Welcome to our community

Be a part of something great, join today!

[SOLVED] A generalized fractional logarithm integral

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I came up with the following integral

\(\displaystyle I(t,a) = \int^t_0 \frac{\log( x^2+a^2)}{1+x}\, dx \)

Here we have an attempt to solve the integral succeeded by chisigma for the particular case \(\displaystyle I(1,1)\) , I don't now whether there is a closed form for the integral we can start by a simplified version \(\displaystyle I(1,a)\) , hopefully by the end of this thread we have what we are seeking for . All attempts what so ever are always appreciated and welcomed .
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Here is an integral I believe will help us on the process

\(\displaystyle
\begin{align*}
\int^x_0 \frac{\log(1+t)}{1-t}\, dt&= \int^{1}_{1-x} \frac{\log(2-t)}{t}\, dt\,\, \,\,\text{For} \,\, 0\leq x<1 \\ &=\int^{1}_{1-x} \frac{\log (2)+\log\left(1-\frac{t}{2}\right)}{t}\, dt \\
&=\int^{1}_{1-x} \frac{\log (2)}{t}+\int^{1}_{1-x} \frac{\log\left(1-\frac{t}{2}\right)}{t}\, dt \\
&=-\log(1-x)\log(2)+\int^{\frac{1}{2}}_{\frac{1-x}{2}} \frac{\log(1-t)}{t}\, dt \\
&=-\log(1-x)\log(2)- \text{Li}_2 \left(\frac{1}{2} \right) +\text{Li}_2 \left(\frac{1-x}{2}\right) \\
\end{align*}
\)

I believe that the process of solving the general formula will involve some complex numbers , I hope someone will cover for me if I make mistakes .
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Here is an attempt for the special case \(\displaystyle I(1,a)\) where $a$ is real and $0\leq a<1$

we start by the following

\(\displaystyle I(a) = \int^1_0 \frac{\log(1+ax)}{1+x}\, dx \)

\(\displaystyle
\begin{align*}
I'(a) = \int^1_0 \frac{x}{(1+x)(1+ax)}\, dx &= \frac{1}{1-a} \left(\int^1_0\frac{1}{(1+ax)}\, dx -\int^1_0 \frac{1}{(1+x)}\right)\\
&= \frac{1}{1-a} \left(\frac{1}{a} \log(1+a)-\log(2) \right)\\
&= \frac{\log(1+a)}{a(1-a)}-\frac{\log(2)}{a-1} \\
&= \frac{\log(1+a)}{1-a}+\frac{\log(1+a)}{a} -\frac{\log(2)}{1-a} \\

\end{align*}
\)

\(\displaystyle I(a)=-\text{Li}_2 \left(\frac{1}{2} \right) +\text{Li}_2 \left(\frac{1-a}{2}\right)-\text{Li}_2(-a)+C\)


using \(\displaystyle I(0)=0\) we obtain \(\displaystyle C=0\)


\(\displaystyle I(a) = \int^1_0 \frac{\log(1+ax)}{1+x}\, dx =- \text{Li}_2 \left(\frac{1}{2} \right) +\text{Li}_2 \left(\frac{1-a}{2}\right) -\text{Li}_2(-a)\,\,\,\, \)


\(\displaystyle I(ia)+I(-ia) = \int^1_0 \frac{\log(1+a^2x^2)}{1+x}\, dx\)

\(\displaystyle \int^1_0 \frac{\log(1+a^2x^2)}{1+x}\, dx = -2\text{Li}_2 \left(\frac{1}{2} \right) +\text{Li}_2 \left(\frac{1-ai}{2}\right)+ \text{Li}_2 \left(\frac{1+ai}{2}\right) -\text{Li}_2(-ai)-\text{Li}_2(ai)\)

First we solve the following

\(\displaystyle
\begin{align*}
\text{Li}_2 \left(\frac{1-ai}{2}\right)+ \text{Li}_2 \left(\frac{1+ai}{2}\right) &= \frac{\pi^2}{6}- \log \left(\frac{1-ai}{2} \right) \log \left(\frac{1+ai}{2} \right)\\
&=\frac{\pi^2}{6}- \left( \frac{1}{2}\log \left( \frac{1+a^2}{4} \right)- i \arctan (a) \right) \left( \frac{1}{2}\log\left( \frac{1+a^2}{4} \right)+ i\arctan(a) \right)\\
&= \frac{\pi^2}{6}-\frac{1}{4}\log^2 \left( \frac{1+a^2}{4} \right) - \arctan^2 ( a)
\end {align*}
\)

Secondly we solve the following

\(\displaystyle \text{Li}_2(-ai)+\text{Li}_2(ai) = \frac{1}{2} \text{Li}_2 \left( -a^2\right)\)


Collecting the results together we obtain


\(\displaystyle \int^1_0 \frac{\log(1+a^2x^2)}{1+x}\, dx =-2\text{Li}_2 \left(\frac{1}{2} \right)+\frac{\pi^2}{6}-\frac{1}{4} \log^2 \left( \frac{1+a^2}{4} \right) - \arctan^2 ( a)-\frac{1}{2} \text{Li}_2 \left( -a^2\right)\)

\(\displaystyle \int^1_0 \frac{\log(a^2+x^2)}{1+x}\, dx =-2 \log(a) \log(2) -2\text{Li}_2 \left(\frac{1}{2} \right)+\frac{\pi^2}{6}-\frac{1}{4}\log^2 \left( \frac{1+a^2}{4a^2} \right) - \arctan^2 \left(\frac{1}{a} \right)-\frac{1}{2} \text{Li}_2 \left( \frac{-1}{a^2}\right)\)

For the special case \(\displaystyle I(1,1)\) we have

\(\displaystyle
\begin{align*}
\int^1_0 \frac{\log(1+x^2)}{1+x}\, dx &=-2\text{Li}_2 \left(\frac{1}{2} \right)+\frac{\pi^2}{6}-\frac{1}{4}\log^2 \left( \frac{1}{2} \right) - \frac{\pi^2}{16}-\frac{1}{2} \text{Li}_2 \left( -1 \right)\\
&= \frac{3}{4}\log^2(2) - \frac{\pi^2}{48}
\end{align*}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
To solve for the general form we need to generalize the integral on #2

\(\displaystyle
\begin{align*}
\int^x_0 \frac{\log(1+at)}{1-t}\, dt&= \int^{1}_{1-x} \frac{\log(1+a-at)}{t}\, dt \\ &=\int^{1}_{1-x} \frac{\log (1+a)+\log\left(1-\frac{at}{1+a}\right)}{t}\, dt \\
&=\int^{1}_{1-x} \frac{\log (1+a)}{t}+\int^{\frac{a}{a+1}}_{\frac{a(1-x)}{a+1}} \frac{\log\left(1-t \right)}{t}\, dt \\

&=-\log(1-x)\log(1+a)- \text{Li}_2 \left( \frac{a}{a+1} \right) +\text{Li}_2 \left(\frac{a-ax}{a+1}\right) \\
\end{align*}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I should now post the full solution , the process is similar to that of post #3


\(\displaystyle I(a) = \int^t_0 \frac{\log(1+ax)}{1+x}\, dx \)

\(\displaystyle
\begin{align*}
I'(a) = \int^t_0 \frac{x}{(1+x)(1+ax)}\, dx &= \frac{1}{1-a} \left(\int^t_0\frac{1}{(1+ax)}\, dx -\int^t_0 \frac{1}{(1+x)}\right)\\
&= \frac{1}{1-a} \left(\frac{1}{a} \log(1+at)-\log(1+t) \right)\\
&= \frac{\log(1+at)}{a(1-a)}-\frac{\log(1+t)}{a-1} \\
&= \frac{\log(1+at)}{1-a}+\frac{\log(1+at)}{a} -\frac{\log(1+t)}{1-a} \\

\end{align*}
\)

\(\displaystyle I(a)=- \text{Li}_2 \left( \frac{t}{t+1} \right) +\text{Li}_2 \left(\frac{t-ta}{t+1}\right)-\text{Li}_2(-at) +C\)


using \(\displaystyle I(0)=0\) we obtain \(\displaystyle C=0\)


\(\displaystyle I(a) = \int^t_0 \frac{\log(1+ax)}{1+x}\, dx = - \text{Li}_2 \left( \frac{t}{t+1} \right) +\text{Li}_2 \left(\frac{t-ta}{t+1}\right)-\text{Li}_2(-at) \)


\(\displaystyle I(ia)+I(-ia) = \int^t_0 \frac{\log(1+a^2x^2)}{1+x}\, dx\)

\(\displaystyle \int^t_0 \frac{\log(1+a^2x^2)}{1+x}\, dx = -2\text{Li}_2 \left( \frac{t}{t+1} \right) +\text{Li}_2 \left(\frac{t-tai}{t+1}\right)+\text{Li}_2 \left(\frac{t+tai}{t+1}\right)-\text{Li}_2(-ait) -\text{Li}_2(ait) \)

we can simplify a little bit to get

\(\displaystyle \int^t_0 \frac{\log(1+a^2x^2)}{1+x}\, dx = -2\text{Li}_2 \left( \frac{t}{t+1} \right) +\text{Li}_2 \left(\frac{t-tai}{t+1}\right)+\text{Li}_2 \left(\frac{t+tai}{t+1}\right)-\frac{1}{2}\text{Li}_2(-a^2t^2) \)


I still don't know if I can simplify

\(\displaystyle \text{Li}_2 \left(\frac{t-tai}{t+1}\right)+\text{Li}_2 \left(\frac{t+tai}{t+1}\right)\)

Actually I am looking for a general formula for the following

\(\displaystyle \text{Li}_2(z) + \text{Li}_2( \bar{z}) \) \(\displaystyle \,\,\, \text{For} \,\,\,0 \leq \text{Re}(z) \leq 1\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
First we shall treat Dilogarithms that involve complex numbers

we start by the following

\(\displaystyle
\begin{align*}
\text{Li}_2(i) &= \sum_{k\geq 1} \frac{i^k}{k^2}\\
&= 1+\frac{i}{2^2}-\frac{1}{3^2}+\frac{1}{4^2}+\cdots \\
&= i-\frac{1}{2^2}-\frac{i}{3^2}+\frac{1}{4^2}+\cdots \\
&= -\frac{1}{4}\left( 1-\frac{1}{2^2}+\frac{1}{4^2}+\cdots \right) +i\left(1-\frac{1}{3^2}+\frac{1}{5^2}+\cdots \right) \\
&=\frac{-1}{4}\sum_{k\geq 1}\frac{(-1)^{k-1}}{k^2}+i \sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^2}\\
&=-\frac{\pi^2}{48}+iG\,\,\,\, \text{where G is the Catalan's constant }
\end{align*}
\)

G : Catalan's constant
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
We shall now evaluate \(\displaystyle \text{Li}_2(1+i)\)

\(\displaystyle
\begin{align*}

\text{Li}_2(1+i)&= -\int^{1+i}_0 \frac{\log(1-x)}{x}\, dx \\
&=-\int^{1}_{-i} \frac{\log(x)}{1-x}\, dx \\
&=\log(-i)\log(1+i)-\int^{1}_{-i}\frac{\log(1-x)}{x}\, dx \\
&=\frac{i \pi}{2}\left( \log(\sqrt{2}+i\frac{\pi}{4}\right)+\text{Li}_2(1)-\text{Li}_2(-i)\\
&=\frac{i \pi \log(2)}{4}-\frac{\pi^2}{8}+\text{Li}_2(1)+\text{Li}_2(i)-\frac{1}{2}\text{Li}_2(-1)\\
&=\frac{i \pi \log(2)}{4}-\frac{\pi^2}{8}+\frac{\pi^2}{6}-\frac{\pi^2}{48}+iG+\frac{\pi^2}{24}\\
&=\frac{\pi^2}{16}+ \left( \frac{\pi \log(2)}{4}+G\right)i
\end{align*}
\)

Similiarliy we conclude that

\(\displaystyle \text{Li}_2(1-i)=\overline{\text{Li}_2(1+i)}=\frac{\pi^2}{16}- \left( \frac{\pi \log(2)}{4}+G\right)i \)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Conjuncture

  • \(\displaystyle \overline{\text{Li}_2(z)}=\text{Li}_2(\bar{z})\)

Corollary

  • \(\displaystyle \frac{\text{Li}_2(z)+\text{Li}_2(\bar{z})}{2} = \text{Re} \left( \text{Li}_2(z)\right)\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
We can prove the conjuncture for purely imaginary complex numbers

Consider the following with \(\displaystyle x\in \mathbb{R}\)

\(\displaystyle
\begin{align*}
\text{Li}_2(ix) &= -\int^{ix}_0 \frac{\log(1-t)}{t}\,dt\\

&= -\int^x_0 \frac{\log(1-it)}{t}\,dt\\
&= -\int^x_0 \frac{\log(\sqrt{1+t^2})}{t}\,dt+ i \int^x_0 \frac{\arctan(t)}{t}\, dt \\

\end{align*}

\)

Similarly we can show that

\(\displaystyle
\begin{align*}
\text{Li}_2(-ix) &= -\int^{-ix}_0 \frac{\log(1-t)}{t}\,dt\\

&= -\int^x_0 \frac{\log(1+it)}{t}\,dt\\
&= -\int^x_0 \frac{\log(\sqrt{1+t^2})}{t}\,dt- i \int^x_0 \frac{\arctan(t)}{t}\, dt \\

\end{align*}

\)

Hence \(\displaystyle \text{Li}_2(ix) = \overline{\text{Li}_2(-ix)}\)

I shall try to generalize for any complex number $z$ in the next post .
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Now to generalize , consider the complex number \(\displaystyle z\)

\(\displaystyle
\begin{align*}

\text{Li}_2(z) &= -\int^z_0 \frac{\log(1-t)}{t}\, dt \\

&= -\int^{|z|^2}_0 \frac{ \log \left(1-\frac{t}{\overline{z}} \right)}{t}\, dt \\

\end{align*}
\)

On the other hand

\(\displaystyle
\begin{align*}

\text{Li}_2(\overline{z}) &= -\int^{\overline{z}}_0 \frac{\log(1-t)}{t}\, dt \\

&= -\int^{|z|^2}_0 \frac{ \log \left(1-\frac{t}{z} \right)}{t}\, dt \\

\end{align*}
\)

Now consider

\(\displaystyle
\begin{align*}

\log \left(1-\frac{t}{\overline{z}} \right) &= \log \left(1-\frac{z t}{|z|^2} \right) \\

\end{align*}
\)


\(\displaystyle
\begin{align*}

\log \left(1-\frac{t}{z} \right) &= \log \left(1-\frac{\overline{z} t}{|z|^2} \right) \\

\end{align*}
\)

Clearly the two complex functions have the same real part and opposite imaginary parts because of the oddness of the \(\displaystyle \arctan \) function which proves our conjecture.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I think posting this concludes the thread

\(\displaystyle \int^t_0 \frac{\log(1+a^2x^2)}{1+x}\, dx = -2\text{Li}_2 \left( \frac{t}{t+1} \right) +2\, \text{Re} \left( \text{Li}_2 \left(\frac{t+tai}{t+1}\right) \right)-\frac{1}{2}\text{Li}_2(-a^2t^2)\)