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The integral does not diverge.

Choose $u = e^x$ so $\mathrm dx = \frac{\mathrm du}{e^x}$ so $\int_0^{\infty}(\frac{e^x}{e^{2x}+1})\mathrm dx$ becomes $\int_1^{\infty}( \frac{\mathrm du}{u^2 +1})$ but it diverges.

What is the fallacy here? I clearly made some mistake in changing the base but I cannot spot it.

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    By the way, it's the *integral* that doesn't diverge. There's no *equation* in sight.2011-08-22

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The fallacy is the claim that $\int_1^{\infty}\frac{du}{u^2+1}$ diverges.

$\begin{align*} \int_1^{\infty}\frac{du}{u^2+1} &= \lim_{t\to\infty}\int_1^t\frac{du}{u^2+1}\\ &= \lim_{t\to\infty}\left(\arctan(u)\Bigm|_{1}^{t}\right)\\ &= \lim_{t\to\infty}\left(\arctan(t) - \arctan(1)\right)\\ &= \lim_{t\to\infty}\left(\arctan(t) - \frac{\pi}{4}\right)\\ &= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \end{align*}$

The second fallacy is believing that Wolframalpha is never wrong (even if you put in the wrong input).

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    I upvoted because of the last sentence... :D2011-08-23