• Support PF! Buy your school textbooks, materials and every day products Here!

Does this look right?

  • Thread starter suspenc3
  • Start date
  • #1
402
0
[tex] \int_0^* \frac {dz}{z^23Z +2}[/tex]

[tex]=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{z^2+3Z+2}[/tex]

[tex]=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{(z+3/2)^2 -1/4}[/tex]

let u=z+3/2
du=dz

[tex]=\lim_{t\rightarrow \*} \int_0^* \frac{du}{(u)^2 -1/4}[/tex]

[tex] = \frac{1}{2(1/2)}ln|\frac{u-1/2}{u+1/2}[/tex]

[tex] = ln |\frac{z+1}{z+2}| \right]_0^*[/tex]

[tex] = -ln \frac{1}{2}[/tex]

this is one of the first of these im doing so bear with me if its horrible wrong :rolleyes:

EDIT:* denotes infinite..PS is there a latex for infinite?
 
Last edited:

Answers and Replies

  • #2
I don't know why you used the lim symbol in the first place, and where the t commes in play, but

[tex]= \int_0^* \frac{dz}{z^2+3Z+2} [/tex]
[tex]= \int_0^* \frac{du}{(u)^2 -1/4}[/tex]

using your algebraic manipulations. After that, you used the partial sum decomposition skillfully and unless I'm mistaking, you are right!

Just one minor thing : [tex] -ln \frac{1}{2} [/tex] = ln(2)
 
Last edited:
  • #3
402
0
yyyea..but what?I already had that is it right?
 
  • #4
0rthodontist
Science Advisor
1,230
0
It's correct except you forgot to change the lower limit of integration. When z goes from 0 to infinity, then u = z + 3/2 goes from what to what?

You do the infinity sign like [tex]\infty[/tex]
 

Related Threads for: Does this look right?

Replies
2
Views
948
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
729
Replies
6
Views
2K
  • Last Post
Replies
6
Views
551
Replies
3
Views
1K
Top