Question

Can any of these integrals be solved by hand?

Answer 1

These are improper integrals so you would know just to take the limit of t-> infinity and put that as the upper limits

but you are right..how are we supposed to find the antiderivatives?

Answer 2

I compared the first integral to arctan(x^2) and said that it converges, but I don't know what to say about the rest

Answer 3

I think if the function approaches 0 as x approaches infinity, it converges

Answer 4

Here's the issue though. I'm supposed to compare them to other known integrals, but I'm not sure how to do that with these ones.

Answer 5

You should recall the direct comparison test for series. We an apply a similar result here. Let \(a \in \mathbb{R}\).

If the improper integral
\(\displaystyle\int\limits_a^\infty f(x)\textrm{ d}x\)
converges and \(0 \leq g(x) \leq f(x)\) for \(x \geq a\), we can use the monotonicity property of the integral to say that
\(\displaystyle\int\limits_a^\infty g(x)\textrm{ d}x \leq \int\limits_a^\infty f(x)\textrm{ d}x,\)
so that the improper integral
\(\displaystyle\int\limits_a^\infty g(x)\textrm{ d}x\)
also converges.

By the same token, if the improper integral
\(\displaystyle\int\limits_a^\infty f(x)\textrm{ d}x\)
diverges and \(0 \leq f(x) \leq g(x)\) for \(x \geq a\), then
\(\displaystyle\int\limits_a^\infty f(x)\textrm{ d}x \leq \int\limits_a^\infty g(x)\textrm{ d}x,\)
which means that the improper integral
\(\displaystyle\int\limits_a^\infty g(x)\textrm{ d}x\)
also diverges.

The point of this exercise it that, for each integrand, you should find some similar function that you can integrate by hand and then compare them. As a hint, you can do A) by arguing that since \(x \geq 1\), we have \(x^4 +1 \geq x^2 + 1 > x^2\). Therefore, we can write:
\(\displaystyle \dfrac{1}{1+x^4} < \dfrac{1}{x^2},\)
which means that:
\(\displaystyle \int\limits_1^\infty \dfrac{\textrm{d}x}{1+x^4} < \int\limits_1^\infty \dfrac{\textrm{d}x}{x^2} = -\lim\limits_{b \to \infty}\dfrac{1}{x}\Big\vert_1^b = -\lim\limits_{b\to\infty} \left(\dfrac{1}{b}-\dfrac{1}{1}\right) = 1.\)

By the test above, the improper integral
\(\displaystyle \int\limits_1^\infty \dfrac{\textrm{d}x}{1+x^4}\)
must converge. Also, note that this is different from actually computing the integral directly, which is \(\underline{\textrm{not}}\) what the exercises asks you to do.

Can you see how to do the others now? (:

Answer 6

Thanks so much dude, it was hard to follow at first because we've never done smth like this before, but I'll give it a shot. I really appreciate it!

Answer 7

No problem! Just let me know if you have any more trouble.

By the way, I didn't refer it earlier, but notice that there are multiple ways to solve the exercise. For instance, your solution for A) is also correct and you could also get exactly the same result by saying \(x^4 +1 > x^4\) and then using a similar argument to the one I showed above. In fact, you can also employ that very trick for B). (: