# Math Help - Improper Integrals

1. ## Improper Integrals

Use the comparison test to determine whether the improper integral converges or diverges:
$\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$

So, since this is true... $\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$ < $\int_{0}^{\infty} \frac{1}{\sqrt{x}}$ I can just use the larger integral to find out whether or not it converges or diverges because it is much simpler.

$\int_{0}^{\infty} x^\frac{-1}{2}$ which equals:

$\lim_{t \to \infty} \int_{0}^{t} x^\frac{-1}{2}$ equals

$\lim_{t \to \infty} \frac {x^\frac{1}{2}}{2}$ from 0 -> $\infty$

So this just turns out to be $\infty - \frac {0^\frac{1}{2}}{2}$ so this is equal to...
$\infty - 0$ which = $\infty$

So because this integral DIVERGES TO $\infty$, this tells us nothing? So what do I do now?

2. Originally Posted by larson
Use the comparison test to determine wehter the improper integral converges or diverges:
$\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$

So, since this is true... $\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$ < $\int_{0}^{\infty} \frac{1}{\sqrt{x}}$ I can just use the larger integral to find out whether or not it converges or diverges because it is much simpler.

$\int_{0}^{\infty} x^\frac{-1}{2}$ which equals:

$\lim_{t \to \infty} \int_{0}^{t} x^\frac{-1}{2}$ equals

$\lim_{t \to \infty} \frac {x^\frac{1}{2}}{2}$ from 0 -> $\infty$

So this just turns out to be $\infty - \frac {0^\frac{1}{2}}{2}$ so this is equal to...
$\infty - 0$ which = $\infty$

So because this integral DIVERGES TO $\infty$, this tells us nothing? So what do I do now?
All that you have said is that the integrals values is LESS THAN or equal to infinity. to show divergence you would need to show that it is greater than or equal to infinity.

Try this comparison

$\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}=\int_0^1 \frac {dx}{\sqrt{x}(1 + x)} +\int_1^\infty \frac {dx}{\sqrt{x}(1 + x)}$

$< \int_{0}^{1}\frac{dx}{\sqrt{x}}+\int_1^{\infty}\fr ac{dx}{x^{3/2}}$

3. Originally Posted by larson
Use the comparison test to determine whether the improper integral converges or diverges:
$\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$
Note it does converges by my above post now, but also it can be integrated with this sub

$u=\sqrt{x} \mbox{ then } du=\frac{dx}{\sqrt{x}}$

Note that the limits of int don't change so we get

$\int_{0}^{\infty}\frac{du}{1+u^2}=\tan^{-1}u|_{0}^{\infty}= \frac{\pi}{2}-0=\pi/2$

I hope this helps.

4. Originally Posted by larson
Use the comparison test to determine whether the improper integral converges or diverges:
$\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$

So, since this is true... $\int_{0}^{\infty} \frac {dx}{\sqrt{x}(1 + x)}$ < $\int_{0}^{\infty} \frac{1}{\sqrt{x}}$ I can just use the larger integral to find out whether or not it converges or diverges because it is much simpler.

$\int_{0}^{\infty} x^\frac{-1}{2}$ which equals:

$\lim_{t \to \infty} \int_{0}^{t} x^\frac{-1}{2}$ equals

$\lim_{t \to \infty} \frac {x^\frac{1}{2}}{2}$ from 0 -> $\infty$

So this just turns out to be $\infty - \frac {0^\frac{1}{2}}{2}$ so this is equal to...
$\infty - 0$ which = $\infty$

So because this integral DIVERGES TO $\infty$, this tells us nothing? So what do I do now?
...maybe I misread the past posts but $\int_0^{\infty}\frac{1}{\sqrt{x}(1+x)}=2arctan(\sq rt{x})|_0^{\infty}=\pi-0=\pi$

5. Originally Posted by Mathstud28
...maybe I misread the past posts but $\int_0^{\infty}\frac{1}{\sqrt{x}(1+x)}=2arctan(\sq rt{x})|_0^{\infty}=\pi-0=\pi$
Oh and to make how this integral was begotten let me write it in a better form $2\int_0^{\infty}\frac{\frac{1}{2\sqrt{x}}}{(1^2+(\ sqrt{x})^2)}dx$

6. Hmm... now I'm just confused...

7. Originally Posted by larson
Hmm... now I'm just confused...
What is confusing?

8. Is what I posted at the first post correct? It's just I should have rather of found an integral that was less than the original integral? Because I didn't really prove anything in the end, right?

9. Originally Posted by larson
Is what I posted at the first post correct? It's just I should have rather of found an integral that was less than the original integral? Because I didn't really prove anything in the end, right?
Try $\frac{1}{x^{\frac{3}{2}}}$ instead

10. Originally Posted by Mathstud28
Try $\frac{1}{x^{\frac{3}{2}}}$ instead
Converges to 0 correct?

11. Originally Posted by larson
Converges to 0 correct?

12. Originally Posted by Mathstud28
Finally I get something correct... lol. Thanks.

13. ## It is double improper

Originally Posted by Mathstud28
Try $\frac{1}{x^{\frac{3}{2}}}$ instead
I am sorry to say that this diverges see post #2

$\int_0^{\infty}\frac{dx}{x^{(3/2)}} \lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\frac{dx}{x^{3/2}}$

$\lim_{a \to 0} \lim_{b \to \infty} -\frac{2}{\sqrt{x}}|_a^{b}=\lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\left( \frac{-2}{\sqrt{b}}+\frac{2}{\sqrt{a}}\right)=0+\infty=\i nfty$

So it diverges

It is double improper

14. Originally Posted by TheEmptySet
I am sorry to say that this diverges see post #2

$\int_0^{\infty}\frac{dx}{x^{(3/2)}} \lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\frac{dx}{x^{3/2}}$

$\lim_{a \to 0} \lim_{b \to \infty} -\frac{2}{\sqrt{x}}|_a^{b}=\lim_{a \to 0} \lim_{b \to \infty}\int_a^{b}\left( \frac{-2}{\sqrt{b}}+\frac{2}{\sqrt{a}}\right)=0+\infty=\i nfty$

So it diverges

It is double improper
Oops...forgot the integral...but you get the idea use one that when you integrate you get a degree less than one

15. I still believe it converges... Mathstud pointed me to use this... $\frac {1}{x^\frac{3}{2}}$ which is greater than the original function. I proved that this $\frac {1}{x^\frac{3}{2}}$ converges, so the original much converge...

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