Ha habido muchos temas en la función de Riemann Zeta, específicamente $\zeta(2)$.$$\zeta(2)=\sum_{n=1}^\infty\frac{1}{n^2}=\int_0^1\int_0^1\frac{1}{1-xy}dA$$This is the Basel Problem. Taking the multivariable calculus approach, one could make the change of variables $(x,y)=(\frac{u-v}{\sqrt2},\frac{u+v}{\sqrt2})$. Following this path, one would come across the following iterated integral: $$\int_0^\sqrt2\int_{|u-\frac{\sqrt2}2|-\frac{\sqrt2}2}^{\frac{\sqrt2}2-|u-\frac{\sqrt2}2|}\frac 2{v^2-u^2+2}dv\;du$$All proofs that I've found online that use multivariable calculus integrate with respect to $v$ then $u$, so I wanted to write a proof integrating with respect to $u$ then $v$. For the sake of maintaining this post's relative brevity, I won't write out the full proof here. Instead, my proof can be found here.
Integrating with respect to $u$ then $v$ would require this iterated integral:$$\int_{-\frac{\sqrt2}2}^{\frac{\sqrt2}2}\int_{|v|}^{\sqrt2-|v|}\frac2{v^2-u^2+2}du\;dv$$My question is this: is there any way to evaluate the following integral analytically?$$\int_0^{\arctan\frac 1 2}\sec\theta\ln{\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}}d\theta=\frac{\pi^2}{12}+\frac{1}2\ln^2{\frac{\sqrt5-1}2}+\frac{1}2\ln^2{\frac{\sqrt5+1}2}$$Rearranging the terms results in$$2\int_0^{\arctan\frac 1 2}\sec\theta\ln{\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}}d\theta-\ln^2{\frac{\sqrt5-1}2}-\ln^2{\frac{\sqrt5+1}2}=\frac{\pi^2}6$$Using Fubini's Theorem, one can see that this is in fact another way of stating the Basel Problem. Is there any other way to prove this result, though?
All of the steps in between are in my link above.
I've spent a week reviewing every number and symbol in my work, and I can guarantee there are no errors. Wolfram Alpha produces an approximation to within $0.000004$ de mi resultado, Aunque no indique el valor exacto.
Respuestas
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¡Ya lo tengo!!! Contrario a mi respuesta más reciente de David H la respuesta, no me terminan usando integración de Lebesgue. En lugar de eso, tuve que aprender por mí mismo las propiedades de polylogarithms en el lapso de dos días (que fue muy divertido). De todos modos, esto va a ser algo de una larga prueba, así que por favor tengan paciencia conmigo. También, puede que tenga que actualizar la página después de hacer clic en los enlaces; Wolfram Alpha ocasionalmente no interpretar la consulta correctamente.
Para reducir la cantidad de desplazamiento necesario, aquí es la identidad que se va a proven:$$\int_0^{\arctan{\frac12}}\sec\theta\ln\left(\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}\right)\mathbb{d}\theta=\frac{\pi^2}{12}+\frac12\ln^2\left({\frac{\sqrt5-1}2}\right)+\frac12\ln^2\left({\frac{\sqrt5+1}2}\right)$$By applying David's change of variable of $t=\tan\frac\theta2$, the integral above becomes$$\int_0^{\sqrt5-2}\frac{\ln(1-t)}{1-t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1+t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(t)}{1-t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(t)}{1+t}\mathbb{d}t+\int_0^{\sqrt5-2}\frac{\ln{(1-t)}}{1+t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1-t}\mathbb{d}t\;.$$Tackling the first integral,$$\int_0^{\sqrt5-2}\frac{\ln(1-t)}{1-t}\mathbb{d}t=-\int_{\sqrt5-2}^0\frac{\ln(1-t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)\text{Li}_0(t)}t\mathbb{d}t$$make the change of variable $k=\text{Li}_1(t)$ such that $\mathbb{d}k=\frac{\text{Li}_0(t)}t\mathbb{d}t$.$$\begin{align}\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)\text{Li}_0(t)}t\mathbb{d}t=\int_{\text{Li}_1(\sqrt5-2)}^0k\,\mathbb{d}k=\frac12\left[k^2\right]_{\text{Li}_1(\sqrt5-2)}^0=&-\frac12\text{Li}_1^2(\sqrt5-2)\\=&-\frac12\ln^2(3-\sqrt5)\end{align}$$Now, focusing on the second integral,$$-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1+t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1+t}\mathbb{d}t$$make the change of variable $m=\ln(1+t)$ such that $\mathbb{d}m=\frac{\mathbb{d}t}{1+t}$.$$\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1+t}\mathbb{d}t=\int_{\ln(\sqrt5-1)}^0m\,\mathbb{d}m=\frac12\left[m^2\right]_{\ln(\sqrt5-1)}^0=-\frac12\ln^2(\sqrt5-1)$$Taking the third integral by the horns,$$-\int_0^{\sqrt5-2}\frac{\ln(t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\ln(t)}{1-t}\mathbb{d}t$$make the change of variable $p=1-t$ such that $\mathbb{d}p=-\mathbb{d}t$.$$\int_{\sqrt5-2}^0\frac{\ln(t)}{1-t}\mathbb{d}t=-\int_{3-\sqrt5}^1\frac{\ln(1-p)}p\mathbb{d}p=\left[\text{Li}_2(p)\right]_{3-\sqrt5}^1=\frac{\pi^2}6-\text{Li}_2(3-\sqrt5)$$Tackling the fourth integral,$$\begin{align}-\int_0^{\sqrt5-2}\frac{\ln(t)}{1+t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\ln(t)}{1+t}\mathbb{d}t=&\left[\ln(t)\ln(1+t)\right]_{\sqrt5-2}^0-\int_{\sqrt5-2}^0\frac{\ln(1+t)}t\mathbb{d}t\\=&\int_0^{\sqrt5-2}\frac{\ln(1+t)}t\mathbb{d}t-\ln(\sqrt5-2)\ln(\sqrt5-1)&\end{align}$$make the change of variable $h=1+t$ such that $\mathbb{d}h=\mathbb{d}t$.$$\begin{align}\int_0^{\sqrt5-2}\frac{\ln(1+t)}t\mathbb{d}t-\ln(\sqrt5-2)\ln(\sqrt5-1)=&\int_1^{\sqrt5-1}\frac{\ln(h)}{h-1}\mathbb{d}h-\ln(\sqrt5-2)\ln(\sqrt5-1)\\=&\int_{\sqrt5-1}^1\frac{\ln(h)}{1-h}\mathbb{d}h-\ln(\sqrt5-2)\ln(\sqrt5-1)&\end{align}$$Now make a second change of variable, $n=1-h$, such that $\mathbb{d}n=-\mathbb{d}h$.$$\begin{align}\int_{\sqrt5-1}^1\frac{\ln(h)}{1-h}\mathbb{d}h-\ln(\sqrt5-2)\ln(\sqrt5-1)=&-\int_{2-\sqrt5}^0\frac{\ln(1-n)}{n}\mathbb{d}n-\ln(\sqrt5-2)\ln(\sqrt5-1)\\=&\left[\text{Li}_2(n)\right]_{2-\sqrt5}^0-\ln(\sqrt5-2)\ln(\sqrt5-1)\\=&{-\text{Li}_2(2-\sqrt5)}-\ln(\sqrt5-2)\ln(\sqrt5-1)\end{align}$$Now, one could evaluate the fifth and sixth integrals separately, but evaluating them together turns out to be much simpler.$$\int_0^{\sqrt5-2}\frac{\ln(1-t)}{1+t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)}{1+t}\mathbb{d}t+\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1-t}\mathbb{d}t\\\begin{align}&=\left[\text{Li}_1(t)\ln(1+t)\right]_{\sqrt5-2}^0-\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1-t}\mathbb{d}t+\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1-t}\mathbb{d}t\\&=\left[\text{Li}_1(t)\ln(1+t)\right]_{\sqrt5-2}^0=-\text{Li}_1(\sqrt5-2)\ln(1+(\sqrt5-2))=\ln(3-\sqrt5)\ln(\sqrt5-1)\end{align}$$Gathering these results yields$$\begin{align}&\int_0^{\arctan{\frac12}}\sec\theta\ln\left(\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}\right)\mathbb{d}\theta\\=&\frac{\pi^2}6-\text{Li}_2(3-\sqrt5)-\text{Li}_2(2-\sqrt5)-\frac12\ln^2(\sqrt5-1)-\frac12\ln^2(3-\sqrt5)+&\\&\ln(\sqrt5-1)\ln(3-\sqrt5)-\ln(\sqrt5-2)\ln(\sqrt5-1)\end{align}$$According to Wolfram Alpha, this is equivalent to the desired identity. In order to verify the alleged congruence, assume that the following is true:$$\begin{align}&\frac{\pi^2}{12}+\frac12\ln^2\left(\frac{\sqrt5-1}2\right)+\frac12\ln^2\left(\frac{\sqrt5+1}2\right)\\=&\frac{\pi^2}6-\text{Li}_2(3-\sqrt5)-\text{Li}_2(2-\sqrt5)-\frac12\ln^2(\sqrt5-1)-\frac12\ln^2(3-\sqrt5)+\\&\ln(\sqrt5-1)\ln(3-\sqrt5)-\ln(\sqrt5-2)\ln(\sqrt5-1)\;.\end{align}$$Using dilogarithmical identities, the right-hand side becomes $$\begin{align}&\frac{\pi^2}{12}+\text{Li}_2(\sqrt5-2)-\text{Li}_2(3-\sqrt5)+\frac12\text{Li}_2(4\sqrt5-8)-\frac12\ln^2(3-\sqrt5)-\frac12\ln^2(\sqrt5-1)+\\&\ln(\sqrt5-2)\ln(3-\sqrt5)+\ln(\sqrt5-1)\ln(3-\sqrt5)\;.\end{align}$$One could find other equivalent expressions, but this one is useful because it, like the original identity, contains the term $\frac{\pi^2}{12}$.
This means the assumed-to-be-true equation can be simplified to$$\frac12\ln^2\left(\frac{\sqrt5-1}2\right)+\frac12\ln^2\left(\frac{\sqrt5+1}2\right)=\text{Li}_2(\sqrt5-2)-\text{Li}_2(3-\sqrt5)+\frac12\text{Li}_2(4\sqrt5-8)-\frac12\ln^2(3-\sqrt5)-\frac12\ln^2(\sqrt5-1)+\ln(\sqrt5-2)\ln(3-\sqrt5)+\ln(\sqrt5-1)\ln(3-\sqrt5)\;.$$I've tried to simplify this further, yet I always seem to end up going in circles. However, there is another way to approach the equation: graphically. Replacing $\sqrt5-1$ with the variable $z$ results in the functional equation $$\frac12\ln^2\left(\frac{z}2\right)+\frac12\ln^2\left(\frac{z+2}2\right)=\text{Li}_2(z-1)-\text{Li}_2(2-z)+\frac12\text{Li}_2(2z-z^2)-\frac12\ln^2(2-z)-\frac12\ln^2(z)+\ln(z-1)\ln(2-z)+\ln(z)\ln(2-z)\;.$$The two graphs aren't very similar, but it's only necessary to check for an intersection at $z=\sqrt5-1$. It's not apparent looking at the two graphs side by side, but moving everything to one side of the equation appears to result in a zero at $z=\sqrt5-1$. Checking this gives a result of $0$, which means that the functions do intersect at $z=\sqrt5-1$ and, therefore,$$\int_0^{\arctan{\frac12}}\sec\theta\ln\left(\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}\right)\mathbb{d}\theta=\frac{\pi^2}{12}+\frac12\ln^2\left({\frac{\sqrt5-1}2}\right)+\frac12\ln^2\left({\frac{\sqrt5+1}2}\right)\;.$$
El uso de la tangente a la mitad de ángulo de sustitución de $t=\tan{\left(\frac{\theta}{2}\right)}$, parcial fracción de descomposición, y lo básico de las propiedades de los logaritmos, la integral en cuestión puede reducirse a una suma de seis conceptos básicos de registros integrales. Para $\alpha\in[0,\frac{\pi}{2}]$, tenemos
$$\begin{align} \mathcal{I}{(\alpha)} &=\int_{0}^{\alpha}\sec{\theta}\ln{\left(\frac{\cos{\theta}-\sin{\theta}+1}{\sin{\theta}-\cos{\theta}+1}\right)}\,\mathrm{d}\theta\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{1+t^2}{1-t^2}\ln{\left(\frac{1-t}{t(t+1)}\right)}\cdot\frac{2\,\mathrm{d}t}{1+t^2}\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{2}{1-t^2}\ln{\left(\frac{1-t}{t(t+1)}\right)}\,\mathrm{d}t\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(\frac{1-t}{t(t+1)}\right)}}{1+t}\,\mathrm{d}t+\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(\frac{1-t}{t(t+1)}\right)}}{1-t}\,\mathrm{d}t\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}-\ln{\left(t\right)}-\ln{\left(t+1\right)}}{1+t}\,\mathrm{d}t\\ &~~~~~ +\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}-\ln{\left(t\right)}-\ln{\left(t+1\right)}}{1-t}\,\mathrm{d}t\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}}{1-t}\,\mathrm{d}t-\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t+1\right)}}{1+t}\,\mathrm{d}t-\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t\right)}}{1-t}\,\mathrm{d}t\\ &~~~~~ -\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t\right)}}{1+t}\,\mathrm{d}t+\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}}{1+t}\,\mathrm{d}t-\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t+1\right)}}{1-t}\,\mathrm{d}t.\\ \end{align}$$
La primera de las dos integrales son primarias, y la tercera es bien conocida la representación integral de la dilogarithm función reflexionó acerca de la unidad de intervalo. Los tres restantes integrales también tienen no muy complicado anti-derivados en términos de los registros y dilogs. A partir de allí, la obtención de una forma cerrada es sólo una cuestión de convertir el álgebra de las bielas.
