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$$\begin{equation} \begin{split} \int\frac{2dx}{e^x+e^{-x}} & = \int\frac{2e^{-x}dx}{1+e^{-2x}} \\ &= \int 2e^{-x} \sum_{k=0}^{\infty}(-1)^ke^{-2kx}\,dx \\ &= 2\sum_{k=0}^{\infty}(-1)^k\int e^{-x(2k+1)}\,dx +c\\ &= -2\sum_{k=0}^{\infty}(-1)^k\frac{ e^{-x(2k+1)}}{(2k+1)}+c \end{split} \end{equation} $$ Note that $\displaystyle \arctan(x)=\sum_{k=0}^{\infty}(-1)^k \frac{x^{2k+1}}{2k+1}$, giving $$\begin{equation} \begin{split} \int\frac{2\,dx}{e^x+e^{-x}} & = -2\arctan(e^{-x})+c \end{split} \end{equation} $$ Wolfram says this integral is $\displaystyle\arctan\left(\tanh(\frac{x}{2})\right)$$\displaystyle=\arctan \left(\frac{1-e^{-x}}{1+e^{-x}}\right)=\arctan(1)+\arctan(-e^{-x})=\frac{\pi}{4}+\arctan(e^x)$

So, is $\displaystyle -2\arctan(e^{-x})=k+\arctan(e^x)$ (if so, why?), or did I do something incorrectly? Normally I'd defer the mistake (if there is one) to my neglect for convergence, but the results are so alike that I think my method must be fundamentally correct.

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  • $\begingroup$ That you have a correct answer cannot be an issue, since differentiating $-2\arctan(e^{-x})$ gives you the right thing. $\endgroup$ – André Nicolas Jul 28 '13 at 14:02
  • $\begingroup$ @AndréNicolas Yes, it boiled down to a algebraic error in the end. $\endgroup$ – Meow Jul 28 '13 at 14:03
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You made two mistakes:

1/ Wolframalpha gives a factor $2$ that you missed.

2/ And this is the correct version of the difference of $\arctan$'s

$$\arctan \left(\frac{1-e^{-x}}{1+e^{-x}}\right)=\arctan(1)-\arctan(e^{-x})$$

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  • $\begingroup$ The second was a typo, sorry. Yes, that helps. $\endgroup$ – Meow Jul 28 '13 at 13:56

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