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Chapter 6: Problem 26

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} \frac{e^{x}}{1+e^{x}} d x $$

Short Answer

Expert verified
The improper integral \(\int_{0}^{\infty} \frac{e^{x}}{1+e^{x}} dx \) diverges. It does not evaluate to a finite value.

Step by step solution

01

Define the Function and the Limit for the Improper Integral

Start by defining the function for which the integral is sought: \(f(x) = \frac{e^{x}}{1+e^{x}}\). Then, define the limits for the improper integral: \(\lim_{{b \to \infty}} \int_{{0}}^{{b}} f(x) dx \). This transformation positions the problem to be solved through limiting processes. Setting it up like this creates the problem: \(\lim_{{b \to \infty}} \int_{{0}}^{{b}} \frac{e^{x}}{1+e^{x}} dx \).
02

Apply Substitution

For the function \( f(x) = \frac{e^{x}}{1+e^{x}}\), let \(u = 1 + e^x\). Then, differentiating gives \( du = e^x dx \). Rewrite the integral in terms of \(u\): \(\lim_{b \to \infty} \int_{1}^{1+e^{b}} \frac{1}{u} du \).
03

Apply Formula for Definite Integral

The definite integral of \(1/u\) from 1 to \(1+e^{b}\) is the natural logarithm of the upper limit divided by the lower limit. Substituting \(1+e^{b}\) back in for \(u\), we get \(\lim_{b \to \infty} [\ln(1+e^{b}) - \ln(1)]\)
04

Simplify and Evaluate the Limit

Simplify the limit to get \(\lim_{b \to \infty} \ln(1+e^{b})\). As \(b \to \infty\), \(1+e^{b}\) also tends to infinity, and thus \(\ln(1+e^{b})\) tends to infinity as well. So the limit, and thus the integral, does not exist or diverges.

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Most popular questions from this chapter

Show that \(\lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}}=0\) for any integer \(n>0\).

Consider the region satisfying the inequalities. (a) Find the area of the region. (b) Find the volume of the solid generated by revolving the region about the \(x\) -axis. (c) Find the volume of the solid generated by revolving the region about the \(y\) -axis. $$ y \leq e^{-x}, y \geq 0, x \geq 0 $$

Prove that \(I_{n}=\left(\frac{n-1}{n+2}\right) I_{n-1},\) where \(I_{n}=\int_{0}^{\infty} \frac{x^{2 n-1}}{\left(x^{2}+1\right)^{n+3}} d x, \quad n \geq 1 .\) Then evaluate each integral. (a) \(\int_{0}^{\infty} \frac{x}{\left(x^{2}+1\right)^{4}} d x\) (b) \(\int_{0}^{\infty} \frac{x^{3}}{\left(x^{2}+1\right)^{5}} d x\) (c) \(\int_{0}^{\infty} \frac{x^{5}}{\left(x^{2}+1\right)^{6}} d x\)

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(a) Let \(f^{\prime}(x)\) be continuous. Show that \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h}=f^{\prime}(x)\) (b) Explain the result of part (a) graphically.
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