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Trigonometric logarithmic integral

Calculus (Integrals, Series)
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Riemann
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Trigonometric logarithmic integral

#1

Post by Riemann » Tue May 23, 2017 2:30 pm

Let $0 \leq \alpha, \beta \leq \pi$ and $\kappa>0$. Prove that

\[\int_0^{\infty} \frac{1}{x} \log \left( \frac{x^2 + 2\kappa x \cos \beta + \kappa^2}{x^2 + 2 \kappa x \cos \alpha + \kappa^2} \right) \, {\rm d}x = \alpha^2 - \beta^2\]
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Trigonometric logarithmic integral

#2

Post by Riemann » Sat Nov 27, 2021 7:02 pm

We note that

$$\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx\cos a+k^2} \right )=\ln \left ( x^2+2kx \cos \alpha +k^2 \right ) \bigg|_{\alpha =a}^{\alpha =b}$$

Hence,

\begin{align*}
\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{\mathrm{d} }{\mathrm{d} \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{align*}
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Trigonometric logarithmic integral

#3

Post by Riemann » Sat Nov 27, 2021 7:07 pm

We are using the following results:

1. $\displaystyle \sum_{n=1}^{\infty}\frac{x^n \cos (na)}{n}=-\frac{1}{2}\ln \left ( x^2-2x \cos a+1 \right ) \; , \; \left | x \right |<1$

2. $\displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\sin n x}{n}=-\frac{x}{2}\; , \; \left | x \right |<\pi$

3. $\displaystyle \sum_{n=1}^{\infty}(-1)^n \frac{\cos nx }{n^2}=\frac{x^2}{4}- \frac{\pi^2}{12} \; , \; \left | x \right |<\pi$

Thus,

\begin{align*} \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 + \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{align*}
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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