Valor de una integral relacionada con la fórmula de Stirling

Question

Considere la siguiente integral incorrecta: $$ I = \ int_1 ^ \ infty \ left (\ {t \} - \ frac {1} {2} \ right) \ frac {dt} {t}. $$

Comparando con la fórmula de Stirling, podemos ver que$I = \ln(\sqrt{2\pi}) - 1$. ¿Hay una forma más directa de calcular esto?

Editar: he dividido mis dos preguntas: la otra parte está aquí .

Answer 1

La Riemann zeta función se define como: $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}.$ Escribir esto como una Riemann Stieltjies integral y aplicando integración por partes tenemos que $$\zeta(s)=\int_{1}^{\infty}x^{-s}d\left[x\right]=\int_{1}^{\infty}x^{-s}dx-\int_{1}^{\infty}x^{-s}d\left\{ x\right\}$$

$$=\frac{s}{s-1}-s\int_{1}^{\infty}\left\{ x\right\} x^{-s-1}dx,$$ and this holds for all $\text{Re}(s)>0.$ Notice that taking the limit as $s\rightarrow0$ yields $\zeta(0)=-\frac{1}{2}.$ Now, since $\frac{s}{2}\int_{1}^{\infty}x^{-s-1}=\frac{1}{2},$ it follows that $$\zeta(s)+\frac{1}{2}=\frac{s}{s-1}-s\int_{1}^{\infty}\left(\left\{ x\right\} -\frac{1}{2}\right)x^{-s-1}dx,$$ and so for $\text{Re}(s)>0,$ we have that $$\int_{1}^{\infty}\left(\left\{ t\right\} -\frac{1}{2}\right)t^{-s-1}dt=\frac{1}{s-1}+\frac{-\zeta(s)-\frac{1}{2}}{s}.$$ Taking the limit as $s\rightarrow0,$ and using the fact that $\zeta(0)=-\frac{1}{2},$ we find that $$\int_{1}^{\infty}\left(\left\{ t\right\} -\frac{1}{2}\right)\frac{dt}{t}=-\zeta^{'}(0)-1.$$ Since $\zeta^{'}(0)=-\frac{1}{2}\log(2\pi)$, this yields $$\int_{1}^{\infty}\left(\left\{ t\right\} -\frac{1}{2}\right)\frac{dt}{t}=\frac{1}{2}\log(2\pi)-1,$$ como se desee.

Demostrando que $\zeta^{'}(0)=-\frac{1}{2}\log(2\pi).$

Recordemos la ecuación funcional para la función zeta, $$\zeta(z)=2^{z}\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\Gamma\left(1-z\right)\zeta\left(1-z\right).$$ Taking the logarithmic derivative, we have that $$\frac{\zeta^{'}(z)}{\zeta(z)}=\log2+\log\pi-\frac{\Gamma^{'}\left(1-z\right)}{\Gamma\left(1-z\right)}+\frac{d}{dz}\log\left(\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)\right).$$ Since $\zeta(1-z)=-\frac{1}{z}+\gamma+O(z),$ and $\sin\left(\frac{\pi z}{2}\right)=\frac{\pi z}{2}+O\left(z^{3}\right),$ we have that $$\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)=\frac{\pi}{2}-\frac{\pi\gamma}{2}z+O\left(z^{2}\right),$$

y así $$\frac{d}{dz}\log\left(\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)\right)=\frac{-\frac{\pi\gamma}{2}+O\left(z\right)}{\frac{\pi}{2}+O\left(z\right)}=-\gamma+O(z).$$ Thus $$\frac{\zeta^{'}(0)}{\zeta(0)}=\log2\pi-\Gamma^{'}\left(1\right)-\gamma.$$ As $\Gamma^{'}(1)=-\gamma,$ and $\zeta(0)=-\frac{1}{2},$ we conclude that $$\zeta^{'}(0)=-\frac{1}{2}\log2\pi.$$

Comentario: Esto me recordó de mi respuesta con respecto a la evaluación de $\Gamma\left(\frac{1}{2}\right).$