I am working with the dilogarithm defined as
$$
Li_2(z) = -\int_0^1 ln(1 – zt)\frac{dt}{t}
\tag1$$
which is valid for $z\in\mathbb{C}$, while $t\in\mathbb{R}$
In this link , it is provided a paper where Eq. (1) is splitted into its real and imaginary parts as
$$
Li_2(z) = -\frac{1}{2}\int_0^1 ln(1 – 2tr\cos\phi + (tr)^2)\frac{dt}{t} + i\int_0^1 \arctan\left(\frac{tr\sin\phi}{1 – tr\cos\phi}\right)\frac{dt}{t}, \quad z = re^{i\phi}\in\mathbb{C},\ r\in\mathbb{R},\ \phi\in[0, 2\pi)
\tag2$$
My question is: if Eq. (1) is valid in $\mathbb{R}$ for being defined in general for $\mathbb{C}$, then for $\phi = 0$ Eq. (2) should be reduced to Eq. (1) with $z = r$, but what ou really get is
$$
Li_2(z = r) = -\frac{1}{2}\int_0^1 ln(1 – tr(2 -tr))\frac{dt}{t}, \quad r\in\mathbb{R}
$$
Isn't this a contradiction? Is maybe Eq. (2) wrong?
Best Answer
I don't know how you get your result. (You don't show your work.) Expanding the original integrand, using $\Re(z)$ for the real part of $z$ and $\Im(z)$ for the imaginary part, \begin{align*} R &= \Re(\ln(1-z t)) \\ &= \frac{1}{2}\ln( 1 - 2t \Re(z) + t^2 \Re(z)^2 + t^2 \Im(x)^2 ) \text{.} \end{align*}
If $z$ is real, $\Im(z) = 0$ and $\Re(z) = z$. So, \begin{align*} R &= \frac{1}{2}\ln( 1 - 2t z + t^2 z^2 ) \text{.} \end{align*} Similarly, if $z$ is nonnegative real and $r \geq 0$, $\phi = 0$ and if $z$ is nonpositive real and $r \leq 0$, $\phi = 0$. Either way, we get your $\phi = 0$, so $\cos \phi = 1$ and also $z = r$. Therefore, \begin{align*} R &= \frac{1}{2}\ln( 1 - 2t r\cos \phi + (t r)^2 ) \end{align*} and we find the real part integrand in eqn. (2) is what is expected form the integrand in eqn. (1) under the assumptions $r = z$ and $\phi = 0$.