Evaluate the integral from 0 to infinity of arctan(x) divided by cosh(x), with respect to x.

Step-by-step solution

1. 1

   Confirm the integral converges before working on it. As $x\to\infty$, $\arctan x\to\tfrac{\pi}{2}$ while $\operatorname{sech}x\sim 2e^{-x}$, so the integrand decays exponentially. Near $0$ it behaves like $x$. The integral is therefore finite - and bounded above by $\tfrac{\pi}{2}\int_0^{\infty}\operatorname{sech}x\,dx=\tfrac{\pi^2}{4}\approx2.467$.

2. 2

   Expand $1/\cosh x$ as an alternating exponential series. For $x>0$,

   $\frac{1}{\cosh x}=\frac{2e^{-x}}{1+e^{-2x}}=2\sum_{n=0}^{\infty}(-1)^n e^{-(2n+1)x}$

   This works because $e^{-2x}<1$ on the whole range, so the geometric series converges everywhere the integral lives.

3. 3

   Integrate term by term. Swapping sum and integral gives

   $I=2\sum_{n=0}^{\infty}(-1)^n\int_0^{\infty}e^{-(2n+1)x}\arctan x\,dx$

4. 4

   Reduce each Laplace-type piece by parts. With $u=\arctan x$ and $dv=e^{-ax}dx$ the boundary term vanishes at both ends, leaving

   $\int_0^{\infty}e^{-ax}\arctan x\,dx=\frac{1}{a}\int_0^{\infty}\frac{e^{-ax}}{1+x^{2}}\,dx$

   so

   $I=2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^{\infty}\frac{e^{-(2n+1)x}}{1+x^{2}}\,dx$

5. 5

   Accept that there is no elementary closed form. The inner integral is an auxiliary cosine-and-sine-integral combination, not an elementary function, so the series above is the honest exact answer. A decimal value has to be produced numerically.

6. 6

   Evaluate numerically by two independent routes. Gauss-Legendre quadrature on $[0,80]$ split into seven panels gives

   $I\approx 1.0986875$

   Summing the series above to $400$ terms gives $1.0986859$, agreeing to six digits ✓. Both sit safely under the bound $\pi^2/4=2.467$ from step 1. (Note $I$ is close to, but not equal to, $\ln 3=1.0986123$.)