How do you determine if the improper integral converges or diverges $\int \frac{e^{x}}{e^{2} x + 1} d x$ $int (e^x)/(e^2x + 1) dx$ from 0 to infinity?

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Answer 1 · 2018-01-30T15:28:01

The integral converges, and:

$int_0^oo \ (e^x)/(e^(2x)+1) \ dx = pi/4$

Assuming a correction, let us first find:

$I = int \ (e^x)/(e^(2x)+1) \ dx$

The utilising a substitution if $u=e^x => (du)/dx=e^x$ yields:

$I = int \ (1)/(u^2+1) \ du$

Which is a standard integral, and so integrating:

$I = arctan (u) + C$

And then reversing the earlier substitution:

$I = arctan (e^x) + C$

If we look at the graph of the function, it would appear as if the bounded area is finite:

graph{(e^x)/(e^(2x)+1) [-10, 10, -2, 2]}

So let us test this prediction analytically. Consider the improper definite integral:

$J = int_0^oo \ (e^x)/(e^(2x)+1) \ dx$

$\ \ = lim_(n rarr oo) [arctan (e^x)]_0^n$

$\ \ = lim_(n rarr oo) (arctan (e^n)) -arctan (e^0)$

Now, as $x rarr oo => e^x rarr oo$ , thus:

$J = lim_(n rarr oo) (arctan n) - arctan (1)$

$\ \ = pi/2-pi/4$

$\ \ = pi/4$

How do you determine if the improper integral converges or diverges int (e^x)/(e^2x + 1) dx∫exe2x+1dxint (e^x)/(e^2x + 1) dx from 0 to infinity?