Inverse tangent integral

The inverse tangent integral is defined by:

${\displaystyle \operatorname {Ti} _{2}(x)=\int _{0}^{x}{\frac {\arctan t}{t}}\,dt}$

The arctangent is taken to be the principal branch; that is, −π/2 < arctan(t) < π/2 for all real t.^[1]

Its power series representation is

${\displaystyle \operatorname {Ti} _{2}(x)=x-{\frac {x^{3}}{3^{2}}}+{\frac {x^{5}}{5^{2}}}-{\frac {x^{7}}{7^{2}}}+\cdots }$

which is absolutely convergent for ${\displaystyle |x|\leq 1.}$^[1]

The inverse tangent integral is closely related to the dilogarithm ${\textstyle \operatorname {Li} _{2}(z)=\sum _{n=1}^{\infty }{\frac {z^{n}}{n^{2}}}}$ and can be expressed simply in terms of it:

${\displaystyle \operatorname {Ti} _{2}(z)={\frac {1}{2i}}\left(\operatorname {Li} _{2}(iz)-\operatorname {Li} _{2}(-iz)\right)}$

That is,

${\displaystyle \operatorname {Ti} _{2}(x)=\operatorname {Im} (\operatorname {Li} _{2}(ix))}$

for all real x.^[1]

The inverse tangent integral is an odd function:^[1]

${\displaystyle \operatorname {Ti} _{2}(-x)=-\operatorname {Ti} _{2}(x)}$

The values of Ti₂(x) and Ti₂(1/x) are related by the identity

${\displaystyle \operatorname {Ti} _{2}(x)-\operatorname {Ti} _{2}\left({\frac {1}{x}}\right)={\frac {\pi }{2}}\log x}$

valid for all x > 0 (or, more generally, for Re(x) > 0). This can be proven by differentiating and using the identity ${\displaystyle \arctan(t)+\arctan(1/t)=\pi /2}$.^[2][3]

The special value Ti₂(1) is Catalan's constant ${\textstyle 1-{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}-{\frac {1}{7^{2}}}+\cdots \approx 0.915966}$.^[3]

Similar to the polylogarithm ${\textstyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}$, the function

${\displaystyle \operatorname {Ti} _{n}(x)=\sum \limits _{k=0}^{\infty }{\frac {(-1)^{k}x^{2k+1}}{\left(2k+1\right)^{n}}}=x-{\frac {x^{3}}{3^{n}}}+{\frac {x^{5}}{5^{n}}}-{\frac {x^{7}}{7^{n}}}+\cdots }$

is defined analogously. This satisfies the recurrence relation:^[4]

${\displaystyle \operatorname {Ti} _{n}(x)=\int _{0}^{x}{\frac {\operatorname {Ti} _{n-1}(t)}{t}}\,dt}$

By this series representation it can be seen that the special values ${\displaystyle \operatorname {Ti} _{n}(1)=\beta (n)}$, where ${\displaystyle \beta (s)}$ represents the Dirichlet beta function.