Real Part of the Dilogarithm
It is well known that
$$frac{x-pi}{2}=-sum_{kgeq 1}frac{sin{kx}}{k}forall xin(0,tau),$$
which gives
$$frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}=sum_{kgeq 1}frac{cos(kx)}{k^2}.$$
Note that
$$textrm{Li}_2(e^{ix})=sum_{kgeq 1}frac{cos(kx)+isin(kx)}{k^2}$$
This means that
$$mathfrak{R}textrm{Li}_2(e^{ix})=frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}$$
unless I'm wrong on one of the above statements. Now, with the previous restrictions on $x$, we have that this formula is valid for all complex numbers lying on the complex unit circle as inputs. My question is: is there a way to define something along these lines (a finite degree polynomial, preferably) for complex inputs of the dilogarithm that don't necessarily lie on the complex unit circle? More specifically,
$$mathfrak{R}textrm{Li}_2(re^{ix})=?$$
complex-analysis fourier-series polylogarithm
asked Oct 9 at 0:58
46andpi
788
add a comment |
It is well known that
$$frac{x-pi}{2}=-sum_{kgeq 1}frac{sin{kx}}{k}forall xin(0,tau),$$
which gives
$$frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}=sum_{kgeq 1}frac{cos(kx)}{k^2}.$$
Note that
$$textrm{Li}_2(e^{ix})=sum_{kgeq 1}frac{cos(kx)+isin(kx)}{k^2}$$
This means that
$$mathfrak{R}textrm{Li}_2(e^{ix})=frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}$$
unless I'm wrong on one of the above statements. Now, with the previous restrictions on $x$, we have that this formula is valid for all complex numbers lying on the complex unit circle as inputs. My question is: is there a way to define something along these lines (a finite degree polynomial, preferably) for complex inputs of the dilogarithm that don't necessarily lie on the complex unit circle? More specifically,
$$mathfrak{R}textrm{Li}_2(re^{ix})=?$$
complex-analysis fourier-series polylogarithm
asked Oct 9 at 0:58
46andpi
788
add a comment |
It is well known that
$$frac{x-pi}{2}=-sum_{kgeq 1}frac{sin{kx}}{k}forall xin(0,tau),$$
which gives
$$frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}=sum_{kgeq 1}frac{cos(kx)}{k^2}.$$
Note that
$$textrm{Li}_2(e^{ix})=sum_{kgeq 1}frac{cos(kx)+isin(kx)}{k^2}$$
This means that
$$mathfrak{R}textrm{Li}_2(e^{ix})=frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}$$
unless I'm wrong on one of the above statements. Now, with the previous restrictions on $x$, we have that this formula is valid for all complex numbers lying on the complex unit circle as inputs. My question is: is there a way to define something along these lines (a finite degree polynomial, preferably) for complex inputs of the dilogarithm that don't necessarily lie on the complex unit circle? More specifically,
$$mathfrak{R}textrm{Li}_2(re^{ix})=?$$
complex-analysis fourier-series polylogarithm
asked Oct 9 at 0:58
46andpi
788
It is well known that
$$frac{x-pi}{2}=-sum_{kgeq 1}frac{sin{kx}}{k}forall xin(0,tau),$$
which gives
$$frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}=sum_{kgeq 1}frac{cos(kx)}{k^2}.$$
Note that
$$textrm{Li}_2(e^{ix})=sum_{kgeq 1}frac{cos(kx)+isin(kx)}{k^2}$$
This means that
$$mathfrak{R}textrm{Li}_2(e^{ix})=frac{x^2}{4}-frac{pi x}{2}+frac{pi^2}{6}$$
unless I'm wrong on one of the above statements. Now, with the previous restrictions on $x$, we have that this formula is valid for all complex numbers lying on the complex unit circle as inputs. My question is: is there a way to define something along these lines (a finite degree polynomial, preferably) for complex inputs of the dilogarithm that don't necessarily lie on the complex unit circle? More specifically,
$$mathfrak{R}textrm{Li}_2(re^{ix})=?$$
complex-analysis fourier-series polylogarithm
complex-analysis fourier-series polylogarithm
asked Oct 9 at 0:58
46andpi
788
asked Oct 9 at 0:58
46andpi
788
asked Oct 9 at 0:58
46andpi
788
asked Oct 9 at 0:58
46andpi
788
asked Oct 9 at 0:58
46andpi
788
788
add a comment |
add a comment |
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