# Determine whether $\int_0^{\infty} \frac{\arctan(x)}{e^x - e}dx$ converges or diverges.

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?

• What happens when $x = 1$? Commented May 7 at 15:57
• Diverges$\ldots$ Commented May 7 at 15:58
• Perhaps you wanted $e^x-1$ in the denominator Commented May 7 at 17:04
• So how, can I prove it diverges? I can split the integral that's true, but both integrals are divergent. Commented May 7 at 18:18
• $${\arctan x\over e^x-e}\approx {\pi\over 4e}{1\over x-1}$$ when $x\to 1.$ Commented May 7 at 20:08

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" $$-1.17955$$.
As noticed in the comments the integral is problematic at $$x=1$$, indeed let $$y=x-1$$ and we have
$$\int_1^{2} \frac{\arctan(x)}{e^x - e}dx=\int_0^{1} \frac{\arctan(1+y)}{e(e^y-1)}dy$$
and at $$y \to 0$$ since $$e^y\sim 1+y$$
$$\frac{\arctan(1+y)}{e(e^y-1)} \sim\frac{\arctan(1)}{e} \frac 1y$$