I'm learning about convergence/divergence of improper integrals and need help with the following problem:
Find for what values of $a$ does the following integrals exists
$$(1) \int_0^\infty\frac{\tanh x}{1 + x^a} \, dx;$$
$$(2) \int_0^\infty\left(\frac{x^a}{1 + x^2}\right)^4 \, dx.$$
Here's my work so far:
$(1)$ For $a > 1$ and since $|\tanh x| \leq 1$ we have:
$$\left|\frac{\tanh x}{1 + x^a}\right| \leq \frac{1}{1 + x^a} \leq \frac{1}{x^a} = x^{-a}.$$
Therefore, a just need to check convergence for
$$\int_0^\infty x^{-a} \, dx$$
but this integral does not converge for $a > 1$.
For $a = 0$ we have:
$$\frac{1}{2}\int_0^\infty\tanh x$$
which also does not converge.
Is my work correct so far? I don't know how I should take care of the case where $0 < a < 1$. I also have difficulties for $(2)$.
Since the integrand is a continuous function over each compact of the form $[0,M]$, $M>0$, then we are left to see what happens as $x \to \infty$, where we have: $$ \frac{\tanh x}{1 + x^a} \sim \frac1{x^a} $$ which gives a finite integral if and only if $a>1$, so the initial integral is convergent if and only if $a>1$.
Since the integrand is a continuous function over each set of the form $(0,M)$, $M>0$, then there are a potential problems as $x \to 0$ and as $x \to \infty$.
As $x \to 0$ we have, $$\left(\frac{x^a}{1 + x^2}\right)^4 \sim x^{4a} $$ which gives a finite integral if and only if $-4a<-1$.
As $x \to \infty$ we have, $$\left(\frac{x^a}{1 + x^2}\right)^4 \sim x^{4a-8} $$ which gives a finite integral if and only if $-4a+8>1$.
Then the initial integral is convergent if and only if