# Integral of $\arctan(2e^{-x-2})$

I need some help with this integral

$$\int\arctan(2e^{-x-2})\,dx$$

I actually stumbled about this function in school and still haven't found a way to solve it properly. I taught myself the basics of integration involving all the common methods like integration by parts or substitution and also know some important substitutions of the latter one. Would appreciate if you guys could give tips for this :) even if they are beyond my horizons

• Pretty sure this integral cannot be expressed in terms of elementary functions. – Simon S Nov 12 '15 at 18:08
• Maple does the answer in terms of dilogarithm. – GEdgar Nov 12 '15 at 18:14
• Let $f(x)=\arctan\left(\alpha e^{-x}\right)$ (in your case $a=2e^{-2}$. Let $u=ae^{-x}$. Then $du=-u\,dx$. Then: $$\int f(x)\,dx = -\int \frac{\arctan(u)}{u}\,du$$ That's probably easier to read, in any event. – Thomas Andrews Nov 12 '15 at 18:19
• this integral is quite "famous" and has it's own name: a legendre-$\chi$ function of imaginary argument. :) – tired Nov 12 '15 at 18:24
• Are you really interested in a primitive or just the integral over $\mathbb{R}^+$ or $[2,+\infty)$? – Jack D'Aurizio Nov 12 '15 at 18:49

## 1 Answer

By setting $x=\log t$, then using integration by parts: $$\int \arctan(2e^{-x})\,dx = \int \arctan\left(\frac{2}{t}\right)\frac{dt}{t}= \log(t)\arctan\left(\frac{2}{t}\right)+\int\frac{2\log(t)}{4+t^2}\,dt.$$ If now we set $t=2u$, the problem boils down to computing: $$\int\frac{\log(u)}{1+u^2}\,du = \frac{d}{d\alpha}\left.\int\frac{u^\alpha}{1+u^2}\,du\right|_{\alpha=0}$$ that is clearly related with the inverse tangent integral and the imaginary part of $\text{Li}_2(iu)$, by partial fraction decomposition.