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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?

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    $\begingroup$ What happens when $x = 1$? $\endgroup$
    – K. Jiang
    Commented May 7 at 15:57
  • $\begingroup$ Diverges$\ldots$ $\endgroup$ Commented May 7 at 15:58
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    $\begingroup$ Perhaps you wanted $e^x-1$ in the denominator $\endgroup$ Commented May 7 at 17:04
  • $\begingroup$ So how, can I prove it diverges? I can split the integral that's true, but both integrals are divergent. $\endgroup$
    – good12
    Commented May 7 at 18:18
  • $\begingroup$ $${\arctan x\over e^x-e}\approx {\pi\over 4e}{1\over x-1}$$ when $x\to 1.$ $\endgroup$ Commented May 7 at 20:08

2 Answers 2

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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" $-1.17955$.

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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 $e^y\sim 1+y$

$$\frac{\arctan(1+y)}{e(e^y-1)} \sim\frac{\arctan(1)}{e} \frac 1y$$

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