# Improper Integrals (Very Basic)

So im having a huge amount of trouble with some basic improper integrals of different types. as they are very basic im going to post the ones i got incorrect. ( my textbook just labels converges or diverges as answers)

1)$\int^{\infty}_{1} \frac{dx}{x(x+3)^{1/2}}$
Clearly the problem occurs at infinity

2)$\int^{\infty}_{3} \frac{(x^{2}-3x-1)dx}{x(x^{2}+2)}$

Again at infinity

3)$\int^{\infty}_{3} \frac{(\sin 4x)dx}{x^{2}-x-2}$

Again at infinity

4)$\int^{\infty}_{1} \tan(\frac{1}{x})dx$

I have the basic methods as well as the ratio test im just not well versed in implementing it these are ones that i thought i had figured out but did incorrectly. if someone could give me a simple function or layout to apply to these i would really appreciate it

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The common thread in these problems is the behaviour for large $x$.

1) Informally, for large $x$ our function behaves like $\frac{1}{x^{3/2}}$. You should know that for $p\gt 1$, $\int_1^\infty \frac{dx}{x^p}$ converges.

So our integral should converge. More formally, we have $$0\lt \frac{1}{x(x+3)^{1/2}}\lt \frac{1}{x^{3/2}},$$ so by the Comparison Test our integral converges.

2) Informally, the top is dominated by the $x^2$ term for large $x$, and the bottom behaves essentially like $x^3$. So the ratio behaves like $\frac{1}{x}$. Recall that $\int_1^\infty \frac{dx}{x}$ diverges. So our integral should diverge.

To make it formal, use the Limit Comparison Test. That is, if $f(x)$ is our function, show that $\lim_{x\to\infty}\frac{f(x)}{x}$ is positive.

3) Note that $|\sin(4x)|\le 1$. So if we can show that the integral of $\frac{1}{x^2-x-2}$ converges, we will be finished.

But informally this function behaves like $\frac{1}{x^2}$ for large $x$. You can make it more formal by using the Limit Comparison Test. Or else you can observe that for large enough $x$, the bottom is bigger than $x^2/2$, and do a Comparison with the integral of $\frac{2}{x^2}$.

4) Note that $\lim_{t\to 0}\frac{\tan t}{t}=1$. So for large $x$, $\tan(1/x)$ behaves like $1/x$. But we know that $\int_1^\infty \frac{dx}{x}$ diverges.

More formally, do a Limit Comparison Test. Or observe that if $x$ is large, then $tan(1/x)\gt \frac{1}{2}\cdot \frac{1}{x}$.

Remark: What was in common here was the family of integrals $\int_1^\infty \frac{dx}{x^p}$. We had to know that there is convergence if $p\gt 1$, and divergence if $p\le 1$.

Then, in these problems, we needed to do some informal thinking about dominant terms ("roughly how big/small is this?"). And finally, for the formal part of the argument, we needed tools such as Comparison and Limit Comparison.

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this is a very good answer its clear cut as to what i am not grasping! Tyvm for your help i have been getting some right some wrong its very frustrating when i don't know why something is incorrect. – Faust7 Mar 31 '13 at 22:18

Here is a start,

1) Use the inequality $x(x+3)^{1/2} > x^{3/2},\, \forall x \geq 1.$

$$\int^{\infty}_{1} \frac{dx}{x(x+3)^{1/2}}dx < \int^{\infty}_{1} \frac{dx}{x^{3/2}}dx < \infty.$$

2) You can compare with the function $\frac{1}{x}$, since $$\frac{x^{2}-3x-1}{x(x^{2}+2)}\sim \frac{x^2}{x^3}=\frac{1}{x}.$$

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