Question

How do I know if this integral converges or diverges?

Answer 1

\[\int\limits_{0}^{+\infty}\frac{ arctanx }{ x ^{2}(2+\sqrt{x}) }dx\]

Answer 2

Over \([0,\infty)\), \(\arctan x\) bounded above by \(\dfrac{\pi}{2}\) and below by \(0\), so you could make the argument that since
\[\frac{\arctan x}{x^2(2+\sqrt x)}\le\frac{\frac{\pi}{2}}{x^2(2+\sqrt x)}\]
then if the right side converges, the left side must also converge.

This is referred to as the comparison test. More info here: http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

Answer 3

Ok, I need to ask you one more thing.

Answer 4

Basically, what you have to do is evaluate the integral,
\[\frac{\pi}{2}\int_0^\infty \frac{dx}{x^2(2+\sqrt x)}\]
Alternatively, you can try making a similar argument as before, that this integrand is bounded by a convergent function, and so on.

Answer 5

How do I apply the comparison test on this integral?\[\int\limits_{2}^{+\infty}\frac{ 2+cosx }{ x ^{3}+1}dx\]The book says that to demonstrat that it converges I should put it like this\[\frac{ 2+cosx }{ x ^{3}+1 }\le \frac{ 3 }{ x ^{3} }\]

Answer 6

Why is that? How do I reason?

Answer 7

If I have this one the book still says to put it like this:\[\int\limits_{\pi}^{+\infty}\frac{ 6+sinx }{ \sqrt{x}-1 }dx\] Why should i put it like this and not the other way round?\[\frac{ 6+sinx }{ \sqrt{x}-1 }\ge \frac{ 5 }{ \sqrt{x} }\]

Answer 8

You know that \(\cos x\) is bounded, right? Specifically, \(-1\le\cos x\le1\), or simply \(|\cos x|\le1\).

So, you can see that, at most, you can have the numerator be equivalent to \(2+\cos x=2+|\cos x|=3\).

Now you have that
\[\frac{2+\cos x}{x^3+1}\le\frac{3}{x^3+1}\]
For the next part, consider the fact that \(\dfrac{1}{3}<\dfrac{1}{2}\). The denominator, 3, is larger, but this make the number on the left smaller.

In terms of what we want, you can say that
\[\frac{3}{x^3+1}<\frac{3}{x^3}\]

Answer 9

where did you get 1/3 and 1/2 from?

Answer 10

1/2 and 1/3 are used as an example to help you show why the next step follows.

For any given integral \(\displaystyle\int_c^\infty f(x)~dx\), if you can find a function \(g(x)\) such that \(0\le f(x)\le g(x)\) and \(\displaystyle\int_c^\infty g(x)~dx\) converges, then \(\displaystyle\int_c^\infty f(x)~dx\) necessarily converges.

Otherwise, you might be able to find \(h(x)\) such that \(0\le h(x)\le f(x)\) and \(\displaystyle\int_c^\infty h(x)~dx\) diverges. Then \(\displaystyle\int_c^\infty f(x)~dx\) must diverge as well.

Answer 11

But why in the latter integration it shows that 6+sinx/...>5/... even though I have to demonstrate that it converges? Couldn't I just aplly the same method of 2+cosx....?

Answer 12

The reason for that is that the integral containing sine happens to diverge.
\[\frac{6+\sin x}{\sqrt x-1}\]
Just like with \(\cos x\), you have that \(|\sin x|\le1\), so consider the smallest value sine can be (-1):
\[\frac{6+\sin x}{\sqrt x-1}\ge\frac{6-1}{\sqrt x-1}=\frac{5}{\sqrt x-1}\]
Now notice that the denominator is less than \(\sqrt x\), which means
\[\frac{\text{whatever}}{\sqrt x-1}\ge\frac{\text{whatever}}{\sqrt x}\]
and thus
\[\frac{6+\sin x}{\sqrt x-1}\ge\frac{5}{\sqrt x}\]

Answer 13

The general strategy with these sorts of problems is to find a similar expression that you can easily evaluate or tell right away that is convergent or divergent.

If you can find something that is convergent, then you have to show that this new expression is bigger than the old one.

Otherwise, if you can find something that diverges, then you show that the new expression is smaller than the old one. Make sense?

Answer 14

Yes, thanks a lot

Answer 15

You're welcome