Determine whether $\int_0^{\infty} \frac{\arctan(x)}{e^x – e}dx$ converges or diverges.
Attempt: I know that $\arctan x\leq \pi/2$, but how can I proceed from here? I also split the integral as:
$\int_0^{\infty} \frac{\arctan(x)}{e^x – e}dx$=$\int_0^{1} \frac{\arctan(x)}{e^x – e}dx$+$\int_1^{\infty} \frac{\arctan(x)}{e^x – e}dx$.
What now?
Best Answer
As @RyszardSzwarc wrote in comments, the Laurent series is $$\frac{\tan ^{-1}(x)}{e^x-e}=\sum_{n=\color{red}{-1}}^\infty a_n\, (x-1)^n$$ where the first coefficients are $$\left\{\frac{\pi }{4 e},\frac{4-\pi }{8 e},\frac{\pi -24}{48 e},\frac{1}{4 e},\frac{-180-\pi }{2880 e},-\frac{3}{160 e},\cdots\right\}$$ Limited to the above terms, the approximation is really good (make a plot).
If you use the above for the integral from $0$ to $0.99$, you will obtain $-1.17645$ to be compared to the "exact" $-1.17955$.