1 + z^2\right)}\right)\,\, \mathrm{d}z = -\int_0^{1} \frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{\left(1 - z\right)^2}{z\left(1 + z^2\right)}\right)\,\, \mathrm{d}z \\&= \int_0^{1} \frac{z^{2\left|a\right| - 1}\left(1 - z\right)}{1 + z^2}\int_0^1 z^p \mathrm{d}p\,\, \mathrm{d}z = \int_0^{1}\int_0^1 \frac{z^{2\left|a\right| + p - 1}\left(1 - z\right)}{1 + z^2} \mathrm{d}p\,\, \mathrm{d}z \\&= \int_0^{1}\int_0^1 \sum_{k=0}^{\infty} \left(-1\right)^k z^{2\left|a\right| + p + 2k - 1}\left(1 - z\right) \mathrm{d}z\,\, \mathrm{d}p \\&= \int_0^{1} \sum_{k=0}^{\infty} \left(-1\right)^k \int_0^1 z^{2\left|a\right| + p + 2k - 1}\left(1 - z\right) \mathrm{d}z\,\, \mathrm{d}p \\&= \int_0^{1} \sum_{k=0}^{\infty} \left(-1\right)^k \left(\frac{1}{2\left|a\right| + p + 2k} - \frac{1}{2\left|a\right| + p + 2k + 1}\right)\, \mathrm{d}p \\&= \frac{1}{2}\int_0^{1} \sum_{k=0}^{\infty} \left(-1\right)^k \left(\frac{1}{k + \frac{2\left|a\right| + p}{2}} - \frac{1}{k + \frac{2\left|a\right| + p + 1}{2}}\right)\, \mathrm{d}p \\&= -\frac{1}{4}\int_0^{1} \left(\psi_0\left(\frac{2\left|a\right| + p}{4}\right) - \psi_0\left(\frac{2\left|a\right| + p}{4} + \frac{1}{2}\right) \right. \\&\qquad\qquad\qquad\left.- \psi_0\left(\frac{2\left|a\right| + p + 1}{4}\right) + \psi_0\left(\frac{2\left|a\right| + p + 1}{4} + \frac{1}{2}\right)\right) \mathrm{d}p \\&= -\log\left(\frac{\Gamma\left(\frac{2\left|a\right| + p}{4}\right)\Gamma\left(\frac{2\left|a\right| + p+ 1}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + p}{4} + \frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right| + p + 1}{4}\right)}\right)\biggr\vert_0^1 \\&= - \log\left(\frac{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)\Gamma\left(\frac{2\left|a\right| + 2}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) + \log\left(\frac{\Gamma\left(\frac{2\left|a\right|}{4}\right)\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right|}{4} + \frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right) \\&= - \log\left(\frac{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)^2\Gamma\left(\frac{2\left|a\right| + 2}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)^2 \Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) + \log\left(\frac{\Gamma\left(\frac{2\left|a\right|}{4}\right)}{\Gamma\left(\frac{2\left|a\right|}{4} + \frac{1}{2}\right)}\right) \\&= -\log\left(\frac{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)^2\Gamma\left(\frac{\left|a\right|}{2} + 1 \right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)^2 \Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) + \log\left(\frac{\Gamma\left(\frac{\left|a\right|}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 2}{4}\right)}\right) \\&= -\log\left(\frac{\left|a\right|\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)^2}{2\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)^2}\right) = 2\log\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\sqrt{\left|a\right|}\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right). \end{align*} Por lo tanto \begin{align} \Delta\left(a\right) &= 2\log\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\sqrt{\left|a\right|}\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right) + \ln{\left|a\right|} \\&= 2\log\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right| + 1}{4} + \frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right| + 1}{4}\right)}\right).\tag{1}\label{rzq} \end{align} Tomando el límite de \eqref {rzq} como $a \to 0$ tenemos $$\int_0^{\infty} \frac{\log\left(x\right)}{\cosh\left(\pi x\right)} \, \mathrm{d}x = \log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right),$$ lo que implica $$ \int_0^{\infty} \ln{x}\mathrm{sech}\left(\pi x\right)\mathrm{d}x= \log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$ $$ \frac{1}{\pi}\int_0^{\infty} \ln\left(\frac{x}{\pi}\right)\mathrm{sech}\left(x\right)\mathrm{d}x = \log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$ $$\int_0^{\infty} \ln\left(\frac{x}{\pi}\right)\mathrm{sech}\left(x\right)\mathrm{d}x = \pi\log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$ $$\int_0^{\infty} \ln{x}\mathrm{sech}\left(x\right)\mathrm{d}x - \ln{\pi}\int_0^{\infty} \mathrm{sech}\left(x\right)\mathrm{d}x = \pi\log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$ $$\int_0^{\infty} \ln{x}\mathrm{sech}\left(x\right)\mathrm{d}x - \frac{\pi}{2}\ln{\pi} = \pi\log\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)$$ $$\int_0^{\infty} \ln{x}\mathrm{sech}\left(x\right)\mathrm{d}x= \pi\log\left(\frac{\sqrt{2}\pi^{\frac{1}{2}}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right) = \pi\log\left(\frac{2\pi^{\frac{3}{2}}}{\Gamma\left(\frac{1}{4}\right)^2}\right).$$

Answer 3

El teorema del residuo da: $$ \text{sech}(z) = \sum_{n\geq 0}\frac{4\pi(-1)^n (2n+1)}{4z^2+(2n+1)^2}\tag{1}$$ por lo tanto:

$$ \int_{0}^{+\infty}x^s \text{sech}(x)\,dx = \pi^{s+1}2^{-s}\sec\left(\frac{\pi s}{2}\right)\sum_{n\geq 0}(-1)^n (2n+1)^s\tag{2} $$ y eso se puede simplificar a: $$\begin{eqnarray*}\int_{0}^{+\infty}x^s \text{sech}(x)\,dx &=& \frac{\Gamma(s+1)}{2^{2s+1}}\left(\zeta(1+s,1/4)-\zeta(1+s,3/4)\right)\\ &=&2\Gamma(s+1)\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{s+1}}\tag{3} \end{eqnarray*}$$ es decir, un producto entre un valor del $\Gamma$ y un valor de una función Dirichlet $L$ -función asociada al carácter no principal $\!\pmod{4}$ . No es difícil comprobar que el límite como $s\to 0$ del LHS de $(3)$ es $\pi/2$ . Para calcular $\frac{d}{ds}$ de $(3)$ recordemos que $f'(s)=f(s)\cdot\frac{d}{ds}\log(f(s))$ y tanto el $\Gamma$ y la función $L$ -que aparece en $(3)$ pueden escribirse de forma agradable como productos infinitos: el producto de Weierstrass para el $\Gamma$ conduce a $\frac{\Gamma'}{\Gamma}(1)=-\gamma$ mientras que el producto de Euler: $$ \sum_{n\geq 0}\frac{(-1)^{n+1}}{(2n+1)^{s+1}}=\prod_{p\equiv 1\pmod{4}}\left(1-\frac{1}{p^{s+1}}\right)^{-1}\cdot \prod_{p\equiv 3\pmod{4}}\left(1+\frac{1}{p^{s+1}}\right)^{-1}\tag{4}$$ debería llevarnos a la respuesta final. Sigo trabajando en esto.