Example for Comparison Test for Improper Integrals
Example: Find out if the following integral is convergent or divergent.
∫^{∞}_{2 }(cos^{2} x) / x^{2 }(dx)
Solution
Let's just take a second and think about how the Comparison Test works. Determine If this integral is convergent then we'll need to discover a larger function that as well converges on similar interval.
Similarly, if this integral is divergent then we'll require to find out a smaller function that as well diverges.
Thus, it seems such as it would be nice to have some idea as to if the integral converges or diverges ahead of time thus we will know whether we will need to have a look for a larger and convergent function or a smaller and divergent function.
To obtain the guess for this function let's notice that the numerator is good and bounded and just won't get too large. Hence, it seems likely that the denominator will ascertain the convergence or divergence of this integral as we know that
∫^{∞}_{2} 1/x^{2} (dx)
converges since converge.
p = 2 > 1 by the fact in the preceding section. Thus now her let's guess that this integral will
Thus we now know that we need to find out a function that is larger than
cos^{2} x / x^{2}
and as well converges. Making a fraction larger is actually a quite simple procedure. We can either create the numerator larger or we can make the denominator smaller. In this type of case we can't do so much about the denominator. Though we can make use of the fact that 0 ≤ cos^{2} x ≤ 1 to make the numerator larger (that is we'll replace the cosine with something we know to be larger that is 1). So,
Cos^{2} x / x^{2} < 1/x^{2}
Here now, as we have already noted
∫^{∞}_{2} 1/x^{2} (dx)
converges and thus by the Comparison Test we know that
∫^{∞}_{2 }cos^{2} x / x^{2} (dx)