For the integrand $\frac{(1 + x^{25q})(1 - x^q)}{((1 + x^q)(1 - x^{25q})}$, I calculated that the dominant term in the numerator is $x^{26q}$ and the dominant term in the denominator is also $x^{26q}$. So the integrand simplifies to 1. What does this mean in terms of convergence and divergence? The answer is that the integral diverges for any real value of q but I don't know what theorems are used to find this.
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If this integral were convergent, you would necessarily get a decay towards infinity. Do you have such decay of the integrand? – Aniruddha Deshmukh May 01 '23 at 04:42
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For $q>0$ the dominant term of the top and bottom for large $x$ are equal as you said, so the $\lim_{x \to \infty} =1$. Since this limit is defined and not $0$, the integral to infinity must diverge.
For $q<0$ the dominant terms on top and bottom are both 1, so again it must diverge by the same logic.
Finally, for $q=0$, we are dividing by zero, so the integrand is not defined.
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I'm confused about the definition of a divergent integral. I thought that a divergent integral is where an improper integral's limit doesn't exist and that if the limit exists and is a real number it's convergent? – Crystal May 01 '23 at 06:37
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@Crystal This is right but if $\lim_{x \to \infty} f(x) = 1$ as discussed above, then the integral won't converge. Think of the divergence test for infinite series. – JayP May 01 '23 at 08:12