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Improper comparisons

mark

Consider the following two improper integrals:

  1. \displaystyle \int_0^{\infty} \frac{1+e^{-x}}{\sqrt{x+1}} \, dx and
  2. \displaystyle \int_0^{\infty} \frac{e^{-x}}{\sqrt{x+1}} \, dx.

One of these improper integrals converges and the other diverges. Use the comparison test to clearly explain which is which and why.

mcollin5

I agree with the post above. I’ll add that when viewing the functions independently
the first problem

\displaystyle \int_0^{\infty} \frac{1+e^{-x}}{\sqrt{x+1}} \, dx

can be rewritten as

\displaystyle \int_0^{\infty} \frac{1+\frac{1}{e^x}}{\sqrt{x+1}} \, dx

knowing that the harmonic series (\frac{1}{n}) is divergent and that

\frac{1+\frac{1}{e^x}}{\sqrt{x+1}} \, > \frac{1}{n}

comparison would become

0 \le\ \frac{1+\frac{1}{e^x}}{\sqrt{x+1}} \ge\ \frac{1}{n}

as apposed to:

0 \le\ f(x) \le\ g(x)

This means that the first problem fails the comparison test and proves to be divergent.

While second problem

\displaystyle \int_0^{\infty} \frac{e^{-x}}{\sqrt{x+1}} \, dx

can be rewritten as

\displaystyle \int_0^{\infty} \frac{{1}}{\sqrt{x+1}\times (e^x)} \, dx

knowing that e^{-x} is convergent, using the comparison test:

0 \le \frac{{1}}{\sqrt{x+1}\times (e^x)} \le \frac{1}{e^{x}}

The test becomes true and shows the problem 2 is convergent.