Evaluate the integral:
$$\int_{0}^{\infty}\frac{1}{x}\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx \cos a+k^2} \right )\,dx $$
whereas $0\leq a, b \leq \pi, \;\; k>0$.
Solution:
1st way : We are transforming the integral into a double one and by invonking Fubini's theorem for interchanging the integrals. Successively we have:
$$\begin{aligned}
\int^{\infty}_{0} \frac{1}{x}\ln\left(\frac{x^2+2kx\cdot \cos b+k^2}{x^2+2k x\cdot \cos a+k^2}\right)\,dx &= \int_{0}^{\infty}\frac{1}{x}\ln \left ( x^2+2k x \cos \alpha +k^2 \right )\bigg|_{\alpha =a}^{\alpha =b}\,dx \\
&= \int_{0}^{\infty}\frac{1}{x}\int_{a}^{b}\frac{\partial }{\partial \alpha }\ln \left ( x^2+2k x \cos \alpha +k^2\right )\,d\alpha \;dx\\
&= -\int_{a}^{b}\int_{0}^{\infty}\frac{2k\sin \alpha }{x^2+2k\cos a x +k^2}\,dx\;d\alpha \\
&=-\int_{a}^{b}\int_{0}^{\infty}\frac{2k\sin \alpha }{\left ( x+k\cos a \right )^2+k^2\sin^2 \alpha }\,dx\;d\alpha \\
&= -2\int_{a}^{b}\tan^{-1}\left ( \frac{x}{k\sin \alpha } +\frac{1}{\tan \alpha }\right )\bigg|_{0}^{\infty}d\alpha \\
&=-2\int_{a}^{b}\left [ \frac{\pi}{2}-\tan^{-1}\left ( \frac{1}{\tan \alpha } \right ) \right ]\,d\alpha \\
&=-2\int_{a}^{b}\tan^{-1}\left ( \tan \alpha \right )\, d\alpha \\
&=-2\int_{a}^{b}\alpha \,d\alpha =a^2-b^2
\end{aligned}$$
2nd way: Using Fourier Series.
We are using the following formulae:
$\blacksquare \displaystyle \; \; \sum_{n=1}^{\infty}\frac{x^n \cos (na)}{n}=-\frac{1}{2}\ln \left ( x^2-2x\cos a +1 \right ), \; |x|<1 $ which is derived pretty easily by subbing $z=xe^{ia}$ into the Taylor's expansion of $\ln (1-z)$.
$\blacksquare \displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\cos nx}{n^2}=\frac{x^2}{4}-\frac{\pi^2}{12}, \;\; \left | x \right |<\pi$ which is derived by integrating the Fourier series $\displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\sin n x}{n}=-\frac{x}{2}, \; \left | x \right |<\pi$ once and subbing $x=0$. Note that:
$$ \begin{aligned}
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} &=-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} \\
&= -\eta (2)\\
&= -\left ( 1-2^{1-2} \right )\zeta(2)\\
&= -\frac{\zeta(2)}{2}=-\frac{\pi^2}{12}
\end{aligned}$$
whereas $\eta$ is the $\eta$ Dirichlet function. Another way of calculating the series is by re-arranging its terms since it converges absolutely.
Hence the original integral is transformed as follows:
$$\begin{aligned}
\int_{0}^{\infty}\frac{1}{x}\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx \cos a+k^2} \right )\,dx &\overset{u=x/k}{=\! =\! =\!}\int_{0}^{\infty}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\
&= \int_{0}^{1}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du + \\
& \quad \quad \quad \quad +\int_{1}^{\infty}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\
&\overset{u \mapsto 1/u}{=\! =\! =\! =\!}2\int_{0}^{1}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\
&=4\int_{0}^{1}\frac{1}{u}\sum_{n=1}^{\infty}(-u)^n \frac{\cos na-\cos nb}{n}\,du \\
&= 4\sum_{n=1}^{\infty}(-1)^n \frac{\cos na-\cos nb}{n}\int_{0}^{1}u^{n-1}\,du \\
&=4\sum_{n=1}^{\infty}(-1)^n \frac{\cos na-\cos nb}{n^2} \\
&=4\left ( \frac{a^2}{4}-\frac{b^2}{4} \right ) \\
&=a^2-b^2
\end{aligned}$$
Source: mathematica.gr
$$\int_{0}^{\infty}\frac{1}{x}\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx \cos a+k^2} \right )\,dx $$
whereas $0\leq a, b \leq \pi, \;\; k>0$.
Solution:
1st way : We are transforming the integral into a double one and by invonking Fubini's theorem for interchanging the integrals. Successively we have:
$$\begin{aligned}
\int^{\infty}_{0} \frac{1}{x}\ln\left(\frac{x^2+2kx\cdot \cos b+k^2}{x^2+2k x\cdot \cos a+k^2}\right)\,dx &= \int_{0}^{\infty}\frac{1}{x}\ln \left ( x^2+2k x \cos \alpha +k^2 \right )\bigg|_{\alpha =a}^{\alpha =b}\,dx \\
&= \int_{0}^{\infty}\frac{1}{x}\int_{a}^{b}\frac{\partial }{\partial \alpha }\ln \left ( x^2+2k x \cos \alpha +k^2\right )\,d\alpha \;dx\\
&= -\int_{a}^{b}\int_{0}^{\infty}\frac{2k\sin \alpha }{x^2+2k\cos a x +k^2}\,dx\;d\alpha \\
&=-\int_{a}^{b}\int_{0}^{\infty}\frac{2k\sin \alpha }{\left ( x+k\cos a \right )^2+k^2\sin^2 \alpha }\,dx\;d\alpha \\
&= -2\int_{a}^{b}\tan^{-1}\left ( \frac{x}{k\sin \alpha } +\frac{1}{\tan \alpha }\right )\bigg|_{0}^{\infty}d\alpha \\
&=-2\int_{a}^{b}\left [ \frac{\pi}{2}-\tan^{-1}\left ( \frac{1}{\tan \alpha } \right ) \right ]\,d\alpha \\
&=-2\int_{a}^{b}\tan^{-1}\left ( \tan \alpha \right )\, d\alpha \\
&=-2\int_{a}^{b}\alpha \,d\alpha =a^2-b^2
\end{aligned}$$
2nd way: Using Fourier Series.
We are using the following formulae:
$\blacksquare \displaystyle \; \; \sum_{n=1}^{\infty}\frac{x^n \cos (na)}{n}=-\frac{1}{2}\ln \left ( x^2-2x\cos a +1 \right ), \; |x|<1 $ which is derived pretty easily by subbing $z=xe^{ia}$ into the Taylor's expansion of $\ln (1-z)$.
$\blacksquare \displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\cos nx}{n^2}=\frac{x^2}{4}-\frac{\pi^2}{12}, \;\; \left | x \right |<\pi$ which is derived by integrating the Fourier series $\displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\sin n x}{n}=-\frac{x}{2}, \; \left | x \right |<\pi$ once and subbing $x=0$. Note that:
$$ \begin{aligned}
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} &=-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} \\
&= -\eta (2)\\
&= -\left ( 1-2^{1-2} \right )\zeta(2)\\
&= -\frac{\zeta(2)}{2}=-\frac{\pi^2}{12}
\end{aligned}$$
whereas $\eta$ is the $\eta$ Dirichlet function. Another way of calculating the series is by re-arranging its terms since it converges absolutely.
Hence the original integral is transformed as follows:
$$\begin{aligned}
\int_{0}^{\infty}\frac{1}{x}\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx \cos a+k^2} \right )\,dx &\overset{u=x/k}{=\! =\! =\!}\int_{0}^{\infty}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\
&= \int_{0}^{1}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du + \\
& \quad \quad \quad \quad +\int_{1}^{\infty}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\
&\overset{u \mapsto 1/u}{=\! =\! =\! =\!}2\int_{0}^{1}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\
&=4\int_{0}^{1}\frac{1}{u}\sum_{n=1}^{\infty}(-u)^n \frac{\cos na-\cos nb}{n}\,du \\
&= 4\sum_{n=1}^{\infty}(-1)^n \frac{\cos na-\cos nb}{n}\int_{0}^{1}u^{n-1}\,du \\
&=4\sum_{n=1}^{\infty}(-1)^n \frac{\cos na-\cos nb}{n^2} \\
&=4\left ( \frac{a^2}{4}-\frac{b^2}{4} \right ) \\
&=a^2-b^2
\end{aligned}$$
Source: mathematica.gr
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