Improper Integrals

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• Apr 19th 2008, 07:12 PM
larson
Improper Integrals
Show that $\int_{0}^{1} cos(x)dx$ converges.

A little bit puzzled here.
• Apr 19th 2008, 07:15 PM
Mathstud28
Quote:

Originally Posted by larson
Show that $\int_{0}^{1} cos(x)dx$ converges.

A little bit puzzled here.

It is defined on $[0,1]$ so just do it $\int_0^{1}cos(x)dx=sin(x)\bigg|_0^{1}=sin(1)-0=sin(1)$ therefore it converges
• Apr 19th 2008, 07:18 PM
larson
Quote:

Originally Posted by Mathstud28
It is defined on $[0,1]$ so just do it $\int_0^{1}cos(x)dx=sin(x)\bigg|_0^{1}=1-0=1$ therefore it converges

crap... i wrote the problem wrong... it's ... $\int_{0}^{1} \frac {cos(x)}{x^2}dx$ show that it CONVERGES.
• Apr 19th 2008, 07:18 PM
TheEmptySet
Quote:

Originally Posted by larson
Show that $\int_{0}^{1} cos(x)dx$ converges.

A little bit puzzled here.

This seems a little weird. Cosine is continous for all real numbers and has an antiderivative.

This type of question is usuall asked when you have a vertical asymptote or somthing.

It would converge to $\sin(1)$

Maybe I missed the point?
• Apr 19th 2008, 07:22 PM
TheEmptySet
Quote:

Originally Posted by larson
crap... i wrote the problem wrong... it's ... $\int_{0}^{1} \frac {cos(x)}{x}dx$

ahh the plot thickens

express cosine as its power series $\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$

so then

$\frac{\cos(x)}{x}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n+1)!}$

integrate term by term and evaluate.

This will converges by the AST.

Good luck. (Rock)
• Apr 19th 2008, 07:22 PM
PaulRS
Quote:

Originally Posted by larson
crap... i wrote the problem wrong... it's ... $\int_{0}^{1} \frac {cos(x)}{x}dx$

That integral diverges, note that $\frac{\cos(x)}{x}\sim{\frac{1}{x}}$ as $x\rightarrow{0}$
• Apr 19th 2008, 07:25 PM
larson
Quote:

Originally Posted by TheEmptySet
ahh the plot thickens

express cosine as its power series $\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$

so then

$\frac{\cos(x)}{x}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n+1)!}$

integrate term by term and evaluate.

This will converges by the AST.

Good luck. (Rock)

I'm a little confused here by this... $\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ is that just general knowledge?
• Apr 19th 2008, 07:26 PM
Mathstud28
Quote:

Originally Posted by larson
crap... i wrote the problem wrong... it's ... $\int_{0}^{1} \frac {cos(x)}{x}dx$

The integral diverges?
• Apr 19th 2008, 07:28 PM
TheEmptySet
My above post is wrong I use the sine power series
express cosine as its power series $\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$

so then

$\frac{\cos(x)}{x}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n-1}}{(2n)!}$

integrate term by term and evaluate and the sum diverges.
• Apr 19th 2008, 07:29 PM
TheEmptySet
Quote:

Originally Posted by larson
I'm a little confused here by this... $\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ is that just general knowledge?

I used the wrong power series. sorry
• Apr 19th 2008, 07:30 PM
larson
Quote:

Originally Posted by Mathstud28
The integral diverges?

I'm so sorry, I'm making this much much more confusing that what I intended it to be. It converges...
• Apr 19th 2008, 07:32 PM
Mathstud28
Quote:

Originally Posted by larson
I'm so sorry, I'm making this much much more confusing that what I intended it to be. It converges...

This integral $\int_0^{1}\frac{cos(x)}{x}dx$ diverges?
• Apr 19th 2008, 07:34 PM
larson
Quote:

Originally Posted by Mathstud28
This integral $\int_0^{1}\frac{cos(x)}{x}dx$ diverges?

Sorry, I wrote it down wrong. The problem is...

$\int_{0}^{1} \frac{cos(x)}{x^2}dx$ explain how this CONVERGES.
• Apr 19th 2008, 07:35 PM
PaulRS
Quote:

Originally Posted by larson
I'm so sorry, I'm making this much much more confusing that what I intended it to be. It converges...

Maybe it is $\int_0^1\frac{1-\cos(x)}{x}dx$ or $\int_0^1\frac{1-\cos(x)}{x^2}dx$
(both converge since $1-\cos(x)\sim{\frac{x^2}{2}}$ as $x\rightarrow{0}$)
• Apr 19th 2008, 07:39 PM
Mathstud28
Quote:

Originally Posted by larson
Sorry, I wrote it down wrong. The problem is...

$\int_{0}^{1} \frac{cos(x)}{x^2}dx$ explain how this CONVERGES.

ok $\int_0^{1}\frac{cos(x)}{x^2}dx=\int_0^{1}\sum_{n=1 }^{\infty}\frac{(-1)^{n}x^{2n-2}}{(2n)!}dx=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n-1}}{(2n-1)\cdot{(2n)!}}$ which diverges from 0 to 1...it still doesn't converge
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