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

Question

Elijah Abram · Accepted Answer

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#

Emma Aldridge · Answer

undefined To determine if the improper integral converges or diverges, we can use the limit comparison test. Let's compare the given integral to a simpler integral with which we're familiar.

Consider the integral of $ e^{-x} $ from 0 to infinity. This integral converges, as it represents an exponential decay function.

Now, let's compare $ \frac{e^x}{e^{2x} + 1} $ to $ e^{-x} $. As $ x $ approaches infinity, the term $ e^{-x} $ dominates $ \frac{e^x}{e^{2x} + 1} $.

So, we'll compare the given integral to $ \int_0^\infty e^{-x} $ using the limit comparison test.

$$ \lim_{x \to \infty} \frac{\frac{e^x}{e^{2x} + 1}}{e^{-x}} = \lim_{x \to \infty} \frac{e^{2x}}{e^{2x} + 1} = 1 $$

Since the limit is a finite positive number, and since $ \int_0^\infty e^{-x} $ converges, by the limit comparison test, $ \int_0^\infty \frac{e^x}{e^{2x} + 1} dx $ also converges.