[SOLVED]A generalized fractional logarithm integral

ZaidAlyafey

Well-known member
MHB Math Helper
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 .

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ZaidAlyafey

Well-known member
MHB Math Helper
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 .

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ZaidAlyafey

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MHB Math Helper
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
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
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
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
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
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
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 .

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ZaidAlyafey

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MHB Math Helper
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
I think posting this concludes the thread

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