Improper integral $\int _{-\infty}^1 \arctan(e^x)\,\mathrm dx$

Question

How can I calculate this integral? $${\int _{-\infty}^{1}\textrm{arctan}(e^x)\,\mathrm dx}$$

It converges. Can I associate some constant to its value?

Answer 1

$$I={\int _{-\infty}^{1}\textrm{arctan}(e^x)\,\mathrm dx}=I_1+I_2\qquad \begin{cases} I_1=\int _{-\infty}^{0}\textrm{arctan}(e^x)\,\mathrm dx \\ I_2=\int _{0}^{1}\textrm{arctan}(e^x)\,\mathrm dx \end{cases}$$ $$I_1=\int _{0}^{\infty}\textrm{arctan}(e^{-x})\,\mathrm dx = \int _{0}^{\infty}\sum_{n=0}^\infty \frac{ (-1)^n (e^{-x})^{2n+1} }{2n+1} dx $$ $$I_1=\sum_{n=0}^\infty\int _{0}^{\infty} \frac{ (-1)^n (e^{-x})^{2n+1} }{2n+1} dx =\sum_{n=0}^\infty \frac{ (-1)^n }{(2n+1)^2} = C \qquad \text{Catalan constant}$$

$$I_2=\int _{0}^{1}\left(\frac{\pi}{2}-\textrm{arccot}(e^x)\right)\,\mathrm dx = \frac{\pi}{2}-\int _{0}^{1}\textrm{arctan}(e^{-x})\,\mathrm dx $$ $$I_2= \frac{\pi}{2}-\int _{0}^{1}\sum_{n=0}^\infty \frac{ (-1)^n (e^{-x})^{2n+1} }{2n+1}dx =\frac{\pi}{2}-\sum_{n=0}^\infty \int _{0}^{1}\frac{ (-1)^n (e^{-x})^{2n+1} }{2n+1}dx $$ $$I_2=\frac{\pi}{2}+\sum_{n=0}^\infty \frac{ (-1)^n (e^{-(2n+1)}-1) }{(2n+1)^2} = \frac{\pi}{2}+\sum_{n=0}^\infty \frac{ (-1)^n e^{-(2n+1)} }{(2n+1)^2}-C$$

The Catalan constant is eliminated in the sum $I_1+I_2$.

$$I={\int _{-\infty}^{1}\textrm{arctan}(e^x)\,\mathrm dx}=I_1+I_2=\frac{\pi}{2}+\sum_{n=0}^\infty \frac{ (-1)^n e^{-(2n+1)} }{(2n+1)^2}$$

$${\int _{-\infty}^{1}\textrm{arctan}(e^x)\,\mathrm dx}=\frac{\pi}{2}+\frac{1}{e}-\frac{1}{9e^3} +\frac{1}{25e^5}-\frac{1}{49e^7} +...$$

The series is quickly convergent. This allows to easily compute good approximates :

$${\int _{-\infty}^{1}\textrm{arctan}(e^x)\,\mathrm dx} \simeq 1.933396177216636$$

In addition :

Thanks to the comment from Claude Leibovici, a closed form can be derived :

$${\int _{-\infty}^{1}\textrm{arctan}(e^x)\,\mathrm dx}=\frac{\pi}{2}+\frac{1}{16e}\Phi(e^{-4}\:,\:2\:,\:1/4)-\frac{1}{16e^3}\Phi(e^{-4}\:,\:2\:,\:3/4)$$ $\Phi$ is the Lerch function. http://mathworld.wolfram.com/LerchTranscendent.html

Answer 2

It's about $1.93$. We get this by using the following methods:

First, we rewrite the $\arctan$ function using logarithms:

$$\int \arctan(e^x) \,dx=\int-\frac{i(\ln(ie^x+1)-\ln(1-ie^x))} 2$$

Then, by applying linearity we obtain:

$$\frac i 2 \int \ln(1-ie^x)\,dx - \frac{i}{2} \int\ln(ie^x+1) \, dx$$

By substituting $u=-ie^x$, we obtain:

$$\int \ln(ie^x+1) \,dx = \int -\frac{\ln(1-u)} u \,du=-\operatorname{Li}_2(u)=-\operatorname{Li}_2(-ie^x)$$

We do the same to $\int\ln(1-ie^x)\,dx$ with the same $u$-substitution:

$$\int ln(1-ie^x) \,dx = -\operatorname{Li}_2(ie^x)$$

Then, we back-substitute:

$$\frac i 2 \int\ln(1-ie^x)\,dx -\frac i 2 \int \ln(ie^x+1)\,dx = \frac{i\operatorname{Li}_2(-ie^x)}2 - \frac{i\operatorname{Li}_2(ie^x)} 2 + C$$

We can rewrite this in a neat way:

$$-\frac{\pi\ln(e^{2x}+1)}{4}-\frac{i\operatorname{Li}_2(ie^x-+)}2 + \frac{i\operatorname{Li}_2(1-ie^x)} 2 + x\arctan(e^x)+C$$

We evaluate at the given bounds, obtaining $\approx1.9334$