Improper Integrals

**larson** · April 20th 2008, 04:15 PM

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?

**TheEmptySet** · April 20th 2008, 04:23 PM

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

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

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

**TheEmptySet** · April 20th 2008, 04:39 PM

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.

**Mathstud28** · April 20th 2008, 05:00 PM

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 $\int_0^{\infty}\frac{1}{\sqrt{x}(1+x)}=2arctan(\sq rt{x})|_0^{\infty}=\pi-0=\pi$

**Mathstud28** · April 20th 2008, 05:01 PM

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 $2\int_0^{\infty}\frac{\frac{1}{2\sqrt{x}}}{(1^2+(\ sqrt{x})^2)}dx$

**larson** · April 20th 2008, 05:06 PM

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

**Mathstud28** · April 20th 2008, 05:08 PM

Originally Posted by larson

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

What is confusing?

**larson** · April 20th 2008, 05:20 PM

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?

**Mathstud28** · April 20th 2008, 05:29 PM

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

**larson** · April 20th 2008, 06:39 PM

Originally Posted by Mathstud28

Try $\frac{1}{x^{\frac{3}{2}}}$ instead

Converges to 0 correct?

**Mathstud28** · April 20th 2008, 06:40 PM

Originally Posted by larson

Converges to 0 correct?

**larson** · April 20th 2008, 06:42 PM

Originally Posted by Mathstud28

Finally I get something correct... lol. Thanks.

**TheEmptySet** · April 20th 2008, 07:02 PM

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

**Mathstud28** · April 20th 2008, 07:05 PM

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

**larson** · April 20th 2008, 07:23 PM

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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