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2 logarithmic Integral

Calculus (Integrals, Series)
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jacks
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2 logarithmic Integral

#1

Post by jacks » Wed Jul 13, 2016 5:41 am

$(a)\;$ Evaluation of $\displaystyle \int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$

$(b)\;$ Evaluation of $\displaystyle \int_{0}^{1}\frac{\ln(1+x^2)}{1+x}dx$
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Re: 2 logarithmic Integral

#2

Post by Grigorios Kostakos » Mon Aug 22, 2016 6:21 pm

jacks wrote:...$(b)\;$ Evaluation of $\displaystyle \int_{0}^{1}\frac{\ln(1+x^2)}{1+x}dx$
For the 2nd integral:

For $a>0$ we have \begin{align*}
I(a)&=\displaystyle{\int_0^1 \frac{\ln (1+ax^2)}{1+x}\, {\rm d}x}\quad \Longrightarrow\\
\frac{d}{da} I(a)&=\displaystyle{\int_0^1 \frac{x^2}{(1+ax^2)(1+x)}\, {\rm d}x}\\
&=\frac{1}{2}\,\frac{\log{1+a)}}{(1+a)a}-\frac{\arctan\big({\sqrt{a}\,}\big)}{(1+a)\sqrt{a}}+\log2\,\frac{1}{1+a}\quad \Longrightarrow\\
I(1)&=\frac{1}{2}\int_{0}^{1}{\frac{\log{(1+a)}}{(1+a)a}\,da}-\int_{0}^{1}{\frac{\arctan\big({\sqrt{a}\,}\big)}{(1+a)\sqrt{a}}\,da}+\int_{0}^{1}{\frac{\log{2}}{1+a}\,da}\\
&=\frac{\pi^2}{24}-\frac{\log^22}{4}-\frac{\pi^2}{16}+\log^22\\
&=\frac{3\log^22}{4}-\frac{\pi^2}{48}\,.
\end{align*}

Note: Evaluating $\displaystyle\int_{0}^{1}{\frac{\log{(1+a)}}{(1+a)a}\,da}\,,\; \int_{0}^{1}{\frac{\arctan\big({\sqrt{a}\,}\big)}{(1+a)\sqrt{a}}\,da}$ is quite easy.
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Tolaso J Kos
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Re: 2 logarithmic Integral

#3

Post by Tolaso J Kos » Mon Aug 22, 2016 6:48 pm

Grigorios Kostakos wrote: Note: Evaluating $\displaystyle\int_{0}^{1}{\frac{\log{(1+a)}}{(1+a)a}\,da}\,,\; \int_{0}^{1}{\frac{\arctan\big({\sqrt{a}\,}\big)}{(1+a)\sqrt{a}}\,da}$ is quite easy.
I am supplementing what Grigoris has left out.

\begin{align*}
\int_{0}^{1}\frac{\log (1+a)}{a(1+a)}\, {\rm d}a &=\int_{0}^{1}\frac{\log (1+a)}{a}\, {\rm d}a - \int_{0}^{1}\frac{\log (1+a)}{1+a}\, {\rm d}a \\
&= \int_{0}^{1}\frac{1}{a}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}a^n}{n}\, {\rm d}a - \int_{1}^{2}\frac{\log a}{a}\, {\rm d}a\\
&= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}- \left [ \frac{\log^2 a}{2} \right ]_1^2\\
&=\left ( 1-2^{1-2} \right )\zeta(2) - \frac{\log^2 2}{2} \\
&=\frac{\zeta(2)}{2} - \frac{\log^2 2}{2}
\end{align*}

For the other integral we substitute $y=\sqrt{a}$ and it comes to the form:

$$2\int_{0}^{1}\frac{\arctan a}{a^2+1}\, {\rm d}a = 2\left [ \frac{\arctan^2 a}{2} \right ]_0^1 = \frac{\pi^2}{16}$$

and it is over. Voilà, les jeux sont faits.
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Re: 2 logarithmic Integral

#4

Post by Tolaso J Kos » Wed Sep 28, 2016 12:21 pm

jacks wrote:$(a)\;$ Evaluation of $\displaystyle \int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$
For the first integral and somewhat quick. The solution will actually involve poylogarithms and basic complex things.

\begin{align*}
\int_{0}^{\sqrt{2}-1} \frac{\ln \left ( 1+x^2 \right )}{1+x} \, {\rm d}x&= \int_{0}^{\sqrt{2}-1} \frac{\ln \left ( 1-i^2 x^2 \right )}{1+x} \, {\rm d}x\\
&= \int_{0}^{\sqrt{2}-1} \frac{\ln (1-ix) + \ln (1+ix)}{1+x} \, {\rm d}x\\
&= \int_{0}^{\sqrt{2}-1} \frac{\ln (1-ix)}{1+x} \, {\rm d}x + \int_{0}^{\sqrt{2}-1} \frac{\ln (1+ix)}{1+x} \, {\rm d}x\\
&=-{\rm Li}_2 \left ( \frac{1-i}{2} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) + \frac{i}{4} \pi \ln \left [ \left ( 1+i \right ) - i \sqrt{2} \right ] \\
& \quad \; -{\rm Li}_2 \left ( \frac{1+i}{2} \right ) + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) - \frac{i}{4} \pi \ln \left [ (1-i) + i \sqrt{2} \right ] \\
&= - \left [ -i \mathcal{G} + \frac{\pi^2}{48} - \frac{1}{2}\left ( -\frac{\ln 2}{2} + \frac{i \pi}{4} \right )^2 \right ] + {\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \\
&\quad \; -\left [ i \mathcal{G} + \frac{\pi^2}{48} - \frac{1}{2} \left ( -\frac{\ln 2}{2} - \frac{i \pi}{4} \right )^2 \right ] +{\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) + \frac{\pi^2}{16} \\
&= \frac{\ln^2 2}{4} - \frac{5 \pi^2}{48} + \frac{\pi^2}{16} + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \\
&= \frac{\ln^2 2}{4} - \frac{\pi^2}{24} + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \approx 0.0173049
\end{align*}

Side note: One interesting identity regarding the dilog function is the following:

$$\Re{ \left[{\rm Li}_{2}\left(\frac{1}{2}+iq\right) \right]}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln^2{\left(\frac{1+4q^2}{4}\right)}}-\frac{{\arctan^2{(2q)}}}{2}$$

where $q \in \mathbb{Q}$ and there is a similar one for the imaginary part which I will post later.

Addendum: In general it holds that:

\begin{align*}
\int_{0}^{1} \frac{\log (1+ax)}{1+x} \, {\rm d}x &\overset{u=1+x}{=\! =\! =\!} \int_{1}^{2} \frac{\log \left ( 1-a+au \right )}{u} \, {\rm d}u \\
&=\int_{1}^{2} \frac{\log \left [ \left ( 1-a \right ) \left ( 1+ \frac{a}{1+a} u \right ) \right ]}{u} \, {\rm d}u \\
&= \log \left ( 1-a \right ) \int_{1}^{2} \frac{{\rm d}u}{u} +\int_{1}^{2} \frac{\log \left ( 1+\frac{a}{1-a} u \right )}{u} \, {\rm d}u \\
&= \log 2 \log (1-a) - \sum_{n=1}^{\infty} \frac{1}{n} \left ( -\frac{a}{1-a} \right )^n\int_{1}^{2} u^{n-1} \, {\rm d}u\\
&= \log 2 \log (1-a) - \sum_{n=1}^{\infty} \frac{1}{n^2} \left ( -\frac{a}{1-a} \right )^n \left ( 2^n-1 \right ) \\
&= \log 2 \log (1-a) -{\rm Li}_2 \left ( -\frac{2a}{1-a} \right ) + {\rm Li}_2 \left ( -\frac{a}{1-a} \right )
\end{align*}

Hence we conclude that:

$$\bbox[blue, 2pt]{{\color{white}{\int_{0}^{1}\frac{\log (1+ax)}{1+x} \, {\rm d}x = \log 2 \log (1-a) -{\rm Li}_2 \left ( -\frac{2a}{1-a} \right ) + {\rm Li}_2 \left ( - \frac{a}{1-a} \right )}}}$$

for $a \neq 1$, while for $a=1$ we get that $\displaystyle \int_0^1 \frac{\log (1+x)}{1+x}\, {\rm d}x = \frac{\log^2 2}{2}$.
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Re: 2 logarithmic Integral

#5

Post by Riemann » Wed Sep 28, 2016 8:13 pm

Maybe the following are worth deriving.

1. $\displaystyle \Im\left [ {\rm Li}_2 (i) \right ] =\mathcal{G}$ since in general

\begin{align*}
{\rm Li}_2 (iz) &= - \int_{0}^{z} \frac{\log (1-it)}{t} \, {\rm d}t \\
&= -\int_{0}^{z} \frac{\log \left [ \left ( 1+t^2 \right )^{1/2} e^{-i \arctan t} \right ]}{t} \, {\rm d}t\\
&= -\frac{1}{2} \int_{0}^{z}\frac{\log \left ( 1+t^2 \right )}{t} \, {\rm d}t + i \int_{0}^{z} \frac{\arctan t}{t} \, {\rm d}t\\
&= \frac{1}{4}{\rm Li}_2 \left ( -z^2 \right ) + i {\rm Ti}_2 (z)
\end{align*}

Now for $z=1$ we have that ${\rm Ti}_2(1)=\mathcal{G}$ and thus the first equation follows.

2. $\displaystyle \Im \left [ {\rm Li}_2 \left ( 1+i \right ) \right ] = \mathcal{G} + \frac{\pi \log 2}{4}$. This pretty much follows from the fundamental equation the dilogarithm function satisfies, i.e

\begin{equation} {\rm Li}_2(z)+ {\rm Li}_2(1-z) = \zeta(2) - \log z \log (1-z) \end{equation}

For $z=-i$ we get the following

\begin{equation} {\rm Li}_2 \left ( -i \right ) + {\rm Li}_2 \left ( 1+i \right ) = \zeta(2) - \log (-i) \log (1+i) \end{equation}

Taking imaginary parts as well as into account the following fact

\begin{align*}
{\rm Li}_2 (-i) + {\rm Li}_2 \left ( 1+i \right ) &= \zeta(2) - \log (-i) \log (1+i) \\
&=\zeta(2) -\frac{\pi^2}{8} +\frac{i \pi \log 2 }{4}
\end{align*}

yields the result , since $\Im \left({\rm Li}_2(-i) \right)=-\mathcal{G}$.

3. $\displaystyle \Im \left [ {\rm Li}_2 \left ( \frac{1+i}{2} \right ) \right ] = \mathcal{G} - \frac{\pi \log 2}{8}$.

Well, we are using another fundamental relation of the dilog, namely:

\begin{equation} {\rm Li}_2 (z) +{\rm Li}_2 \left ( -\frac{z}{1-z} \right ) = - \frac{1}{2} \log^2 \left ( 1-z \right ) \; , \quad z \notin [1, +\infty) \end{equation}

as well as the trivial results

\begin{align}
{\rm Li}_2 \left ( e^{i \theta} \right ) &= {\rm Sl}_2 (\theta) + i {\rm Cl}_2 (\theta) \\
{\rm Sl}_2(\theta)&= \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^2} \\
{\rm Cl}_2(\theta)&= \sum_{n=1}^{\infty} \frac{\cos n \theta}{n^2} = \zeta(2) - \frac{\pi \theta }{2} + \frac{\theta^2}{4}
\end{align}

(Note: Equation $(6)$ is just a Fourier series. Well, $\displaystyle \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \frac{\pi-\theta}{2} , \; \theta \in (0, 2\pi)$. Since the convergence of that series is uniform we can integrate and get the result. )

Now for $z=\frac{1}{2} + \frac{i}{2}$ we get that

\begin{align*}
{\rm Li}_2 \left ( \frac{1+i}{2} \right ) + {\rm Li}_2 (-i) &= -\frac{1}{2} \log^2 \left ( 1-\frac{1+i}{2} \right )\\
&=\frac{\pi^2}{32} - \frac{\log^2 2}{8} - \frac{i \pi \log 2}{8}
\end{align*}

Taking imaginary part yields the result. As a side note $\displaystyle \Im \left [ {\rm Li}_2 \left ( \frac{1-i}{2} \right ) \right ] = -\mathcal{G} + \frac{\pi \log 2}{8}$.

Note: Did you think that these special values had no real part? Well, you're mistaken. Here there are:

\begin{align}
\Re \left ( {\rm Li}_2 (i) \right ) &= - \frac{\pi^2}{48}\\
\Re \left ( {\rm Li}_2(-i) \right )&= -\frac{\pi^2}{48}\\
\Re \left [ {\rm Li}_2 \left ( \frac{1+i}{2} \right ) \right ]&= \frac{5\pi^2}{96} - \frac{\log^2 2}{8} \\
\Re \left [ {\rm Li}_2 \left ( \frac{1-i}{2} \right ) \right ] &=\frac{5 \pi^2}{96} - \frac{\log^2 2}{8}
\end{align}

Maybe make a tutorial on this and complete it with the special trilog. values.
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