Findinf residue of 1/5+4sin(deta)

  • Thread starter Thread starter doey
  • Start date Start date
  • Tags Tags
    Residue
doey
Messages
27
Reaction score
0

Homework Statement


it ask me to integral from 0 to 2pi for 1/[5+4sin(deta)]

Homework Equations


i knoe i nid to find out the pole bu convert sin(deta) into exponential fomat

The Attempt at a Solution


after i solve it until 2z^2+5iz-2 i stucked,can i juz use [-b+-squarot[b^2-4ac]]/2a formula to solve this by juz let my b=5i?
and i get the answer is in negative value,if i do that.i think i might wrong ,any one can help me?
 
Physics news on Phys.org
doey said:

Homework Statement


it ask me to integral from 0 to 2pi for 1/[5+4sin(deta)]


Homework Equations


i knoe i nid to find out the pole bu convert sin(deta) into exponential fomat

The Attempt at a Solution


after i solve it until 2z^2+5iz-2 i stucked,can i juz use [-b+-squarot[b^2-4ac]]/2a formula to solve this by juz let my b=5i?
and i get the answer is in negative value,if i do that.i think i might wrong ,any one can help me?

Yes. You can use the quadratic equation to solve something like that. But I don't think you've got the right quadratic and I don't know what "i get the answer is in negative value". You really need to show your work.
 
Dick said:
Yes. You can use the quadratic equation to solve something like that. But I don't think you've got the right quadratic and I don't know what "i get the answer is in negative value". You really need to show your work.

∫1/5+4sin(deta) dθ =∫(dz/iz)/[5+4(z/2i-1/2iz)]
= ∫(dz)/[5iz+4(iz^2/2i-iz/2iz)]
= ∫(dz)/[5iz+(4iz^2)/(2i)-4iz/2iz)]
=∫(dz)/[5iz+2z^2-2]

then i nid solve 5iz+2z^2-2 to get my pole,am i wrong? then i will get my root in -0.5i and -2i.
 
doey said:
∫1/5+4sin(deta) dθ =∫(dz/iz)/[5+4(z/2i-1/2iz)]
= ∫(dz)/[5iz+4(iz^2/2i-iz/2iz)]
= ∫(dz)/[5iz+(4iz^2)/(2i)-4iz/2iz)]
=∫(dz)/[5iz+2z^2-2]

then i nid solve 5iz+2z^2-2 to get my pole,am i wrong? then i will get my root in -0.5i and -2i.

Yep, that's what I get. I made a mistake but I caught it thanks you to showing what you did.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top