How would one prove that:

$$\Im({\rm Li}_2(iz))={\rm Ti}_2(z)$$

where ${\rm Ti}_2$ is the inverse tangent integral.

Board index **‹** Special Functions **‹** An imaginary part
## An imaginary part

How would one prove that:

$$\Im({\rm Li}_2(iz))={\rm Ti}_2(z)$$

where ${\rm Ti}_2$ is the inverse tangent integral.

Hi VJKey,

It follows pretty much from the definition for $z \in \mathbb{R}$: \begin{align*}\operatorname{Li}_2(iz) &= -\int_0^{z} \frac{\log (1-ix)}{x}\,dx \\&= -\int_0^{z} \frac{\log \left((1+x^2)^{1/2}e^{-i\arctan x}\right)}{x}\,dx \\&= -\frac{1}{2}\int_0^{z} \frac{\log \left(1+x^2\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\&= -\frac{1}{4}\int_0^{-z^2} \frac{\log \left(1-x\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\&= \frac{1}{4}\operatorname{Li}_2(-z^2) + i\operatorname{Ti}_2(z)\end{align*}

**Moderator:** Shobhit

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$$\Im({\rm Li}_2(iz))={\rm Ti}_2(z)$$

where ${\rm Ti}_2$ is the inverse tangent integral.

It follows pretty much from the definition for $z \in \mathbb{R}$: \begin{align*}\operatorname{Li}_2(iz) &= -\int_0^{z} \frac{\log (1-ix)}{x}\,dx \\&= -\int_0^{z} \frac{\log \left((1+x^2)^{1/2}e^{-i\arctan x}\right)}{x}\,dx \\&= -\frac{1}{2}\int_0^{z} \frac{\log \left(1+x^2\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\&= -\frac{1}{4}\int_0^{-z^2} \frac{\log \left(1-x\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\&= \frac{1}{4}\operatorname{Li}_2(-z^2) + i\operatorname{Ti}_2(z)\end{align*}

Nicely done r9m. I had not thought that ${\rm Li}_2(iz)$ would have a real part. Thank you.

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