Ολοκλήρωμα

Tolaso J Kos · #1

Να δειχθεί ότι:

\int_{0}^{\infty} \left( \cos x - \cos \frac{x}{2} \right) \frac{\ln x}{x} = \gamma \ln 2 - \frac{\ln^2 2}{2}

Dimessi · #2

Είναι

\textstyle { \displaystyle { \int \limits_{0}^{+\infty} \frac{\cos 2y-\cos y}{y}\ln y \, dy=\lim_{\varepsilon \to 0^{+}} \int \limits_{0}^{+\infty} e^{-\varepsilon y}\frac{\cos 2y-\cos y}{y}\left(\int \limits_{0}^{+\infty} \frac{e^{-x}-e^{-xy}}{x} \, dx\right) \, dy, } }

κι αφού

\textstyle { \displaystyle { \int \limits_{0}^{+\infty} \int \limits_{0}^{+\infty}\left|e^{-\varepsilon y}\cdot \frac{\cos 2y-\cos y}{y}\cdot \frac{e^{-x}-e^{-xy}}{x}\right| \, dx \, dy\le M \int \limits_{0}^{+\infty}e^{-\varepsilon y}\left|\ln y\right| \, dy<+\infty, } }

από Θεώρημα Fubini αφήνοντας $\varepsilon \rightarrow 0^{+}$ είναι

\textstyle { \displaystyle { \int \limits_{0}^{+\infty} \frac{\cos 2y-\cos y}{y}\ln y \, dy=\int \limits_{0}^{+\infty}\frac{\cos 2y-\cos y}{y}\left(\int \limits_{0}^{+\infty}\frac{e^{-\vartheta}-e^{-\vartheta y}}{\vartheta } \, d\vartheta\right) \, dy= } }

\textstyle \displaystyle =\int \limits_{0}^{+\infty} \frac{1}{\vartheta}\left[e^{-\vartheta}\int \limits_{0}^{+\infty}\frac{\cos 2y-\cos y}{y} \, dy-\int \limits_{0}^{+\infty} e^{-\vartheta y}\frac{\cos 2y-\cos y}{y} \, dy\right] \, d\vartheta =

\textstyle { \displaystyle { \int \limits_{0}^{+\infty} \frac{1}{\vartheta}\left[-e^{-\vartheta}\ln 2-\int \limits_{0}^{+\infty}e^{-\vartheta y}\sum_{n=1}^{+\infty} \frac{\left(-1\right)^{n}\left(2^{2n}-1\right)}{\left(2n\right)!}y^{2n-1} \, dy\right] \, d\vartheta }= }

\textstyle { \displaystyle { =\int \limits_{0}^{+\infty}\frac{1}{\vartheta}\left[-e^{-\vartheta}\ln 2-\sum_{n=1}^{+\infty}\frac{\left(-1\right)^{n}\left(2^{2n}-1\right)}{\left(2n\right)!}\int \limits_{0}^{+\infty}y^{2n-1}e^{-\vartheta y} \, dy \right] \, d\vartheta=\int \limits_{0}^{+\infty}\frac{1}{\vartheta}\left[-e^{-\vartheta}\ln 2-\sum_{n=1}^{+\infty}\frac{\left(-1\right)^{n}\left(2^{2n}-1\right)}{\left(2n\right)!}\frac{\Gamma\left(2n\right)}{\vartheta ^{2n}} \right] \, d\vartheta } }=

\textstyle { \displaystyle { =-\ln 2\int \limits_{0}^{+\infty}e^{-\vartheta} \, d\left(\ln \vartheta\right)-\frac{1}{2}\int \limits_{0}^{+\infty} \ln \left(\frac{1+\vartheta ^{2}}{4+\vartheta ^{2}}\right) \, d\left(\ln \vartheta\right)=\gamma \ln 2+\frac{\ln^22}{2}. } }

Αφού

\textstyle { \displaystyle { \int \limits_{0}^{+\infty} \frac{\cos 2y-\cos y}{y} \, dy=-\ln 2 , } }

είναι

\textstyle { \displaystyle { \int \limits_{0}^{+\infty}\left(\cos x-\cos \frac{x}{2}\right)\frac{\ln x}{x} \, dx\overset{x=2y}=\int \limits_{0}^{+\infty} \left(\cos 2y-\cos y\right)\cdot \frac{\ln 2+\ln y}{y} \, dy=\ln 2\int \limits_{0}^{+\infty}\frac{\cos 2y-\cos y}{y} \, dy+\int \limits_{0}^{+\infty} \frac{\cos 2y-\cos y}{y}\ln y \, dy } }

\textstyle \displaystyle =\ln 2\int \limits_{0}^{+\infty}\frac{\cos 2y-\cos y}{y} \, dy+\int \limits_{0}^{+\infty} \frac{\cos 2y-\cos y}{y}\ln y \, dy=-\ln ^{2}2+\left(\gamma \ln 2+\frac{\ln ^{2}2}{2}\right)=\gamma \ln 2-\frac{\ln ^{2}2}{2}.