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I'm interested in this integral: $\int_{{\frac {\pi}{8}}}^{{\frac {7\,\pi}{8}}}\!{\frac {\ln \left( 1- \cos \left( t \right) \right) }{\sin \left( t \right) }}\,{\rm d}t$

I found this particular closed form with Maple and finally : $-{\frac { \left( \ln \left( 2-\sqrt {2+\sqrt {2}} \right) \right) ^{ 2}}{2}}-{\frac {{\pi}^{2}}{12}}+{\frac {11\, \left( \ln \left( 2 \right) \right) ^{2}}{8}}-{\frac { \left( \ln \left( 1+\sqrt {2} \right) \right) ^{2}}{2}}+{\frac {3\, \left( \ln \left( 2+\sqrt {2} \right) \right) ^{2}}{4}}+{\it {Li_2}} \left( -{\frac {\sqrt {2+ \sqrt {2}}}{4}}+{\frac{1}{2}} \right) $, where \Li_2 is the dilogarithm function.

Please, can someone prove it ?

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  • $\begingroup$ Start with the property of definite integrals that $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ $\endgroup$
    – RAHUL
    Nov 8, 2021 at 8:23

1 Answer 1

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$$I=\int{\frac {\log \left( 1- \cos \left( t \right) \right) }{\sin \left( t \right) }}\,dt=\int \sin(t){\frac {\log \left( 1- \cos \left( t \right) \right) }{1-\cos^2\left( t \right) }}\,dt$$

Let $$x=\cos(t)\implies I=\int \frac{\log(1-x)}{x^2-1} \,dx=-\frac12\int\frac{\log(1-x)}{ 1-x}\,dx-\frac12\int\frac{\log(1-x)}{ x+1}\,dx$$

$$\int\frac{\log(1-x)}{ 1-x}\,dx=-\frac{1}{2} \log ^2(1-x)$$ For the second integral, integration by parts gives $$\int\frac{\log(1-x)}{x+1}\,dx=\text{Li}_2\left(\frac{1-x}{2}\right)+\log (1-x) \log \left(\frac{x+1}{2}\right)$$ Combining the results

$$I=\int \frac{\log(1-x)}{x^2-1} \,dx=\frac{1}{4} \left(\log (1-x) (\log (4(1-x))-2 \log (x+1))-2 \text{Li}_2\left(\frac{1-x}{2}\right)\right)$$ Go back to $t$ if you wish and use bounds.

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  • $\begingroup$ So great and thanks ! $\endgroup$
    – Dens
    Nov 8, 2021 at 10:40
  • $\begingroup$ @Dens. Glad to help ! In fact the problem is simple if you think about the first trick $$\frac 1 {\sin(x)}=\frac {\sin(x)}{\sin^2(x)}=\frac {\sin(x)}{1-\cos^2(x)}$$ $\endgroup$ Nov 8, 2021 at 10:49
  • $\begingroup$ Merci bien pour le conseil ! $\endgroup$
    – Dens
    Nov 8, 2021 at 10:50
  • $\begingroup$ If I may ask, where are you located ? I am in Pau. Cheers :-) $\endgroup$ Nov 8, 2021 at 10:52
  • $\begingroup$ Je suis francais et je me trouve en Saône et Loire.D'ailleurs, je vous avais écris au sujet d'une publication sur Vixra << values of barnes function >>. C'est moi ! $\endgroup$
    – Dens
    Nov 8, 2021 at 10:55

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