How to Prove the Divergence of a Definite Integral

In this article, we will explore the methods to prove the divergence of a definite integral. Specifically, we will use two methods: the infinite series approach and the comparison test to show that the integral I

Method I: Infinite Series

To demonstrate the divergence of the integral (I int_1^{sqrt{2}}dfrac{1}{(4-x^4)^2}, dx), we first express the integrand as a binomial series. The general form of the binomial series is given by:

[ frac{1}{(4-x^4)^2} frac{1}{16} lim_{n to infty} sum_{k0}^n k1 left(frac{x^4}{4}right)^k ]

Interchanging the integration and summation, we get:

[ I frac{1}{16} lim_{n to infty} sum_{k0}^n frac{k1}{4^k} int_1^{sqrt{2}} x^{4k} , dx ]

Further simplifying, we obtain:

[ I frac{1}{16} lim_{n to infty} sum_{k0}^n frac{k1}{4^k (4k 1)} left(4^k sqrt{2} - 1^{4k - 1} right) ]

Now, we need to check the divergence of the series. Notice:

[ lim_{n to infty} frac{n1}{4n 1} left(sqrt{2} - left(frac{1}{4}right)^n right) frac{sqrt{2}}{4} eq 0 ]

Since the series does not satisfy the necessary condition for convergence (i.e., the limit of the terms is not zero), the integral diverges. This proves the divergence of the integral by using the infinite series approach.

Method II: Comparison Test

In the alternative method, we use the comparison test. We rewrite the integrand in a factored form as:

[ f(x) frac{1}{(4-x^4)^2} frac{1}{(2x^2)^2 (sqrt{2}x)^2 (sqrt{2}-x)^2} ]

Let's define another function (g(x)) as follows:

[ g(x) frac{1}{(2sqrt{2})^2 (2sqrt{2})^2 (sqrt{2}-x)^2} frac{1}{128 (sqrt{2}-x)^2} ]

Notice that for x in the interval [1, sqrt{2}], we have f(x) ( geq ) g(x). We can see that the integral of g(x) over the limits diverges. To confirm, let's calculate:

[ int g(x) , dx int frac{dx}{128 (sqrt{2}-x)^2} frac{1}{128 (sqrt{2}-x)} C ]

Using the definition of an improper integral, we calculate:

[ J int_1^{sqrt{2}} g(x) , dx lim_{R to (sqrt{2})^-} left[ frac{1}{128 (sqrt{2}-R)} - frac{1}{128 (sqrt{2}-1)} right] ]

As the limit does not exist, the improper integral (J) diverges. Therefore, by the comparison test, the original integral (I) also diverges.

Conclusion

We have successfully demonstrated the divergence of the integral I through two methods: the infinite series approach and the comparison test. The divergence of the integral can be proven by showing that the necessary conditions for convergence are not met.