Math: Is my reasoning valid (integrals, determining whether it is convergent or divergent)?
So, I want to determine if the integral for sin^2(x)/x^4 for the interval minus infinity to infinity is convergent or divergent.
I start by identifying that the function is even, so it is the same thing as 2 times the integral from 0 to infinity.
Instead of doing the integral, I do an easier integral for 1/x^4 and get that it approaches infinity.
Identifying that 1/x^4 = (sin^2(x)+cos^2(x))/x^4 = sin^2(x)/x^4 + cos^2(x)/x^4, so I know that the sum of the integrals for both this expressions is equal to infinity. Since both of them should behave similarly as both sin^2(x) and cos^2(x) oscillate between 0 and 1 and are both divided by x^4, when they approach x=0 they will both approach y=infinity as x^4 is the dominant factor, and when they approach x=0 they approach y=0, since again x^4 is the dominant factor, and they should therefor have integrals that are similar to each other for the said intervall.
Since they both together are equal to infinity, they have to each be approaching infinity. Therefor the integral for sin^2(x)/x^6 from -infinity to infinity is divergent.
I feel like I might have made some false assumption or mistake somewhere. Is what I did okay, or does this problem have to be done some other way?
- JOHNLv 78 months agoFavorite Answer
Hope the below is readable
- rotchmLv 78 months ago
Too long to read. So I propose another way to show that it diverges:
sin^2(x)/x^4 = (sin(x)/x)² * 1/x².
But (sin(x)/x)² is bounded below & above near x=0 [why]?
Thus the integral of (sin(x)/x)² * 1/x² ... [left for you].