# For what values of $a$ does $\int_0^\infty\left(\frac{x^a}{1 + x^2}\right)^4 \, dx$ converge?

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 $$(2)$$.

• Your work is correct so far, but because you are comparing to a divergent integral, you cannot conclude anything for $a>1$, nor for $a=0$. You need to do more work before you have answered any of these cases :). One strategy that often helps is to break apart the integral into two intervals (and compare them to different functions): you'll find that $x^{-a}$ sometimes converges on $\int_0^1$ and sometimes converges on $\int_1^\infty$ but never both at the same time. – Erick Wong Jun 17 '16 at 19:29

1. 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 $a>1$.

2. 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

$$\dfrac14<a<\dfrac74.$$

• Thank you for your explanation. I do have some questions: for $(1)$, I don't understand how you were able to conclude that the integral is convergent for any $a > 1$. It seems to me that the integral $\int_0^\infty\frac{1}{x^a}$ is divergent for any $a > 1$. Also for $(2)$, there is no problem at the point $x=0$ in my opinion. Could you please explain this? – glpsx Jun 17 '16 at 5:58
• @Von Kar You are welcome. The integrand is continuous over $[0,M]$, thus there is no problem with $\int_0^M f(x) dx$. As written we just have to consider $\int_M^\infty f(x) dx$ ($M>0$) which is of the same nature of $\int_M^\infty \frac1{x^a} dx$, the latter is convergent if and only if $a>1$ (see $p$integral) **For the second integral**, since there is no hypothesis that $a\geq 0$, we can have for example the integrand equal to $\frac1{x^2(1+x^2)^4}$ ($a=-1/8$) and the latter function is not continuous at $x=0$. – Olivier Oloa Jun 17 '16 at 12:28
• This makes perfect sense now! Thanks again for your explanation and your time, much appreciated. – glpsx Jun 17 '16 at 19:31