$\int_{0}^{+\infty}{\log({x+\frac{1}{x}})\big/(1+x^2)\,dx}$

Calculus (Integrals, Series)
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Grigorios Kostakos
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$\int_{0}^{+\infty}{\log({x+\frac{1}{x}})\big/(1+x^2)\,dx}$

#1

Post by Grigorios Kostakos »

Evaluate

\[ \displaystyle\int_{0}^{+\infty}{\frac{\log\bigl({x+\tfrac{1}{x}}\bigr)}{1+x^2}\,dx}\,. \]
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Tolaso J Kos
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Re: $\int_0^{+\infty}\log(x+\frac{1}{x})\big/(1+x^2)\,dx$

#2

Post by Tolaso J Kos »

Grigorios Kostakos wrote:Evaluate

\[ \displaystyle\int_{0}^{+\infty}{\frac{\log\bigl({x+\tfrac{1}{x}}\bigr)}{1+x^2}\,dx}\,. \]
Here is a solution to the proposed integral, although I have a huntch that it can be easily generalized.

We are making use of two lemmas:

Lemma 1: \( \displaystyle \int_0^{\pi/2} \ln \tan \theta \, {\rm d}\theta =0 \). This holds since:

$$\begin{aligned}
\int_{0}^{\pi/2} \ln \tan \theta \, {\rm d}\theta &= \int_{0}^{\pi/2} \ln \sin \theta \, {\rm d}\theta - \int_{0}^{\pi/2} \ln \cos \theta \, {\rm d}\theta \\ &\overset{\theta= \pi/2-u}{=\! =\! =\! =\! =\!}\require{cancel} \cancel{\int_{0}^{\pi/2} \ln \cos \theta \, {\rm d}\theta - \int_{0}^{\pi/2} \ln \cos \theta \, {\rm d}\theta} \\
& =0 \end{aligned}$$

Lemma 2: \( \displaystyle \int_{0}^{\pi/2}\ln \sin \theta \, {\rm d}\theta = - \frac{\pi \ln 2}{2} \). The proof of it can be found .... everywhere.

For our proposed integral we have successively:

$$\begin{aligned}
\int_{0}^{\infty}\frac{\log \left ( x+ \frac{1}{x} \right )}{1+x^2}\, {\rm d}x &\overset{x=\tan \theta}{=\! =\! =\! =\!}\int_{0}^{\pi/2}\frac{\log \left ( \tan \theta + \frac{1}{\tan \theta} \right )}{1+\tan^2 \theta} \sec^2 \theta \, {\rm d}\theta\\
&= \int_{0}^{\pi/2} \log \left ( \tan \theta + \frac{1}{\tan \theta} \right )\, {\rm d}\theta\\
&= \int_0^{\pi/2} \log \left [ \tan \theta \left ( \cot^2 \theta +1 \right ) \right ]\, {\rm d}\theta\\
&= \cancelto{0}{\int_0^{\pi/2} \log \tan \theta \, {\rm d}\theta} + \int_0^{\pi/2} \log \left ( \cot^2 +1 \right )\, {\rm d}\theta \\
&= \int_0^{\pi/2} \log \csc^2 \theta \, {\rm d}\theta \\
&= - 2 \int_0^{\pi/2} \log \sin \theta \, {\rm d}\theta \\
&= \pi \ln 2
\end{aligned}$$

ending the calculations.

P.S 1: I also evaluate that:

$$\int_{0}^{\infty}\frac{\ln^2 \left ( x+ \frac{1}{x} \right )}{x^2+1}= \frac{5\pi^3}{24}+ \pi \ln^2 2$$

P.S 2: The reason I actually suspect that this can be generalized is due to the fact that:

$$\int_{0}^{\pi/2} \ln^{2m} \tan \theta \, {\rm d}\theta = \left(\frac{\pi}{2} \right)^{2m+1}(-1)^m {\rm E}_{2m}$$

and the integral \( \displaystyle \int_0^{\pi/2} \ln^n \sin \theta \, {\rm d} \theta \) can actually be derived by differentiating \( n \) times the Beta function. I have not been able to crack down the last integral or at least write it in a much "human" form that consists of polygamma functions. We'll see what results research can get.
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Re: $\int_0^{+\infty}\log(x+\frac{1}{x})\big/(1+x^2)\,dx$

#3

Post by Grigorios Kostakos »

Thank you Tolis. Allow me to suggest a shortcut:

\begin{align*}
\displaystyle\int_{0}^{+\infty}{\frac{\log\bigl({x+\tfrac{1}{x}}\bigr)}{1+x^2}\,dx}&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{x\,=\,\tan\theta}\\
{\frac{1}{1+x^2}\,dx\,=\,d\theta}\\
\end{subarray}}\,\int_{0}^{\frac{\pi}{2}}{\log\bigl({\tfrac{\sin{\theta}}{\cos\theta}+\tfrac{\cos\theta}{\sin{\theta}}}\bigr)\,d\theta}\\
&=\int_{0}^{\frac{\pi}{2}}{\log\bigl({\tfrac{1}{\sin{\theta}\cos\theta}}\bigr)\,d\theta}\\
&=-\int_{0}^{\frac{\pi}{2}}{\log({\sin{\theta}})\,d\theta}-\int_{0}^{\frac{\pi}{2}}{\log({\cos{\theta}})\,d\theta}\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{\theta\,=\,\frac{\pi}{2}-\varphi}\\
{d\theta\,=\,-d\varphi}\\
\end{subarray}}\,-\int_{0}^{\frac{\pi}{2}}{\log({\sin{\theta}})\,d\theta}+\int_{\frac{\pi}{2}}^{0}{\log\big(\cos\big(\tfrac{\pi}{2}-\varphi\big)\big)\,d\varphi}\\
&=-\int_{0}^{\frac{\pi}{2}}{\log({\sin{\theta}})\,d\theta}-\int_{0}^{\frac{\pi}{2}}{\log({\sin{\varphi}})\,d\varphi}\\
&=-2\int_{0}^{\frac{\pi}{2}}{\log({\sin{\theta}})\,d\theta}\\
&=\ldots
\end{align*}
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Re: $\int_0^{+\infty}\log(x+\frac{1}{x})\big/(1+x^2)\,dx$

#4

Post by Tolaso J Kos »

Grigoris,

let me go beyond this integral and present the calculations of the \( \displaystyle \int_0^\infty \frac{\log^2 (x+\frac{1}{x})}{x^2+1}\, {\rm d}x \).

We are using two lemmas:

Lemma 1: \( \displaystyle \int_0^{\pi/2} \ln^2 \tan \theta \, {\rm d}x = \frac{\pi^3}{8} \).
Lemma 2: \( \displaystyle \int_0^{\pi/2} \ln^2 \sin \theta \, {\rm d}\theta = \frac{\pi^3}{24}+\frac{\pi \ln^2 2}{2} \)

Both of them have been done in the site. Search them in the Calculus section.

Hence, for our integral we have successively:

\begin{align*}
\int_{0}^{\infty}\frac{\log^2 \left ( x+ \frac{1}{x} \right )}{x^2+1}\, {\rm d}x &\overset{x=\tan \theta}{=\! =\! =\! =\! =\!}\int_{0}^{\pi/2}\frac{\log^2 \left ( \tan \theta + \frac{1}{
\tan \theta} \right )}{\tan^2 \theta+1}\sec^2 \theta \, {\rm d}\theta \\
&= \int_{0}^{\pi/2}\log^2 \left ( \tan \theta + \frac{1}{\tan \theta} \right )\, {\rm d}\theta\\
&= \int_{0}^{\pi}\log^2 \left [ \tan \theta \left ( \cot^2 \theta +1 \right ) \right ]\, {\rm d}\theta\\
&= \int_{0}^{\pi/2}\log^2 \tan \theta \, {\rm d}\theta + \int_{0}^{\pi/2} \log^2 \left ( \left ( \cot^2 \theta +1 \right ) \right )\, {\rm d}\theta \\
&=\frac{\pi^3}{8}+ \int_{0}^{\pi/2}\log^2 \csc^2 \theta \, {\rm d}\theta \\
&= \frac{\pi^3}{8}+ 2 \int_{0}^{\pi/2} \log^2 \sin \theta \, {\rm d} \theta \\
&= \frac{\pi^3}{8}+ \frac{\pi^3}{12}+ \pi \log^2 2 \\
&= \frac{5\pi^3}{24}+ \pi \log^2 2
\end{align*}
Imagination is much more important than knowledge.
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