MHB Logarithmic Integral on Stack Exchange - author unknown

Mathematics news on Phys.org
Consider the generalized parametric case where $$0 < z \le 1$$:$$\mathcal{I}(z)=\int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx$$Substitute $$y=1+x$$ to obtain:$$\int_1^{1+z}\frac{\log y\log[1-(y-1)]}{y}\,dy=\int_1^{1+z}\frac{\log y\log(2-y)}{y}\,dy=$$$$\int_1^{1+z}\frac{\log y\log\left[2 \left(1-\frac{y}{2} \right) \right]}{y}\,dy=$$$$\log 2\, \int_1^{1+z}\frac{\log y}{y}\,dy+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$Performing an integration by parts on that first integral gives:$$\frac{1}{2}\log 2\, (\log y)^2\, \Bigg|_1^{1+z}=\frac{1}{2}\log 2\log^2(1+z)$$So$$\mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$Next, we split that last integral into two:$$\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=
\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy-\int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$For $$0 < z \le 1$$, Polylogarithms of order $$m \ge 1$$ have the integral representation:$$\text{Li}_m(z)=\frac{(-1)^{m-1}}{(m-2)!}\int_0^1\frac{(\log x)^{m-2}\log(1-zx)}{x}\,dx$$Hence$$\int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=\text{Li}_3\left( \tfrac{1}{2}\right) $$So$$\mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) +\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$For that final integral, apply the substitution $$y=(1+z)\, x$$ to change it into:$$\int_0^1\frac{\log[(1+z)\, x]\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx=$$$$\log(1+z)\, \int_0^1\frac{\log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx+\int_0^1\frac{\log x\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx$$By the integral representation for arbitrary (non-zero) order Polylogs given above, this equates to:$$-\log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)$$
------------------------------
General parametric solution:

$$\int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx=$$$$\frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) - \log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)$$For $$\, \, 0 < z \le 1$$
 
Nice solution . I always consider differentiating the hypergeometric function to be an alternative solution even though not easy .
 
ZaidAlyafey said:
Nice solution . I always consider differentiating the hypergeometric function to be an alternative solution even though not easy .
Thanks Zaid! (Sun)

It can get a bit - erm - 'hairy' when your dealing with higher order integrals of the type above, but all integrals of the form $$\int_0^z\frac{\log^m(1+x)\log(1-x)}{(1+x)},dx$$and $$\int_0^z\frac{\log^m(1-x)\log(1+x)}{(1-x)},dx$$can be done in the same way. The closed form involves some pretty annoying (finite!) double sums (possibly a triple sum too)... I'll add this onto the Logarithmic Integrals thread at some point (soon). It's been "in the post" for a while... ;)
 
Last edited:
That seems interesting , I'll be waiting to see that .
 
DreamWeaver said:
Thanks Zaid! (Sun)

It can get a bit - erm - 'hairy' when your dealing with higher order integrals of the type above, but all integrals of the form $$\int_0^z\frac{\log^m(1+x)\log(1-x)}{(1+x)},dx$$and $$\int_0^z\frac{\log^m(1-x)\log(1+x)}{(1-x)},dx$$can be done in the same way. The closed form involves some pretty annoying (finite!) double sums (possibly a triple sum too)... I'll add this onto the Logarithmic Integrals thread at some point (soon). It's been "in the post" for a while... ;)

ZaidAlyafey said:
That seems interesting , I'll be waiting to see that .

Hi DreamWeaver,

I see it now that you have got a fan, congrats! (Sun)And I envy you!:p
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top