# Thread: Improper Integrals

1. ## Improper Integrals

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

A little bit puzzled here.

2. 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

3. 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.

4. 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?

5. 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.

6. 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 $x\rightarrow{0}$

7. 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.
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?

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

9. ## 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.

10. 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

11. 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...

12. 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?

13. 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.

14. 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 $x\rightarrow{0}$)

15. 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

Page 1 of 2 12 Last