How do you evaluate the integral $int e^x/(1+e^(2x))dx$ from $-oo$ to $oo$ ?

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Answer 1 · 2016-08-22T08:38:44

$pi$

Making $y = e^x$ we have $dy = e^x dx = y dx$

$int e^x/(1+e^(2x))dx equiv int dy/(1+y^2) = arctan(y)$

but

$lim_{x->oo}e^x=lim_{y->oo}=oo$

so

$lim_{a->oo} int_{-a}^a dy/(1+y^2) = pi/2-(-pi/2) = pi$

How do you evaluate the integral int e^x/(1+e^(2x))dxint e^x/(1+e^(2x))dx from -oo-oo to oooo?